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Algebra I
Hartmut Laue
Mathematisches Seminar der Universitat Kiel 2014
Contents
1. Roots, radicals and zeros of polynomials 5
2. Algebraic structures and the extension principle 25
3. Cosets and homomorphisms 47
4. Field extensions 65
A. Appendix
On irreducibility of polynomials over Q 83
Prerequisites
It is assumed that the reader is familiar with basic concepts of Linear Algebra: vec-tor spaces over fields, bases, linear dependence and independence, linear mappings andmatrices. Furthermore, it will be helpful if the reader has taken notice of the notions ofa ring and of a group, but no deeper results about these are expected. Recall, however,the so-called
Subgroup Criterion: A subset H of a group G is a subgroup of G if and only if H 6= ∅and xy−1 ∈ H for all x, y ∈ H.
Basic terminology (terms like homomorphism, monomorphism, epimorphism, isomorph-ism, automorphism, endomorphism, equivalence relation, partition, operations, compos-itions of mappings) should be known although even these are briefly recalled in duecourse.
Some notational hints will be useful as these may differ from other authors’s style:
N = {1, 2, 3, 4, . . .}, N0 = N ∪ {0},n := {k|k ∈ N, 1 ≤ k ≤ n} for all n ∈ N0,
K := K r {0K} if K is a ring.
The image of an element x under a function f is in this course generally denoted byxf (with “inevitable” exceptions for historic reasons like, e. g., f = exp). Other knownchoices of notation are the traditional f(x) or xf . The traditional left-hand notation ofmappings has the doubtless advantage of being widely spread, but as it seems only forpurely historic reasons. If the choice of notation is oriented by principles like consistencywith foundations and smoothness of presentation from a systematic point of view – inparticular, regarding advanced theories in non-commutative algebra like Group Theory,Combinatorial Algebra –, a closer look shows that the right-hand notation is doubtlesslypreferable. Compositions of functions are a fundamental tool, and it is here where thosedifferences in notation have noticeable consequences. In accordance with the aforemen-tioned choice of the right-hand notation, the notation fg for the composition of mappingsf, g has the meaning to apply first f , then g to an element x of the domain of f . Hence
x(fg) = (xf)g.
(Brackets are not needed here so that we may also just write xfg.)
3
1. Roots, radicals and zeros of
polynomials
Let f = (fn)n∈N0 , g = (gn)n∈N0 be number sequences. We set
(fn)n∈N0 † (gn)n∈N0:= (fn + gn)n∈N0,
(fn)n∈N0 • (gn)n∈N0:= (
n∑
k=0
fkgn−k)n∈N0 (Cauchy product).
This defines two operations on the set Π of all number sequences, and (Π, †, •) is aring, i. e., (Π, †) is an abelian group, • is associative, and both distributive laws hold.The sequence e where e0 = 1, en = 0 for all n ∈ N0, is neutral with respect to •: If(fn)n∈N0 ∈ Π, n ∈ N0, then
n∑
k=0
ekfn−k = e0fn−0 + e1fn−1 + · · ·+ enf0 = fn + 0 + · · ·+ 0 = fn,
likewise∑n
k=0 fken−k = fn, hence e • (fn)n∈N0 = (fn)n∈N0 = (fn)n∈N0 • e. A ring whichcontains a multiplicatively neutral element (also called unit element) is called unitary ifthis is different from the zero element.1 Thus (Π,+, •) is a unitary ring. Put
Π0 := {f |f ∈ Π, ∃n ∈ N0 ∀m ∈ N>n fm = 0}.The defining condition means that almost all sequence values are zero. The structure(Π0, †) is an abelian group, and Π0 is closed with respect to • . Moreover, e ∈ Π0.Therefore Π0 is a subring of Π with a unit element, and the latter coincides with theunit element of Π.2 A unitary subring of a unitary ring is called unitalin this course ifboth rings have the same unit element.
The element t := (0, 1, 0, 0, . . . ) of Π plays a special role. We calculate its powers andobtain from the definition of •
t2 = t • t = (0, 0, 1, 0, 0, 0, . . . )t3 = t2 • t = (0, 0, 0, 1, 0, 0, . . . )
...tk = tk−1 • t = (0, . . . , 0︸ ︷︷ ︸
k
, 1, 0, 0, . . . )
1If the additively neutral element is at the same time multiplicatively neutral, it turns out easily thatit is the only ring element, therefore the ring truely “trivial”.
2This is by no means always the case. As a simple example consider the ring Z⊕Z with componentwiseaddition and multiplication. The element (1, 1) is multiplicatively neutral. The subring Z ⊕ {0} isunitary, but its unit element (1, 0) is distinct from (1, 1).
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for all k ∈ N. The first power t1 equals t. The power t0 finally is a special case of aso-called empty product, a product without any factor. In every structure with a neutralelement, the empty product (in case of additive notation, the empty sum) is defined tobe that element. Thus we complete our list by
t1 = (0, 1, 0, 0, . . . )t0 = (1, 0, 0, 0, . . . )
Let K be the domain of numbers over which the set Π of sequences was considered.Defining
∀c ∈ K ∀f ∈ Π c · f := (cfn)n∈N0,
we have for all nonzero f ∈ Π0 the finite additive decomposition
f = (f0, f1, f2, . . . ) = f0(1, 0, 0, 0, . . . ) + f1(0, 1, 0, 0, . . . ) + · · ·+ fn(0, . . . , 0︸ ︷︷ ︸n
, 1, 0, 0, . . . )
where n := max{k|k ∈ N0, fn 6= 0}. Thus every f ∈ Π0 is a K-linear combination ofpowers of t. If c0, c1, . . . , cn ∈ K, cn 6= 0 and c0t
0 + c1t1 + · · ·+ cnt
n = f , then c0 = f0,c1 = f1,. . . , cn = fn, for the left hand side equals (c0, . . . , cn, 0, 0, . . . ). It follows thatthe representation of f ∈ Π0 as a K-linear combination of powers of t with scalars 6= 0is unique. Note that the zero element of Π0 is represented as the empty sum.
1.1 Definition. Let K be a commutative unitary ring, V an abelian group, · : K×V →V an (external) operation such that the laws “known from vector spaces” hold, i. e.,(c + d) · v = c · v + d · v, c · (u + v) = c · u + c · v, (cd) · v = c · (d · v), 1K · v = vfor all c, d ∈ K, u, v ∈ V . Then V is called a K-space.3 A subset T of V is called aK-generating system of V if every element of V is a K-linear combination of elementsof T . If for every finite subset S of T and for all choices of cv, dv ∈ K (v ∈ S) theimplication ∑
v∈Scvv =
∑
v∈Sdvv ⇒ ∀v ∈ S cv = dv
holds true, then T is called K-linearly independent. A K-linearly independent K-gener-ating system of V is called a K-basis of V .
An inspection of all foregoing steps shows that we may start from an arbitrary commut-ative unitary ring K which need not necessarily be a domain of numbers and obtain ourK-spaces Π, Π0 in this more general setting. Furthermore, the set T := {t0, t1, t2, . . . }is a K-basis of Π0. Clearly, Π0 is a K-subspace of ΠS, where we borrow the meaning ofthis term in the obvious way from vector space theory.
Now suppose that V a K-space. Let • be an operation on V with the properties
(i) ∀u, v, w ∈ V u • (v + w) = u • v + u • w, (v + w) • u = v • u+ w • u,3Usually, the point · for the multiplication between “scalars” from K and “vectors” from V is omitted.In this definition, however, we used it in order to make a distinction between that multiplicationand the internal one within K.
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(ii) ∀c ∈ K ∀u, v ∈ V c · (u • v) = (c · u) • v = u • (c · v).Then V (more precisely, (V,+, •)) is called a K-algebra. It is called associative (commut-ative resp.) if • is associative (commutative resp.). In every K-algebra V we write V forthe set of all nonzero elements of V .
Examples. (1) Let L be a unitary commutative ring containing K as a unital subring.Then L is an associative and commutative unitary K-algebra. The trivial case ofL = K shows that K itself may be viewed as a K-algebra.
(2) For every n ∈ N, the set of all n × n matrices over K (for which we shall writeKn×n throughout) is an associative unitary K-algebra with respect to the usualmatrix operations as addition and multiplication (within Kn×n) and componentwisemultiplication by a scalar as product between K and Kn×n. For n > 1 it is notcommutative.
(3) Π, Π0 (as constructed above) are associative and commutative unitary K-algebras.
Let V be an associative unitary K-algebra which contains an element t such that theelements tj (j ∈ N0) are mutually distinct and form a K-basis of V . Then V is calleda polynomial ring over K in the indeterminate (or variable) t. A polynomial over K isan element of a polynomial ring over K. A polynomial of the form ctj where c ∈ K,j ∈ N0, is called a monomial in t. It follows from the definition that polynomial ringsare commutative.
We have shown that Π0 is a polynomial ring over K. It does not contain K as aunital subring, but the special sequences (c, 0, 0, . . . ) ∈ Π form a substructure in whichcalculations take place as in K: For any c, d ∈ K we have
(c, 0, 0, . . . )†(d, 0, 0, . . . ) = (c+d, 0, 0, . . . ), (c, 0, 0, . . . )•(d, 0, 0, . . . ) = (cd, 0, 0, . . . ).
In other words, the mapping ι : K → Π0, c 7→ (c, 0, 0, . . . ), is an injective homomorph-ism with respect to addition, internal multiplication, and external multiplication (“scalartimes vector”). Note that when K is considered as a K-algebra, both scalars and vec-tors are the elements of K and there is no difference in calculating an internal or anexternal product. An injective homomorphism is called a monomorphism. A K-algebrahomomorphism from a K-algebra into another structure with two internal operations(“addition” and “multiplication”) and an external multiplication with elements of K isa mapping which is a homomorphism with respect to the two additions and with re-spect to the two multiplications, and which satisfies the homogeneity rule known fromK-linear mappings in vector space theory.4 For ι, we have for all c, d ∈ K
c · d = cd 7→ι(cd, 0, 0, . . . ) = (cd, c0, c0, . . . ) = c · (d, 0, 0, . . . ),
4The second algebraic structure need not be assumed to be a K-algebra to apply this definition. Inmany applications, however, this will be the case. As we shall see later (3.11), there is a generalreason for the fact that the second structure necessarily at least involves (contains) a K-algebrawhich turns out to be closely related to the initially given one: the set of all images under ϕ.
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and indeed (d, 0, 0, . . . ) is the image of d under ι.
In this course, usually the right notation for mappings will be used, i. e., the image ofan element x under a mapping ϕ will be denoted by xϕ. Thus we may express thehomogeneity of ι as follows: ∀c, d ∈ K (c · d)ι = c · dι.The elements of Π are called power series over K, Π a power series algebra over K inthe indeterminate (or variable) t. The set-theoretic situation which we are studying,including the K-algebra monomorphism ι, may be visualized by the following diagram:
K Kι
Π0
Π
✲ι
·1K ·e
·t
Now let V be an arbitrary polynomial ring over K in the indeterminate t. The elementsof V are uniquely given in the form
∑nj=0 cjt
j where n ∈ N0, cj ∈ K for all j, cn 6= 0.5
Note that the zero element of V is represented by the empty sum. For every elementf ∈ V , the number n in the representation f =
∑nj=0 cnt
n with c0, . . . , cn−1 ∈ K, cn ∈ K,is called its degree with respect to t. We use the notation deg f for the degree of a nonzeroelement f ∈ V (degt f if the reference to t is not obvious from the context). If cn = 1K ,f is called normed.
1.1.1. deg fg ≤ deg f + deg g for all f, g ∈ V with fg 6= 0V .
Proof. Let n,m ∈ N0, c0, . . . , cn, d0, . . . , dm ∈ K such that cn 6= 0K , dm 6= 0K andf =
∑nj=0 cjt
j, g =∑m
j=0 djtj. Then fg is a K-linear combination of powers tk of t with
k ≤ n+m. The claim follows.
It should be noted that the exponent n+m of t arises only if cntn and dmt
m are multiplied:cnt
n dmtm = cndmt
n+m. We know that cn, dm 6= 0K , but in general it may happen thatstill cndm = 0K .
6 An element a ∈ K is called a zero divisor if there exists an elementb ∈ K such that ab = 0K . In many rings K, however, 0K is the only zero divisor, i. e.,a product of nonzero elements is never zero. For these rings, the inequality in 1.1.1thus becomes an equality because the coefficient of the term with the highest possible
5The elements of the power series algebra Π are similarly written in the form∑
n∈N0ckt
k. But note
that this is just a formal notation which has not the meaning of a sum while∑n
k=0 cktk is indeed
an ordinary sum of n + 1 monomials. The notation for power series follows the pattern given bypolynomials but should not give rise to the inadequate idea of a “sum of infinitely many monomials”!
6 For example, if K = Z⊕Z with componentwise addition and multiplication, cn = (1, 0), dm = (0, 1).Clearly, this phenomenon does not occur if cn = 1K or dm = 1K . Hence we have equality in 1.1.1 iff or g is normed.
8
exponent n + m cannot vanish. A commutative unitary ring K is called an integraldomain if K is multiplicatively closed, which is obviously just a reformulation of theaforementioned property. We therefore have
1.1.2. If K is an integral domain, then so is the polynomial ring V , and deg fg =deg f + deg g for all f, g ∈ V . �
Considering deg as a function from V into N0, the contents of 1.1.2 may also be ex-pressed in the following form, observing that every n ∈ N0 occurs as the degree of somepolynomial (e. g., of tn):
1.1.2’ If K is an integral domain, then the function deg is an epimomorphism7 of (V , ·)onto (N0,+). �
Every field, more generally every unital subring of a field, is an integral domain. Atypical example of an integral domain which is not a field is the familiar ring Z. Thering Z⊕ Z is commutative and unitary but not an integral domain (cf. footnote 6).
Let f =∑n
j=0 cjtj ∈ V (where c0, . . . , cn ∈ K, cn 6= 0K), and let b be an element of a
unitary associative K-algebra A. We set
f(b) :=
n∑
j=0
cjbj
and call the mapping Fb : V → A, f 7→ f(b), the substitution (mapping) of b. Iff(b) = 0A, then b is called a zero (note, however, footnote 9) of f in A. In other words,the zeros of f in A are the solutions in A of the equation
∑nk=0 cjx
j = 0A.
1.1.3. Let A be a unitary associative K-algebra, b ∈ A. Then Fb is a unital K-algebrahomomorphism.
Proof. If f, g ∈ V , c ∈ K, f =∑n
j=0 cjtj, g =
∑mi=0 dit
i (where n,m ∈ N0, cj , di ∈ K) we
trivially have (f†cg)(b) = f(b)+cf(b), and moreover (fg)(b) =∑m+n
h=0
(∑hj=0 cjdh−j
)bh =∑n
j=0 cjbj∑m
i=0 dibi = f(b)g(b) (where cj := 0K if j > n, di = 0K if i > m).
We write K[t] for a polynomial ring over K in the indeterminate t which contains Kas a unital subring.8 An element f ∈ K[t] is called pure if there exist an n ∈ N andan a0 ∈ K such that f = tn + a0. Assume that K is a unital subalgebra of a unitaryassociative K-algebra A. An element r ∈ A is called a radical over K if r 6= 0K and thereexists a pure polynomial f ∈ K[t] such that f(r) = 0K . By definition, there exists thena positive integer n such that rn ∈ K. Any such n will be called an exponent for r (withrespect to K). The most important case in this course is that of a field A containing Kas a subfield.
7Recall that a surjective homomorphism is called an epimorphism.8Hence t0 = 1K . The polynomial ring Π0 constructed above does not have this property so that in thismoment the existence of a polynomial ring containing K is unclear. But this is only a temporarygap as will be seen in due course (2.9.2).
9
For example, roots 6= 0K are special radicals. Recall from basic Analysis what a root is:If c ∈ R≥0, n ∈ N, then there exists a unique non-negative real number the n-th powerof which equals c. This number is called the n-th root of c and denoted by n
√c.9
In this course, the notation n
√
c is used only if c is a non-negative real number.
Then n√c is a zero of the pure polynomial tn− c, thus for c 6= 0 it is a radical over every
subfield K of R with c ∈ K. (If c 6∈ K, tn − c is not a polynomial over K!)
Examples. (1) t2 − 3 has two zeros in R, the square root of 3,√3, and its additive
inverse −√3 which of course does not satisfy the definition of a root of a real number.
Both numbers are radicals over Q, and every even number is an exponent for them.
(2) The complex number i 3√2 is a radical over Q of exponent 6, because it is a zero of
the pure polynomial t6 + 4 ∈ Q[t]. Over R, 2 is an exponent for it as it is a zero ofthe pure polynomial t2 + 3
√4 ∈ R[t].
(3) 1+√2 is not a radical over Q: A trivial induction shows that for all n ∈ N there exist
a, b ∈ N such that (1 +√2)n = a + b
√2. Assuming (1 +
√2)n ∈ Q for some n ∈ N
we thus would have a, b ∈ N, c ∈ Q such that a + b√2 = c, hence
√2 = c−a
b∈ Q, a
contradiction.
(4) The roots of the polynomial t2 + t + 1 in C are radicals over Q, because in Q[t] wehave the equation
(t2 + t + 1)(t− 1) = t3 − 1.
We conclude that every zero of t2 + t+ 1 is also a zero of t3 − 1, hence a radical ofexponent 3 over Q.
The equation in (4) is of course the special case n = 3 of the general statement that, forall n ∈ N,
(tn−1 + · · ·+ t+ 1K)(t− 1K) = tn − 1K ,
for any unitary commutative ring K. The zeros of tn−1K in K are called the n-th rootsof unity in K.10 If K is not specified, the term refers to the case K = C.
1.1.4. Let K be a unital subring of an integral domain L. Let f, g, h ∈ K[t], f = gh,b ∈ L and f(b) = 0K. Then g(b) = 0K or h(b) = 0K.
Proof. If f(b) = 0K , then g(b)h(b) = 0K , hence the claim as L has no zero divisors.
9 The terminology introduced here is not used in the same way by all authors. It is common use to talkabout “roots of a polynomial” instead of zeros. In order to prevent any confusion with roots of realnumbers we avoid this usage of the term “root” throughout this course. (Exception: footnote 10.)
10 Here we deviate from our convention about the usage of the term “root” in favour of the traditionalterminology.
10
It follows that the n-th roots of unity in an integral domain L are 1K and the zeros oftn−1 + · · ·+ t + 1 in L. The 3rd roots of unity 6= 1 in C (see Example (4) on p. 10) arethe complex solutions of the equation
0 = x2 + x+ 1 =(x+
1
2
)2+
3
4.
This familiar transformation reduces the determination of the zeros in the form a + biwith a, b ∈ R to the treatment of the simple case of a pure polynomial of degree 2(namely, t2 + 3
4). Thus we obtain the two solutions
x1 = −1
2+
1
2i√3, x2 = −
1
2− 1
2i√3.
The case n = 4 is easier as we have t4 − 1 = (t2 + 1)(t + 1)(t− 1) so that the complex4th roots of unity are 1,−1, i,−i. But it is obvious that we will not always be luckyif the degree n increases and we want to describe the n-th roots of unity, as above, asan R-linear combination of 1 and i. There is, however, a different idea which quicklyleads to this standard representation. Recall the fundamental equation of the complexexponential function: exp(z + z′) = exp z exp z′ for all z, z′ ∈ C, which means, in otherwords:
1.1.5. exp is a homomorphism of (C,+) into (C, ·). �
In particular, by a standard induction we obtain from 1.1.5
1.1.6. exp(nz) = (exp z)n for all n ∈ N0, z ∈ C. �
The replacement of z by iz leads to Moivre’s formula:
cosnz + i sinnz = (cos z + i sin z)n for all n ∈ N0, z ∈ C,
making use of Euler’s formula, the fundamental connection between exp, sin, cos:
exp(iz) = cos z + i sin z for all z ∈ C. 11
We exploit this equation for real values of z and obtain for arbitrary a, b ∈ R the chainof equivalences
exp(a+ bi) = 1⇔ ea(cos b+ i sin b) = 1⇔ ea cos b = 1, eai sin b = 0⇔ b ∈ Z · 2π, a = 0
The kernel of a group homomorphism f , denoted by ker f , is the set of all group elementswhich have the same image under f as the neutral element. In the case of the homo-morphism exp (see 1.1.5), we therefore have z ∈ ker exp ⇔ exp z = exp 0 ⇔ exp z = 1which is equivalent to z ∈ Z · 2πi by what we have just shown. This proves:
11Substituting t by iz in the formal power series∑
n∈N0
tn
n!(which converges everywhere in the complex
plane and defines the complex exponential function), we obtain
exp(iz) =
∞∑
n=0
(iz)n
n!=
∞∑
k=0
(−1)kz2k(2k)!
+ i
∞∑
k=1
(−1)k−1z2k−1
(2k − 1)!= cos z + i sin z.
In particular, exp(2πi) = 1, which is the famous Eulerian equation, combining in the form e2πi = 1“the five most important numbers of Mathematics” 1, 2, e, π, i.
11
1.1.7. ker exp = Z · 2πi. �
It is easily seen that the mapping κ : ]0, 2π] → K(0, 1), c 7→ exp(ci), is a bijection,commonly called the trigonometric parameterization of the unit circle. 12
Now put for every n ∈ N
Rn(1) := {w|w ∈ C, wn = 1}, R(1) :=⋃
n∈NRn(1).
For all w ∈ Rn(1) we have 1 = |wn| = |w|n, hence |w| = 1. Thus Rn(1) ⊆ K(0, 1). Nowlet z ∈ K(0, 1) and c ∈]0, 2π] such that z = exp(ci). Then, by 1.1.6 and 1.1.7,
z ∈ Rn(1)⇔ exp(nci) = 1⇔ nc ∈ Z · 2π ⇔ ∃k ∈ Z z = expk · 2πin
The values exp k·2πin
where k > n or k < 1 are just repetitions of the values where k ∈ n.Therefore,
1.1.8. ∀n ∈ N Rn(1) = {expk · 2πin| k ∈ n}. �
From this description of Rn(1) we obtain a beautiful geometric interpretation for then-th roots of unity: Recall that each point P of the unit circle determines an angle,given by the line segment connecting it with the origin and the positive real axis.Its radian measure is given by the length of the corresponding arc on the unit circle(between P and the point representing 1). If P represents an n-th root of unityexp k·2πi
n(= cos k·2πi
n+ i sin k·2πi
n), the corresponding angle corresponds to k
n-th of the
complete periphery which is of length 2π. Therefore the n-th roots of unity are rep-resented by the vertices of the regular n-gon inscribed in the unit circle such that thepoint representing 1 is one of them.
1
in = 3 :
2π3
∼ 120◦
1
in = 4 :
2π4
∼ 90◦
1
in = 5 :
2π5
∼ 72◦
. . .
We are going to give a specifically algebraic interpretation of Rn(1) and start with thefollowing definition of the most general of all algebraic structures:
12κ is injective as cκ = c′κ (where 0 < c < c′ ≤ 2π) would imply that exp((c′ − c)i
)= 1, hence
c′ − c ∈ Z · 2π by 1.1.7, a contradiction because 0 < c′ − c < 2π. Furthermore, let a, b ∈ R suchthat a+ bi ∈ K(0, 1). Then a2 + b2 = 1 so that |a|, |b| ≤ 1. There are exactly two values c ∈]0, 2π]such that cos c = a, and for both of them we have b2 = 1 − (cos c)2 = (sin c)2, hence b = sin c orb = − sin c. As κ is injective, the sine values for those two elements c cannot coincide. Hence wemay choose c ∈]0, 2π] such that cos c = a, sin c = b. It follows that cκ = a+ bi. Thus κ is bijective.
12
1.2 Definition. A magma is a pair (X, ◦) where X is any set and ◦ is any operation onX . A submagma of (X, ◦) is then a subset of X which is closed with respect to ◦. If Sis a submagma of (X, ◦) such that (S, ◦) is a semigroup (i. e., such that the associativelaw holds in (S, ◦) ), then S is called a subsemigroup of (X, ◦). If there exist neutralelements eX of (X, ◦), eS of (S, ◦), we say that S is a unital submagma if eX = eS. Asubgroup of (X, ◦) is a submagma S such that (S, ◦) is a group.13 A semigroup with aneutral element is called a monoid.
The group (Z,+) has many submagmas which are no subgroups, for example, any subsetZ≥y where y ∈ Z. For every integer a, the set of its multiples aZ is a subgroup of (Z,+)and, as is easily seen, the smallest subgroup containing a. If a 6= 0, the mapping Z→ aZ,z 7→ az, is an isomorphism. If b ∈ Z, then aZ = bZ if and only if a|b and b|a, i. e., if andonly if b ∈ {a,−a}. It is not hard to prove that every subgroup of (Z,+) has the formnZ for a uniquely determined non-negative integer n, where the uniqueness is obviousfrom what we just observed:14
1.2.1. Let H be a subgroup of (Z,+). Then there exists a unique non-negative integern such that H = nZ.
Proof. If H = {0} the claim is trivial. Otherwise H contains a positive integer. Letn := minH ∩ N. Then nZ ⊆ H . Assuming that nZ 6= H , let h ∈ H r nZ. Consideringthe largest multiple of n which is smaller than h exhibits an integer z such that 0 <h− nz < n, a contradiction because h− nz ∈ H ∩ N. Hence nZ = H .
We define for all U, V ⊆ X the product set of U and V by
U ◦V := {u ◦ v|u ∈ U, v ∈ V }.
Then (P(X), ◦) is a magma and f : X → P(X), x 7→ {x}, is a monomorphism of (X, ◦)into (P(X), ◦) as f is injective and (x ◦ x′)f = {x ◦ x′} = {x}◦{x′} = xf ◦ x′f for allx, x′ ∈ X . If (X, ◦) has a neutral element e, then X ◦X = X because X ◦X ⊆ X =X ◦{e} ⊆ X ◦X .
13Note that it is not required in this definition that (X, ◦) be a group. For example, our definitionallows us to say that, for a field K, K is a subgroup of (K, ·), etc. It may happen that (X, ◦) has noneutral element, and if it has one, it may happen that it does not coincide with that of a subgroupS. On the other hand, if (X, ◦) is a group, then it is easily seen (and should be well known) thatall of its subgroups have the same neutral element. Every magma has at most one neutral element.Thus, in a magma with neutral element e, no explicit reference to e is needed when one talks aboutan inverse element y of an element x, i. e., an element with the property that x ◦ y = e = y ◦ x.One should bear in mind that, in general, such an element need not exist, and if it does, there maybe more than one inverse of an invertible element. Again, if (X, ·) is a group, then every elementhas a uniquely determined inverse. Therefore, given a group, it is reasonable to choose a notationfor the (!) inverse of x which directly refers to x. Multiplicatively, x−1 is the common notation forit, additively −x. But note that this notation generally makes no sense if applied outside a grouptheoretic context where neither existence nor uniqueness of an inverse may be assumed.
14Thus the group (Z,+) has the remarkable property that all of its subgroups 6= {0} are isomorphic to(Z,+).
13
1.3 Proposition. Let (X, ◦), (Y, ·) be magmas and f a homomorphism of (X, ◦) into(Y, ·).(1) Xf is a submagma of (Y, ·).
(2) If (X, ◦) is a semigroup, then so is (Xf, ·).
(3) If e is neutral in (X, ◦), then ef is neutral in (Xf, ·). If x, x′ ∈ X such thatx ◦ x′ = x′ ◦ x, then xf · x′f = x′f · xf . If x′ is an inverse of x in (X, ◦), then x′fis an inverse of xf in (Xf, ·).
(4) If (X, ◦) is a group, then Xf is a subgroup of (Y, ·), and 15
ker f := {x|x ∈ X, xf = ef}
is a subgroup of (X, ◦) (where e denotes the neutral element of (X, ◦)).Proof. (1) Let y, y′ ∈ Xf . Then there exist x, x′ ∈ X such that y = xf , y′ = x′f , andy · y′ = xf · x′f = (x ◦ x′)f ∈ Xf.(2) Let y, y′, y′′ ∈ Xf . Then there exist x, x′, x′′ ∈ X such that y = xf , y′ = x′f ,y′′ = x′′f , and (y · y′) · y′′ = (xf · x′f) · x′′f = (x ◦ x′)f · x′′f =
((x ◦ x′) ◦ x′′
)f =(
x ◦ (x′ ◦ x′′))f = xf · (x′ ◦ x′′)f = xf · (x′f · x′′f) = y · (y′ · y′′).
(3) Let y ∈ Xf . Then there exists an x ∈ X such that y = xf , and y · ef = xf · ef =(x ◦ e)f = xf = y, analogously ef · y = y. Clearly x ◦ x′ = x′ ◦ x implies xf · x′f =(x ◦ x′)f = (x′ ◦ x)f = x′f · xf . This holds, in particular, if x′ is an inverse of x. In thiscase the mid terms equal ef .
(4) The first claim follows from (1), (2), (3). Since (X, ◦) is a group, we may applythe subgroup criterion for the second claim: Clearly, e ∈ ker f . If x, x′ ∈ ker f , then(xf)−1 = (ef)−1 = ef , the latter being neutral, hence the inverse of itself. It follows by(3) that (x−1◦x′)f = x−1f ·x′f = (xf)−1·x′f = (ef)−1 ·ef = ef , thus x−1◦x′ ∈ ker f.
Let us recall one of the most important examples of a group, the symmetric group on aset X , consisting of all permutations of X , endowed with the composition of mappingsas operation. It is usually denoted by SX . If (X, ◦) is a magma, the set Aut (X, ◦)(commonly written simply as AutX) of all automorphisms of (X, ◦) clearly is a subsetof SX . In a sense, the following simple remark is of fundamental importance, as it showsthat studying an arbitrary algebraic structure by means of their automorphisms alwaysmeans that a group comes into play:
1.3.1. Let (X, ◦) be any magma. Then AutX is a subgroup of SX .Proof. We just have to observe that idX ∈ AutX (which is trivial) and α, β ∈ AutXimplies that αβ ∈ AutX and β−1 ∈ AutX . Clearly, αβ, β−1 ∈ SX . Compositions ofhomomorphisms are homomorphisms, and for all x, y ∈ X we have
((xβ−1)◦(yβ−1)
)β =
x ◦ y, hence (xβ−1) ◦ (yβ−1) = (x ◦ y)β−1.
15Clearly, if (Y, ·) is a group and eY its neutral element, then ker f = {x|x ∈ X, xf = eY } as thenef = eY .
14
There is a second way to look at AutX , observing first that EndX , the set of allendomorphisms of (X, ◦), is a monoid where the operation is given by composition, andthen that AutX is the set of invertible elements of the monoid EndX . It is easily seenthat in every monoid, the set of invertible elements is a subgroup. Thus AutX arises asan example of this general remark.
If (G, ◦) is a group, we write H ≤ G to say that H is a subgroup of (G, ◦) if due tothe context there is no doubt about the operation ◦ to which this statement refers. Wewant to illustrate in which way 1.3(4) can be instrumentalized, by showing that, withrespect to multiplication,
1.3.2. Rd(1) ≤(d)Rn(1) ≤
(c)R(1) ≤
(b)K(0, 1) ≤
(a)C where d, n ∈ N such that d|n.
Proof. (a) f : C→ R, z 7→ |z|, is a multiplicative homomorphism and ker f = K(0, 1).Hence K(0, 1) ≤ C. – (b) f : Q → C, x 7→ exp(x · 2πi), is a homomorphism of (Q,+)into (C, ·) and Qf = R(1). Hence R(1) ≤ C. – (c), (d) f : C → C, z 7→ zn, is amultiplicative homomorphism and ker f = Rn(1). Hence Rn(1) ≤ C. If d|n, then againof course Rd(1) ≤ C, and additionally Rd(1) ⊆ Rn(1) as z
d = 1 implies zn = 1.
For a second proof of (c), observe that fn : Z→ C, k 7→ exp(k·2πin
), is a homomorphism
of (Z,+) into (C, ·) and Zfn = Rn(1). Hence fn is an epimorphism of (Z,+) onto(Rn(1), ·), implying (c). Furthermore, 1.1.7 immediately allows to describe the kernel offn as the set of all multiples of n. Summarizing, we have
1.3.3. fn is an epimorphism of (Z,+) onto (Rn(1), ·), and ker fn = nZ. �
The subgroups Rn(1) where n ∈ N are in fact the only finite subgroups of (C, ·) :
1.3.4. Let U be a finite subgroup of (C, ·). Then U = Rn(1) where n = |U |.
Proof. 16 As |U | = n it suffices to prove that U ⊆ Rn(1). Let x ∈ U . Then U = xU ,implying
∏U =
∏u∈U(xu) = xn
(∏U). Hence xn = 1, i. e., x ∈ Rn(1).
A multiplication of n-th roots of unity may be visualized (thanks to 1.3.3) by adding theirdefining angles (given counter-clockwise between the connecting line with the origin andthe positive real axis). Therefore the special n-th root of unity w1 = exp 2πi
n, visualized
as the “first upper neighbour” of the trivial root of unity 1, has the property that itspowers range over all of Rn(1), equivalently, that all powers w
k1 where k ∈ n are distinct.
The same holds for the “first lower neighbour” of 1 but certainly not for all n-th rootsof unity. For example, 1 never has this property unless n = 1. For n = 4, the powers ofi and −i range over all 4-th roots of unity while the powers of 1, −1 do not.
1.4 Definition. Let n ∈ N, w ∈ Rn(1). Then w is called primitive if w,w2, . . . , wn aremutually distinct.
1.4.1. w is primitive if and only if wl 6= 1 for all l ∈ n− 1.
16The proof is valid for arbitrary fields in place of C if we adjust the meaning of Rn(1) accordingly.
15
Proof. We have wn = 1, thus trivially wl 6= 1 for all l ∈ n− 1 if w is primitive. On theother hand, if there are j, k ∈ n such that j < k and wj = wk, it follows that wk−j = 1and k − j ∈ n− 1.
We write Pn for the set of all primitive n-th roots of unity and put ϕ(n) := |Pn|. Thenϕ is a mapping of N into N, called the Eulerian function. For example, P4 = {i,−i},ϕ(4) = 2. It is easily seen that P3 = R3(1) r {1}, P5 = R5(1) r {1}, thus ϕ(3) = 2,ϕ(5) = 4.
1.4.2. Let k ∈ n, w = exp k·2πin
. Then w ∈ Pn if and only if gcd(k, n) = 1.
Proof. Suppose first that gcd(k, n) = 1. Let l ∈ n such that 1 = wl = exp lk·2πin
. Then,by 1.1.7, lk
n∈ Z. But k and n have no common prime divisor, hence17 n|l which means
that n = l. By 1.4.1, it follows that w ∈ Pn.Now let k ∈ n arbitrary, put c := gcd(k, n), d := n
c, k′ := k
c. Clearly then gcd(k′, d) = 1,
hence
expk · 2πin
= expk′ · 2πid
∈ Pd
by our first part. In particular, if c 6= 1, then d is a proper divisor of n, hence, making
use of the first (and here strict) inclusion in 1.3.2,(exp k·2πi
n
)j ≤ Rd(1) < Rn(1).
Let Tn := {d|d ∈ N, d|n}. We have proved:
1.4.3. Every n-th root of unity is a primitive d-th root of unity for a (unique) d ∈ Tn.�
As Rd(1) ⊆ Rn(1) if d ∈ Tn we conclude
1.4.4. Rn(1) =⋃d∈TnPd, n =
∑d∈Tn ϕ(d).
Here the dot above the sign for the union means that the union is disjoint. The secondpart of 1.4.4 follows from the first as n = |Rn(1)| =
∑d∈Tn |Pd| =
∑d∈Tn ϕ(d). �
We have seen that 1.4.3 and its converse hold. Hence ϕ(n) = |Pn| is the number ofall k ∈ n such that gcd(k, n) = 1 which may therefore be viewed as a purely number-theoretic definition of ϕ(n). The second part of 1.4.4 is in fact a classical result inelementary number theory (see [HW, Theorem63]) for which we have given an algebraicproof. It happens frequently and is thus a pleasant general component of abstract algebrathat a structural line of reasoning has an interesting consequence in number theory whichwould not suggest an algebraic background at first glance.18
17It should not be overlooked, however, that with this conclusion we have made use of a basic result ofelementary number theory which has not been dealt with as a topic in this chapter (see Example (1)on p. 28 and the condition (∗) on p. 29).
18“Count, for every divisor of a positive integer n, how many numbers between 1 and the divisor areprime to the divisor. If you sum up these numbers, you get : n.” The charme is that you donot need more than a chapter from the teaching curriculum for 12-year-olds to understand and –empirically – enjoy this, while the general proof makes use of properties of algebraic structures!
16
The discussion of radicals has opened up a subtopic of independent interest, that of rootsof unity. These play an important role in many branches of algebra and mathematics ingeneral. We won’t leave this subject without mentioning a famous problem concerningtheir geometric visualisation in the complex plane which leads to regular n-gons as wehave seen on p. 12:
Problem. Let n ∈ N. Starting from a line segment of length 1, is it possible to constructthe n-th roots of unity in the complex plane by ruler and compass?
The complete answer makes use of the theory to be developed in this course, hence ofabstract field theory! It is easily seen that the n-th roots of unity are constructible forn = 1, 2, 3, 4, 6, 8, but why is the answer “yes” for n = 5?, why “no” for n = 7, 9?The problem is famous because C. F.Gauss (1777-1855) settled the case n = 17 (in theaffirmative) ingeniously at the age of 19: not by giving an explicit construction, but bydeveloping a line of reasoning that such a construction exists. Clearly, a quality like thisgives far more evidence of an outstanding mathematical talent than any technical abilityof concrete constructing. Why should Gauss execute for real a complicate constructionwhen there would be no further gain in insight by doing so?19
Having discussed complex roots of unity, it is a very simple step to treat general pureequations over C:
1.5 Proposition. Let a ∈ C, n ∈ N. Write Rn(a) for the set of all complex solutionsof the equation xn = a. Then |Rn(a)| = n. If r ∈ R such that a = |a| exp(ri), w ∈ Pnand c := n
√|a| exp ri
n, then Rn(a) = {c, cw, . . . , cwn−1}.
Proof. We have c 6= 0 as a 6= 0, and |{c, cw, . . . , cwn−1}| = n as w ∈ Pn. Furthermore,(cwk)n = cnwkn = |a| exp(ri) = a for every k ∈ N0. This proves the inclusion “⊇”. Toobtain equality it is sufficient to observe that the polynomial tn − a can have at most nzeros in C.
Instead of giving a special argument for the last statement in this proof we show thefollowing proposition the final part of which is much more general than what is neededfor 1.5.
1.6 Proposition. Let K be a commutative unitary ring, f ∈ K[t].
(1) Let g ∈ K[t] be normed. Then there exist h, r ∈ K[t] such that 20
(i) f = gh+ r, (ii) r = 0K or deg r < deg g.
19The first explicit construction was published, almost 30 years after Gauss’s discovery, by someoneelse.
20Slightly more generally, it suffices to assume that the coefficient b of the term of highest degree in gbe invertible in K. If this is the case, we may apply 1.6(1) to the polynomial b−1g which is normed.Then it suffices to observe that f = (b−1g)h + r = g(b−1h) + r, deg g = deg(b−1g). In particular,the claim in 1.6(1) holds for every nonzero g ∈ K[t] if K is a field. – The commutativity of Kis essential for 1.6. Even if K satisfies all axioms of a field apart from the commutative law ofmultiplication, a polynomial over such a structure (called skew field) can have infinitely many zerosin K.
17
(2) If b ∈ K is a zero of f , there exists a polynomial h ∈ K[t] such that f = (t− b)h.
(3) If K is an integral domain, f 6= 0K, n := deg f , then there exist at most n zeros off in K.
Proof. (1) Put m := deg g. If f = 0K , the claim is trivial (h := 0K =: r). For f 6= 0Kwe proceed by induction on deg f , observing first that the claim is trivial wheneverm > deg f as then we may set h := 0K , r := f . Thus for f ∈ K we must only considerthe case m = 0 which means that g = 1K as g is normed. Then again the claim is trivialas it suffices to set h := f , r := 0K .
For the inductive step suppose that f is of degree n > 0 and the claim holds for allnonzero polynomials of degree < n in place of f . Let c0, . . . , cn ∈ K and f =
∑nj=0 cjt
j ,
cn 6= 0K . Then the polynomial f := f−gcntn−m is either 0K or of degree < n. If f = 0K ,the claim is clear as then f = gcnt
n−m and we may put r := 0K . But if f ∈ K[t] we knowinductively that f = gh + r for polynomials h, r ∈ K[t] where r = 0K or deg r < m.Then f = gcnt
n−m + gh+ r, hence the claim, putting h = cntn−m + h.
(2) We apply (1) with g := t − b. It follows that there exist h ∈ K[t], r ∈ K such thatf = (t− b)h+ r. As b is a zero of f this implies that r = 0K , hence f = (t− b)h.(3) is again proved by induction on n: The case where there is no zero of f in K istrivial, in particular the case n = 0. Now let n > 0 and suppose there exists a zerob ∈ K of f . Then (2) implies that f = (t − b)h where h ∈ K[t], deg h < n by 1.1.2.Assuming inductively that h has at most n− 1 zeros in K, our claim follows: By 1.1.4,the zeros of f in K are b and the zeros of h, hence their number is at most n.
1.7 Definition. Let K be an integral domain, b ∈ K, f ∈ K[t], j ∈ N0. Then b is calleda j-fold zero of f if there exists a polynomial h ∈ K[t] such that f = (t−b)jh, h(b) 6= 0K .The number j, called the multiplicity of b with respect to f , is uniquely determined: Theassumption (t − b)jh = f = (t − b)ih where 0 ≤ i < j, h, h ∈ K[t], h(b) 6= 0K 6= h(b)would imply that (t − b)i((t − b)j−ih − h) = 0K , hence (t − b)j−ih − h = 0K , i. e.,(t− b)j−ih = h. Now the substitution of b gives a contradiction. The element b is calleda multiple zero of f if j ≥ 2. We want to derive a criterion for an element to be amultiple zero and introduce the following formal differential calculus for this purpose:For all n ∈ N0, c0, . . . , cn ∈ K we set
( n∑
j=0
cjtj
)′:=
n∑
j=1
jcjtj−1.
The mapping d : K[t]→ K[t], f 7→ f ′, is obviously K-linear. If m,n ∈ N, we have
(tm · tn)′ = (tm+n)′ = (m+ n)tm+n−1 = mt(m−1)+n + ntm+(n−1) = (tm)′tn + tm(tn)′,
and the rule (tm · tn)′ = (tm)′tn + tm(tn)′ holds trivially if m = 0 or n = 0. Exploitingthe K-linearity of d we conclude easily the following product rule, otherwise known fromclassical differential calculus:
18
1.7.1. ∀f, g ∈ K[t] (fg)′ = f ′g + fg′ �
1.8 Proposition. Let K be an integral domain, f ∈ K[t], b ∈ K. The following areequivalent:
(i) b is a multiple zero of f ,
(ii) b is a zero of f and a zero of f ′.
Proof. Let j ∈ N0, h ∈ K[t] such that f = (t − b)jh, h(b) 6= 0K . If (i) or (ii) holds wehave f(b) = 0K , hence j ≥ 1 by 1.6(2). It follows from 1.7.1 that
f ′ = j(t− b)j−1h+ (t− b)jh′ = (t− b)j−1(jh+ (t− b)h′
),
and we see that for j = 1 the element b is not a zero of this polynomial because h(b) 6= 0K .Therefore f ′(b) = 0K if and only if j ≥ 2.
As an example, we consider the case of a pure polynomial (cf. 1.5) over an arbitraryintegral domain K. Let a ∈ K, f = tn − a for some n ∈ N. We define n1K :=1K + · · ·+ 1K︸ ︷︷ ︸
n
. Then f ′ = ntn−1 = tn−1 + · · ·+ tn−1
︸ ︷︷ ︸n
= (n1K)tn−1, by the distributive
law. Hence, by 1.8, either n1K = 0K or f has no multiple zero in K. As a consequencewe note
1.8.1. Let K be a field containing Q as a subfield, a ∈ K, n ∈ N. Then tn − a has nomultiple zero in K. �
Having considered pure polynomials in some detail, we now turn to polynomials of amore complicated nature over a field K. Under the hypothesis that k, n ∈ N, n1K 6= 0K ,we set k
n:= k(n1K)
−1. The case of a polynomial of degree 2 over a field K should be wellknown if K = R, and the solution in this case is easily generalized to arbitrary fieldsprovided that 1K + 1K 6= 0K : Let a0, a1 ∈ K. In K[t] we have the equation
(∗) t2 + a1t+ a0 =(t+
a12
)2 −(a214− a0
).
Instead of seeking immediately a solution of the equation x2 + a1x + a0 = 0K we first
approach the pure equation z2 − a = 0 where a =a214− a0. If K contains a solution r of
this equation, it suffices to put b := r − a12to obtain a zero of the polynomial originally
given. Explicitly, the zero is then given in the form
−a12
+ r where r2 =a214− a0,
hence expressed by means of (a simple additive shift of) a radical of exponent 2 over K.
The tricky transformation given by (∗) is thus to be understood as a reduction of theproblem of solving an arbitrary equation of degree 2 to the case of a pure equation ofdegree 2. Finding a solution in the general case is reduced to finding a radical : Thisis the message of (∗).
19
Although likewise true, it should not simply be considered as a step in deriving “a for-mula for the solutions of a quadratic equation” – which generations of students had tomemorize as if it were of major importance for everybody’s lifelong happyness. (For themajority of them, the opposite holds.)
The fundamental question which emanates from this success in treating the quadraticcase is now of course the following: Is it possible to reduce the problem of solvingan arbitrary equation of any degree to the problem of solving pure equa-tions? Is it always possible to express general solutions in terms of radicals?
We observe that (∗) is a special case of a simplifying step for a polynomial of arbitrarydegree n ≥ 2 which, however, will not immediately lead to a pure polynomial if n > 2:
1.9 Proposition. Let n ∈ N>1, K a field such that n1K 6= 0K . Let a0, . . . , an−1 ∈ Kand put s := t+ an−1
nin K[t]. Then there exist b0, . . . , bn−2 ∈ K such that
tn + an−1tn−1 + an−2t
n−2 + · · ·+ a1t+ a0 = sn + bn−2sn−2 + · · ·+ b1s+ b0.
Proof. We just have to make standard use of the binomial theorem to find
tn + an−1tn−1 +
n−2∑
j=0
ajtj =
(s− an−1
n
)n+ an−1
(s− an−1
n
)n−1+
n−2∑
j=0
aj(s− an−1
n
)j
= sn − an−1sn−1 + an−1s
n−1 + u
for some u ∈ 〈1K, s, . . . , sn−2〉K . Hence tn +∑n−1
j=0 ajtj = sn + u, proving the claim.
Thus, given any equation, we can always in a first step get rid of the term of secondhighest degree if the hypothesis on K in 1.9 is satisfied. For n = 3 this means: We cansolve an arbitrary equation of degree 3 over a field K with 1K +1K +1K 6= 0K if we cansolve an arbitrary equation of the form
(∗∗) x3 + b1x+ b0 = 0K (where b0, b1 ∈ K).
We shall now show that the latter always allows indeed a reduction to finding appropriateradicals if K is a field where 1K + 1K + 1K 6= 0K 6= 1K + 1K . This hypothesis will beassumed in the sequel. The solubility by radicals of cubic equations may be viewed as acornerstone in the development of modern Algebra. The result dates back to the firsthalf of the 16th century and and was not found, but finally published by GerolamoCardano in 1545. We present it in a form which emphasizes the character of a reductionto radicals in the same spirit as in the case of an equation of degree 2.
Case 1: b1 = 0K . Then the equation is pure, its solutions are radicals by definition.
Case 2: b1 6= 0K . Let z ∈ K be a solution of the pure quadratic equation
x2 −( b3127
+b204
)= 0K .
20
Let y ∈ K be a solution of the pure cubic equation
x3 +(b02− z) = 0K .
Then y 6= 0K and y − b13y
is a solution of (∗∗).
(Note that the coefficient of the pure cubic equation involves the solution z of theforegoing quadratic equation!) In Case 2, we first show that y 6= 0K . Otherwise b0
2−z =
0K , hence z = b02, implying that
(b02
)2 −( b3127
+b204
)= 0K . This means that b1 = 0K ,
contrary to the hypothesis. Thus we have y 6= 0K , z2 =
b3127
+b204, y3 = z − b0
2. Hence
(y − b1
3y
)3+ b1
(y − b1
3y
)+ b0 = y3 − b1y +
b213y− b31
27y3+ b1y −
b213y
+ b0
= z − b02− b31
27(z − b02)+ b0 = z +
b02− z2 − b20
4
z − b02
= 0K .
In the case of a quadratic equation, the field K obviously contains a solution if andonly if the radical used in its description belongs to K. In the case of a cubic equation,however, K may contain a solution even if the pure equations of degree 2 and 3 (seeCase 2 above) are not soluble in K. The solubility of these is only sufficient but notnecessary in order to find a solution in K. At this point it is not even clear if, in thecase where those required radicals do not exist in K, there exists a solution by radicalswhich is different from Cardano’s. Thus we can see that the case of a cubic equationquickly leads into rather intricate problems. We may summarize these by formulatingthe question:
Problem. What has to be done if K does not contain the required radicals z, y?
Furthermore, there is another problem to be noted with respect to Cardano’s solutionby radicals: In the classical case of K = C, the solutions are still far from the standardrepresentation in the form a + bi with a, b ∈ R. In the example x3 − 2x + 1 = 0, theattempt to transform the solutions obtained by radicals into the standard representationwill turn out to be a hard piece of work. But 1 is clearly a zero of the polynomialt3 − 2t + 1 = (t − 1)(t2 + t − 1), and the other two zeros turn out to be −1
2+ 1
2
√5,
−12− 1
2
√5. In particular, all zeros are real numbers which is not evident from the Car-
dano solution. The polynomial t3 − 2t + 1 has rational coefficients, but in Q there isjust one zero. Still it makes sense to talk about zeros in R or C. While we know thesefields, what would we have to do if an arbitrary field K is given? May we expect that,even if K does not contain zeros of a polynomial, there will exist a larger field in whichwe find the requested zeros?
Problem. What has to be done ifK does not contain any solution of a given polynomialequation?
Cardano knew already that not only the equations of degree 3, but also those of degree 4
21
are soluble by radicals. A suitable reduction to a cubic polynomial was found so that thestep from degree 3 to degree 4 is much less exciting than that from degree 2 to degree 3. Ittook about 300 years after the discovery of the solution by radicals of the cubic equationuntil the cases of a degree > 4 could be settled – in the negative: It was proved (Abel1826) that a general solution by radicals does not exist for equations of degree > 4. Evenworse, there exist equations of degree 5 with rational coefficients which are in no waysoluble by radicals (in C). Examples of this kind are x5− 4x+2 = 0, 2x5− 10x+5 = 0.In 1829, Galois characterized the equations which are soluble by radicals. The maininstrument to obtain this characterization is the famous Galois Theory which will bedeveloped in this course. It is of utmost importance for countless developments anddirections of modern Algebra which may be devoted to questions which are not at allconnected with its historical origin.
1.10 Definition. Let K be a subfield of a field L, f ∈ K[t] normed. L is called asplitting field of f over K if there exist b1, . . . , bn ∈ L (not necessarily distinct) such that
(i) f = (t− b1) · · · (t− bn),
(ii) If L is a subfield of L such that K ∪ {b1, . . . , bn} ⊆ L, then L = L.
Then b1, . . . , bn are the zeros of f in K as f(b) = 0K for some b ∈ K if and only if atleast one of the differences b − bj vanishes, i. e., if and only if b ∈ {b1, . . . , bn}. As asimple example, consider f = t2 − 2 ∈ Q[t]. The subring Q[
√2] of R is a field 21 and
clearly does not have a proper subfield containing Q ∪ {√2,−√2}. (In fact there is no
proper subfield 6= Q in Q[√2] at all!).
Problem. Given a normed polynomial f over a field K, does there exist a splitting fieldof f over K?
We shall need some more preparation to see that the answer is positive. For a subfieldK of C, this may be concluded from the following famous result:
Fundamental Theorem of Algebra (Gauss, doctoral dissertation 1799) Every poly-nomial in C[t]rC splits in C[t] into linear factors.
A field L such that every polynomial in L[t]rL splits in L[t] into linear factors is calledalgebraically closed. Thus C is closed. The following important result will not be provedin this course:
Theorem For every field K there exists an algebraically closed field L which containsK as a subfield.
Proofs may be found in numerous textbooks, see e. g. [St, 3.24]. Th same book containstwo proofs of the fundamental theorem of algebra, in its appendix §,A1. If f is a normedpolynomial over a subfield K of C, n its degree, then the fundamental theorem of algebraensures that f = (t − b1) · · · (t − bn) for suitable complex numbers b1, . . . , bn. Thus C
21If a, b ∈ Q, not both 0, then a2 − 2b2 6= 0 as 2 is not a square in Q, and(a+ b
√2)−1 = a
a2−2b2− b
a2−2b2
√2 ∈ Q[
√2].
22
satisfies condition (i) in the definition of a splitting field. The other condition, however,will not hold in general for C itself but for some subfield of C, by an application of thefollowing remark:
1.10.1. LetM be a field and X a nonempty set of subfields ofM . Then⋂X is a subfield
of M .
Proof. Clearly,⋂X is an additive subgroup of M , and
⋂X r {0M} is a multiplicativesubgroup of M . The claim follows.
If f = (t − b1) · · · (t − bn) (where b1, . . . , bn ∈ C), let X be the set of all subfields of Ccontaining K ∪{b1, . . . , bn}. Then C ∈ X , and the field L :=
⋂X is a splitting field of fover K. Since Gauss’s first proof of the fundamental theorem of algebra several othershave been found, also by Gauss himself. To some extent, they all make use of someanalytical properties of the individual field C. The general theorem on the existence ofa splitting field of a given polynomial over an arbitrary field K would follow in the sameway by making use of the quoted theorem on the existence of an embedding of K intoan algebraically closed field. It will, however, be obtained in a much simpler way in duecourse (4.3).
Let us now assume that f splits completely into linear factors t− bj in K[t]. We have
f = (t− b1) · · · (t− bn)= tn − b1tn−1 − b2tn−1 − · · · − bntn−1
+ b1b2tn−2 + b1b3t
n−2 + · · ·+ bn−1bntn−2
− b1b2b3tn−3 − b1b2b4tn−3 − · · · − bn−2bn−1bntn−3
...
+ (−1)nb1 · · · bn= tn − · · ·+ (−1)k(. . .+ bi1bi2 · · · bik + . . . )tn−k + · · ·+ (−1)nb1 · · · bn
where the sum in brackets extends over all products of k factors, i1 < i2 < · · · < ik. Wewrite sk(b1, . . . , bn) for this sum, i. e., we set
sk(b1, . . . , bn) :=∑
ij∈n
i1<i2<···<ik
bi1bi2 · · · bik for all k ∈ n,
and additionally s0(b1, . . . , bn) := 1K . The result of our calculation may then be ex-pressed as follows:
1.11 Theorem. (Viete) Let K be a field, n ∈ N, b1, . . . , bn ∈ K. Then
(t− b1) · · · (t− bn) =n∑
k=0
(−1)ksk(b1, . . . , bn) tn−k.
�
23
As an example, let b1, . . . , bn be the n-th roots of unity. Then (t−b1) · · · (t−bn) = tn−1so that 1.11 implies
sk(b1, . . . , bn) = 0 for all k ∈ n− 1, sn(b1, . . . , bn) = b1 · · · bn = (−1)n.
Now let m := ϕ(n) and b1, . . . , bm be the primitive n-th roots of unity. The polynomialφn :=
∏i∈m(t−bi) is called the n-th cyclotomic polynomial, being related to the problem
of dividing the unit circle into n equal parts (cf. p.12). Clearly, degφn = ϕ(n). Apartfrom the case n = 1 we have φn(0) = 1 because Pn is closed under taking multiplicativeinverses (and, for n = 2, the assertion is obvious). We will prove in A.1.1 that φn ∈ Z[t].For example, for n = 1, 2, 3, 4, 5 we obtain ϕ(n) = 1, 1, 2, 2, 4 and the polynomialsφn = t− 1, t+ 1, t2 + t+ 1, t2 + 1, t4 + t3 + t2 + t+ 1 resp.
The value sk(b1, . . . , bn) does not depend on the order of the elements bi, i. e.,
1.11.1. ∀π ∈ Sn ∀b1, . . . , bn ∈ K sk(b1, . . . , bn) = sk(b1π, . . . , bnπ)
where Sn := Sn.
1.12 Definition. Let B,A be sets, n ∈ N. A mapping g of Bn to A is called symmetricif (b1, . . . , bn)g = (b1π, . . . , bnπ)g for all b1, . . . , bn ∈ B. Then the contents of 1.11.1 maybe expressed as follows: For all k ∈ n, the mapping
Kn → K, (b1, . . . , bn) 7→ sk(b1, . . . , bn),
is symmetric. It is called the k-th elementary symmetric function. Viete’s theorem 1.11essentially means that, up to the sign, the coefficients of a normed polynomial f over afield are the values of the elementary functions for the tuple of its zeros in a splittingfield. In an intricate but most regular way, the coefficients of f are composed by them:
1.11’ Let K be a field, n ∈ N, a0, . . . , an−1 ∈ K, f = tn+an−1tn−1+· · ·+a1t+a0. Suppose
that we are given a splitting field of f so that f splits into linear factors t−b1, . . . , t−bn.Then
an−k = (−1)ksk(b1, . . . , bn) for all k ∈ n.
Although we know, by Viete’s theorem, in which way the zeros of a polynomial (in asplitting field) are transformed into its coefficients by means of the elementary symmetricfunctions, there is no visible general way back to “decipher this code” and reveal, viceversa, the zeros from those.
24
2. Algebraic structures and the
extension principle
For the sake of a clear orientation let us briefly recall some fundamental algebraic ter-minology:
(I) Algebraic structures with one operation
a) Magma : Any set with an operation (see 1.2).Examples: (N,∧) where a ∧ b := ab; (Z,−).
b) Semigroup : Associative magma.Examples: (X,⊣) where X is any set, a ⊣ b := b; (N,+).
c) Monoid : Semigroup with neutral element.Examples: (N0,+); (N, ·); (XX , �) where X is any set and � denotes thecomposition of mappings, i. e., f � g maps x ∈ X to (xf)g. Usually we justwrite fg and use this notation in its natural sense, i. e., the mapping writtenat the first place is the mapping which is applied first.
d) Group : Monoid in which every element is invertible.Examples: (SX , �), the group of invertible elements of the monoid (XX , �)where X is any set; (Z,+).
(II) Algebraic structures with two operations
a) Double magma22: Any set with two operations.
b) Ring : Double magma (R,+, ·) where (R,+) is an abelian23 group, (R, ·) isa semigroup, and both distributive laws hold.24
Example: (EndA, †, �) where A is an abelian group.
c) Integral domain : Commutative unitary ring (R,+, ·) such that R is multi-plicatively closed.Example: (Z,+, ·).
22This term is not commonly used in the literature.23for groups, synonymous with “commutative”24With regard to the notion of a ring, confusion of terminology has been created and spread by a certain
influential book on Algebra. Therefore we have reason to emphasize that a ring does not necessarilycontain a unit element by definition. If there is one and distinct from the additively neutral element,the ring is called unitary in this text.
25
d) Field : Integral domain (R,+, ·) such that R is multiplicatively a group.Examples: (Q,+, ·), (C,+, ·), (Q[
√2],+, ·) (see 1.10); ({, },+, ·) where
+, · are defined as follows:
+
·
(III) Algebraic structures with a set of operators (which we assume here to be a com-mutative unitary ring K)
a) K-magma, K-double magma resp. : Magma, double magma resp., M forwhich there is given some “outer product” K ×M → M .
b) K-space : see 1.1. A K-space where K is a field is called a vector space.
c) K-algebra : see 1.1.Examples: Kn×n; EndKV for a K-space V ; any polynomial ring over K.
We have constructed a polynomial ring over K at the beginning of chapter 1 and in-troduced the notation K[t] for a polynomial ring (in the indeterminate t) over K whichcontains K as a unital subring (see p. 9), but we have not yet established its existence.This will be an immediate consequence of the following general proposition:
2.1 Proposition (Extension Principle). Let B, M be sets and ϕ an injective mappingof B into M . Then there exists a set B containing B and a bijection ϕ of B onto Msuch that ϕ|B = ϕ.
✲
✲
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B
B Bϕ
M
ϕ
ϕ
Supplement. Let ◦ be an operation on B, · an operation on M such that ϕ is ahomomorphism. Then there exists an extension ◦ of ◦ to an operation on B such thatϕ is an isomorphism of (B, ◦) onto (M, ·).25
Given a commutative unitary ringK, we have a monomorphism ofK into the polynomialring Π0 over K (see p. 1.1). Applying 2.1 and its supplement to both operations onK, weobtain that ι extends to an isomorphism of a double magma containing K onto Π0. Thisdouble magma is therefore a polynomial ring over K, containing K as a unital subring.We have thus justified our usage of the notation K[t]. A considerable generalization willbe considered in 2.9.
Proof of 2.1. We will make use of the following purely set theoretic (not completely
25Similarly, if B and M are K-magmas, B may be made into a K-magma by extending the definitionof the product between K and B such that ϕ is an isomorphism of the K-magmas B and M .
26
trivial) statement:
Given any two sets A, B, there exists a set A′ which is equipotent with A and disjointfrom B.26
We apply this to A := M r Bϕ, set B := B∪A′ and choose a bijection ψ of A′ onto A.Then the mapping ϕ := ϕ ∪ ψ is a bijection27, and ϕ|B = ϕ.
As for the supplement, we define the operation ◦ on B by x◦x′ := (xϕ · x′ϕ)ϕ−1 for allx, x′ ∈ B. For all x, x′ ∈ B then x◦x′ = (x◦x′)ϕϕ−1 = (x◦x′)ϕϕ−1 = x◦x′, i. e., ◦ is anextension of ◦. By definition of ◦, (x◦x′)ϕ = (xϕ · x′ϕ)ϕ−1ϕ = xϕ · x′ϕ for all x, x′ ∈ B.Thus ϕ is a homomorphism. Being bijective, it is an isomorphism. �
2.2 Definition. A submonoid of a magma (M, ·) is a subset S of M such that (S, ·)is a monoid. Assume in the following that (M, ·) is a monoid. As usual, we call asubmonoid S unital if 1M = 1S. Frequently, only unital submonoids are of interest in agiven context.28 The following neat rule plays an important role in parts of group theorybut hold under more general hypotheses as follows:
2.2.1. (Generalized “Dedekind’s law”) Let S be a unital submonoid of (M, ·), U a set ofinvertible elements of (S, ·). Then for all T ⊆M
U(T ∩ S) = (UT ) ∩ S, (S ∩ T )U = S ∩ (TU)
(The iuxtaposition here stands for the subset product induced by · as introduced in 1.2.)
Proof. We prove the first equation, the second is obtained similarly. Obviously U(T ∩S)is a subset of both UT and S. Let s ∈ (UT ) ∩ S. Then there exist u ∈ U , t ∈ T suchthat s = u · t. It follows that t = (u−1 · u) · t = u−1 · (u · t) ∈ S.
Clearly, we have
2.2.2. Let X be a nonempty set of unital submonoids of (M, ·). Then⋂X is a unital
submonoid of (M, ·). �
In particular, for any T ⊆ M the intersection of all unital submonoids containing T isa unital submonoid of (M, ·). It is obviously uniquely determined as the smallest unitalsubmonoid containing T . It is called the unital submonoid generated by T . If it equalsM , T is called a generating system of (M, ·). The multiplicative closure of T ,
{m|m ∈M, ∃l ∈ N0 ∃t1, . . . , tl ∈ T m = t1 · · · tl}26Sketch of a proof: Let X := {x|x ∈ ⋃
B, x /∈ x}, A′ :={{a,X}|a ∈ A
}, f : A→ A′, a 7→ {a,X}.
Then f is a bijection and A′ ∩B = ∅. (Observe first that {x|x ∈M, x /∈ x} /∈M for every set M .)
27explicitly, ϕ : B →M, x 7→{xϕ if x ∈ Bxψ if x ∈ A′
.
28Therefore, many authors find it convenient to define a submonoid as unital, thus exclude non-unitalsubmonoid structures right from the beginning. If the neutral element of (M, ·) is the only idempotentelement of M , i. e., satisfying the equation x2 = x, then clearly every submonoid is automaticallyunital. This is the case if (M, ·) is a group, but also in other important monoids like (N, ·), (N0,+).
27
is a unital submonoid and contained in every unital submonoid containing T . Hence itis the unital submonoid generated by T .
Now let · be commutative. Then every element of the unital submonoid generated by Tmay be written in the form
(∗) tj11 · · · tjnn where n ∈ N0, j1 . . . , jn ∈ N, t1, . . . , tn mutually distinct.
The order of the elements t1, . . . , tn is arbitrary as · is commutative. The set T is calledindependent if changing the order of the factors in (∗) is the only possibility to repres-ent an element of the unital submonoid generated by T as a product of powers 6= 1Mof mutually distinct elements of T , i. e., if for all n, n′ ∈ N0, t1, . . . , tn, s1, . . . , sn′ ∈ T ,j1 . . . , jn, i1, . . . , in′ ∈ N the following holds: If t1, . . . , tn are mutually distinct, s1, . . . , sn′
are mutually distinct, and tj11 · · · tjnn = si11 · · · sin′
n′ , then n = n′, and there exists a per-mutation π of n such that tk = skπ, jk = ikπ for all k ∈ n. An independent generatingsystem of (M, ·) is called a basis of the commutative monoid (M, ·).29 If T is a basisof (M, ·) and t ∈ T , obviously the powers t, t2, t3, . . . must be mutually distinct. Inparticular, no basis of (M, ·) contains 1M , and furthermore
2.2.3. If a commutative monoid (M, ·) with |M | 6= 1 has a basis, then M is infinite. �
A commutative monoid is called free if it has a basis. If T is a basis, then the monoidis called free over T or freely generated by T . In contrast with the notion of a basis invector space theory, we have
2.2.4. A commutative monoid has at most one basis. If T is a basis, U any generatingsystem, then T ⊆ U .
Proof. It suffices to prove the last assertion because it implies that any two bases containeach other, hence conincide. Let t ∈ T . There exist u1, . . . , uk ∈ U such that t =u1 · · ·uk. Every uj has a representation as a term in (∗) so that t turns out to be aproduct of such terms. Now the definition of independence implies that k = 1, t = u1,hence t ∈ U .
For example, the commutative monoid (N0,+) is free over {1}, (N0×N0,+) is free over{(1, 0), (0, 1)}, but the commutative monoid (Z,+) is not free: Any generating systemU of the monoid (Z,+) must contain at least one positive integer n and one negativeinteger z. But for all k ∈ N we have (k|z|)n+ (kn)z = kn(−z + z) = 0. Hence U is notindependent. We mention without proof two important examples of free commutativemonoids:
Examples. (1) (N, ·) is a free commutative monoid.
(2) Let K be a field, V a polynomial ring over K and N the set of all normed elementsof V . Then (N, ·) is a free commutative monoid.
29We have chosen the multiplicative notation here. Of course, in an additively written context, productsmust be replaced by sums, powers by multiples, so “tj11 · · · tjnn ” becomes “j1t1 + · · ·+ jntn”, etc.
28
In fact, (1) is just a different way of formulating the so-called fundamental theoremof arithmetic about existence and uniqueness of a prime decomposition of positive in-tegers.30 The (by 2.2.4 unique) basis of (N, ·) is the set of primes. This example showshow closely some algebraic and number-theoretic notions are related. The familiar no-tion of divisibility of integers has a natural generalization for arbitrary commutativemonoids which provides a unified approach to examples like (1), (2): Let (M, ·) be acommutative monoid, a, b ∈ M . We write b |
M
a (“b divides a in M”) if there exists an
element x ∈ M such that bx = a, in which case we call b a divisor of a in M . The rela-tion |
M
is reflexive and transitive. The divisors of 1M are clearly the invertible elements
(also called units) of M and form a multiplicative (unital) subgroup of (M, ·) (cf. thecomments on 1.3.1).31
2.2.5. u |M
b for all b ∈ M and units u of M ,
because ux = 1R implies, for all b ∈M , that u(xb) = b. �
An element q ∈ M is called indecomposable if q is not a unit of M and q = ab witha, b ∈ M implies that a is a unit or b is a unit of M . In other words, indecomposableelements are those non-units which admit a product decomposition only by means ofa unit.32 Indecomposable elements are also called irreducible. In particular, this is thestandard term used in the case of the multiplicative structure of a polynomial ring overa commutative unitary ring. In the examples (1), (2), it is readily proved by a standardinduction that the considered monoid (M, ·) is generated by the set T of indecomposableelements. The independence of T is a consequence of the following nontrivial propertyfor arbitrary elements q ∈ T :
(∗) ∀a, b ∈M (q |M
ab⇒ q |M
a or q |M
b).
If t1, . . . , tn, likewise s1, . . . , sn′, are mutually distinct elements of T , and tj11 · · · tjnn =
si11 · · · sin′
n′ , n ≥ 1, then (∗) implies (after a standard inductive extension to finitely manyfactors) that tn |
M
sj for some j ∈ n′. As tn is a non-unit and sj is indecomposable, we
obtain – not in general, but for the monoids considered in (1), (2) – that tn = sj , hence
tj11 · · · tjn−1
n−1 tjn−1n = si11 · · · s
ij−1j · · · sin′
n′ . Now the number of factors on both sides is lessthan before which shows in which way (∗) opens the way for an induction argument fora proof of the independence of T . Its further details are standard and left to the reader.
30See [HW, Theorem 2].31Note, however, that a zero divisor b in a commutative ring K (recall 1.1) satisfies an equation
bx = 0K for some x ∈ K. The condition that x 6= 0K is of course essential as b |K
0K holds for every
b ∈ K: b0K = 0K . Furthermore, the reader should be aware of the difference between the term“unit element” as a synonym for “neutral element” (see p. 5) and the term “unit” as a synonym for“invertible element”.
32Clearly, for every unit u we have u(u−1b) = b for all b ∈ M so that product decompositions whereone of the factors is a unit exist always.
29
The main argument for a proof of (∗) may be found on p. 63.
We introduce a notation for an “essentially finite” product (sum resp.) in a commutativemonoid (M, ·) ((M,+) resp.): Given a set T and a mapping T → M , t 7→ xt, such thatxt = 1X (xt = 0X resp.) for almost all t ∈ T 33, we set
∏
t∈T
′xt :=
∏
t∈T, xt 6=1M
xt ,( ∑
t∈T
′xt :=
∑
t∈T, xt 6=0M
xt resp.)
Using this notation, we may express the basis property of a subset T of a (multiplicativelywritten) commutative monoid (M, ·) as follows: T is a monoid basis if and only if forevery m ∈ M there exists a unique mapping T → N0, j 7→ jt, such that jt = 0 foralmost all t ∈ T and m =
∏′t∈T t
jt. In the sequel, we simply write M instead of (M, ·) ifthe operation · is either generic or well-known from the context. In general, we use themultiplicative notation unless otherwise specified.
2.3 Proposition (The universal property). Let M be a free commutative monoid, T itsbasis. Let N be any commutative monoid and f0 a mapping of T into N . Then thereexists a unique extension of f0 to a unital homomorphism of M into N .
Proof. 34 Uniqueness: Let f, g be homomorphisms of M into N such that f |T = f0 =g|T . Let m ∈ M , represented in the form
∏′t∈T t
jt as discussed above. Then mf =∏′t∈T (tf0)
jt = mg. Hence f = g.
Existence: If m =∏′
t∈T tjt ∈ M , the independence of T allows us to define without
ambiguity mf :=∏′
t∈T (tf0)jt. Thus we obtain a mapping f of M into N which is an
extension of f0. As 1M is the empty product inM , 1N the empty product in N , we have1Mf = 1N . Let m
′ =∏′
t∈T tj′t ∈M . Then
(mm′)f =(∏
t∈T
′tjt+j
′t
)f =
∏
t∈T
′(tf0)
jt∏
t∈T
′(tf0)
j′t = mf ·m′f
We make a first application, to be seen as a model case, of the Extension Principle 2.1:
2.4 Proposition. Let T be a set.
(1) There is a commutative monoid which is freely generated by T .
33A mapping with this property is commonly called “of finite support”. If the condition holds for allt ∈ T , ∏′
t∈T xt becomes the empty product in (M, ·) (the empty sum in (M,+) resp.) which equals1M (0M resp.)
34A proof of an assertion of existence and uniqueness of a certain object may frequently be approachedcomfortably be considering first the uniqueness part and afterwards the existence part: Provinguniqueness means to show that there is just one single candidate for the object in question. Inmany cases, this line of reasoning sheds enough light on that unique candidate to define it then asis needed for the existence part.
30
(2) Let β be a bijection of T onto a set T ′. Suppose thatM ,M ′ are commutative monoidswhich are freely generated by T , T ′ resp. Then there exists a unique extension of βto an isomorphism of M onto M ′.In particular, a commutative monoid which is freely generated by T is unique up toisomorphism.
Proof. (1) For all f, g ∈ NT0 we put
f † g : T → N0, t 7→ tf + tg.
Then (NT0 , †) is a commutative monoid, o : T → N0, t 7→ 0, its neutral element. For
every s ∈ T define
δs : T → N0, t 7→{1 if t = s
0 otherwise
and set S := {δs|s ∈ T}. We claim:
(∗) S is an independent subset of NT0
Let n, n′ ∈ N0, t1, . . . , tn ∈ T mutually distinct, likewise s1, . . . , sn′, moreover j1, . . . , jn,i1, . . . , in′ ∈ N such that j1ft1 † · · · † jnftn = i1fs1 † · · · † in′fsn′ . For every k ∈ n we havetk(j1ft1 †· · ·†jnftn) = jk 6= 0, hence there exists l ∈ n′ such that tk = sl. As the elementssi are mutually distinct this implies tk(i1fs1 † · · · † in′fsn′ ) = il, hence jk = il. Therefore,for each k ∈ n there exists a unique l ∈ n′ such that tk = sl, jk = il. This also holdsreversely. We conclude that n = n′ and there is a permutation π of n such that tk = skπ,jk = ikπ for all k ∈ n. This shows (∗).Thus the submonoid S of NT
0 generated by S is free over S. This puts us in a positionto apply 2.1: The mapping β : T → S, t 7→ δt, is a bijection. Hence there exists a setT containing T , an extension of β to a bijection β of T onto S, and an operation on Tsuch that β is an isomorphism. It follows that T is a commutative monoid and T is abasis of T because S is a basis of S.
✲
✲ S
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T
T S
NT
β
β
(2) By the universal property 2.3, there exists an extension of β to a homomorphism β ofM into M ′. On the other hand, applying 2.3 to β−1, we obtain an extension of β−1 to a
homomorphism β−1 of M ′ into M . We have β β−1|T = idT , hence β β−1 = idM , likewise
31
β−1β = idM ′. It follows that β is bijective, hence an isomorphism. Homomorphismswhich coincide on T are equal as M is the multiplicative closure of T . It follows thatthe homomorphic extension β of β is unique.
We write T • for the (up to isomorphism) unique free commutative monoid over the setT . Its elements may be assumed in the form
tj11 · · · tjkk where k ∈ N0, t1, . . . , tk ∈ T mutually distinct, j1, . . . , jk ∈ N.
Let K be a commutative unitary ring, V a K-space, • an operation on V . Observethat the two conditions in 1.1 which make V into a K-algebra may be replaced bythe requirement that the operation • be a K-bilinear mapping V × V → V . Thus aK-algebra is a special case of the notion defined in the following:
2.5 Definition. Let K be a commutative unitary ring, A, V K-spaces, ◦ a bilinearmapping of A×V into V . Then V 35 is called an A-left module over K. If ◦ is a bilinearmapping of V ×A into V , V is called an A-right module over K.
We write T ≤AV if T is a K-subspace of V such that a ◦ t ∈ T (t ◦ a ∈ T resp.) for all
a ∈ A, t ∈ T , called an A-submodule of V , also an A-invariant K-subspace.
Now let (A,+, •) be an associative unitary K-algebra. Let ◦ be a bilinear mapping ofA× V into V . Then V is called an A-algebra left module over K if
∀a, b ∈ A ∀v ∈ V (a • b) ◦ v = a ◦ (b ◦ v).
If, moreover, 1A ◦ v = v for all v ∈ V , the A-module V is called unital. (Analogously, if◦ is a bilinear mapping of V ×A into V such that v ◦ (a • b) = (v ◦ a) ◦ b for all a, b ∈ A,v ∈ V , we call V an A-algebra right module over K, and unital if v◦1A = v for all v ∈ V .)The investigations of unital A-algebra modules for a given associative unitary K-algebraA is one of the topics of major interest in current algebraic research, the area of Repres-entation Theory. A systematic approach to the fundamentals on modules and associativealgebras may be found in [Pi]. The simplest example of an A-module is given by V = A,◦ = •. Then we have two actions of A on itself: left multiplication, thus making Ainto an A-algebra left module, and right multiplication, making A into an A-algebraright module. A submodule with respect to left multiplication is called a left ideal, withrespect to right multiplication a right ideal. A one-sided ideal of A is a subspace of Awhich is a left ideal or a right ideal.
Examples. (1) Let n ∈ N, A := Kn×n. Then
T :=
{
c1 . . . cn0K . . . 0K...
...0K . . . 0K
| ci ∈ K
}
35More precisely: the pair (V, δ) where δ is the mapping of A into EndKV which assigns to every a ∈ Athe endomorphism of V induced by a via ◦. This K-linear mapping δ is called the action of A on V .
32
is a right ideal of A while the set of corresponding transposes,
tT :=
{c1 0K . . . 0K...
......
cn 0K . . . 0K
| ci ∈ K
}
is a left ideal of A.36 – Although true in the special case of a one-sided ideal, oneshould not have the intuitive idea that an A-module is “something which is smallerthan A”. In the next example, quite the opposite is the case although, again, themodule action is given by the multiplication in an algebra:
(2) Let A be a unital subalgebra of an associative unitary K-algebra (B,+, •). Then Bbecomes a unital A-algebra left module (right module resp.) if we define
∀a ∈ A ∀b ∈ B a ◦ b := a • b (b ◦ a := b • a resp.)
(i. e., ◦ is the restriction of • to A× B (B × A resp.).) In particular, if the algebraA is a field, the A-module properties of B amount to saying that B is an A-vectorspace.
(3) Consider the case A = K. Given a K-space V , which actions ◦ of K exist to makeV into a unital K-module? For any such action we must have c ◦ v = (c · 1K) ◦ v =1K ◦ (cv) = cv, exploiting (very weakly) that ◦ is bilinear and the module is unital.Hence the only possible definition of ◦ is given by the K-space structure of V westarted from: K-modules (over K) are nothing else than K-spaces.
(4) Let (A,+, •) be a unitary associative K-algebra, X a set, V := AX . We make Vinto a K-space by defining, for all f, g ∈ V ,
f † g : X → A, x 7→ xf + xg, c · f : X → A, x 7→ c(xf) for all c ∈ K.
Now we set a◦f : X → A, x 7→ a•(xf), for all a ∈ A, f ∈ V . Then the distributivelaws and the K-space property of A imply that ◦ is bilinear. Furthermore, exploitingthat A is associative we obtain for all a, b ∈ A, f ∈ V , x ∈ X
x((a • b) ◦ f
)= (a • b) xf = a •
(b • (xf)
)= a •
(x(b ◦ f)
)= x
(a ◦ (b ◦ f)
),
hence (a•b)◦f = a◦(b◦f). Finally, x(1A◦f) = 1A•(xf) = xf , i. e., 1A◦f = f. Thus Vis a unital A-algebra left module. – Note that in the case of X = n (for some n ∈ N)we have the familiar notation of the elements of V as n-tuples over A, and the action◦ of A on V (= An) is just given by the rule a ◦ (a1, . . . , an) := (a • a1, . . . , a • an) forany a, a1, . . . an ∈ A. It simply extends left multiplication within A to multiplyingn-tuples over A from the left. Clearly, this would work analogously from the right.
36Instead of the first, any other row (column resp.) could have been chosen in this definition whichthus supplies n right ideals and n left ideals of A. In the case of a field K it is an easy exercise toshow that these one-sided ideals are minimal in the sense that they are 6= {0A} and do not containany proper one-sided ideal of A apart from {0A}. This shows that the K-algebra Kn×n is a direct
sum of n minimal left ideals and a direct sum of n minimal right ideals.
33
The following remark is straightforward. In particular, we see that a K-space for ahomomorphic unital image K of K is, in the described natural way, also a K-space:
2.5.1. Suppose that A is a K-algebra, V an A-left module (by a bilinear mapping ◦ ofA× V into V ), and a K-algebra homomorphism of A into A, a 7→ a, is given. Then Vbecomes an A-left module by defining a ◦ v := a ◦ v for all a ∈ A, v ∈ V. �
Let A be a unitary associative K-algebra, V a unital A-algebra left module. A subsetY of V is called A-independent if for every finite subset U of Y and for every mappingU → A, u 7→ au, the equation
∑u∈U au ◦u = 0V implies that au = 0A for all u ∈ U . This
generalizes the familiar notion of linear independence in vector spaces. Clearly, {0V } isnot contained in any A-independent subset of V .
We write 〈Y 〉A for the additive closure of the set of all elements a ◦ y where a ∈ A,y ∈ Y . This is obviously the smallest A-submodule of V containing Y and called thesubmodule generated by Y . If 〈Y 〉A = V , V is called an A-generating system of V . AnA-basis of V is an A-independent A-generating system of V .
Let V ′ be a further A-left module, given by a bilinear mapping ◦′ of A × V into V .A mapping ϕ of V into V ′ is called an A-module homomorphism (or, in short, an A-homomorphism) if it is K-linear and if (a ◦ v)ϕ = a ◦′ vϕ for all a ∈ A, v ∈ V .37
2.5.2 (The universal property). Let V,W be unital A-algebra modules, Y an A-basis ofV . Then every mapping of Y into W extends uniquely to an A-homomorphism of Vinto W .
Proof. Uniqueness: Let f0 be a mapping of Y into W . Let f be an extension of f0 toan A-homomorphism of V into W . Then, for any mapping of finite support Y → A,y 7→ ay, we have
(∑
y∈Y
′ay ◦ y
)f =
∑
y∈Y
′ay ◦′ (yf) =
∑
y∈Y
′ay ◦′ (yf0).
Hence f is uniquely determined by f0.
Existence: As Y is an A-basis of V , for every v ∈ V there is a unique mapping offinite support Y → A, y 7→ a
(v)y , such that v =
∑′y∈Y a
(v)y ◦ y. Therefore we may define
f : V → W by vf :=∑′
y∈Y a(v)y ◦′ (yf0) which obviously is a K-linear extension of f0.
Moreover, (a ◦ v)f =∑′
y∈Y (a • a(v)y )(yf0) =
∑′y∈Y a ◦′ (a
(v)y ◦′ yf0) = a ◦′ (vf) for all
a ∈ A, v ∈ V .
37From this it should also be read off what an A-mono/-epi/-iso/-endo/-automorphism is: an A-homomorphism which is injective/surjective/bijective/from V into V /from V into V and bijective.
34
2.6 Proposition. Let K be a commutative unitary ring, B a unitary associative K-algebra, A a unital subalgebra of B, V a unital B-algebra left module, Z ⊆ B, Y ⊆ V .Set Z ◦ Y := {z ◦ y|z ∈ Z, y ∈ Y }.(1) If 〈Z〉A = B, 〈Y 〉B = V , then 〈Z ◦ Y 〉A = V .
(2) If Z is A-independent, Y is B-independent, then the mapping Z × Y → Z ◦ Y ,(z, y) 7→ z ◦ y is bijective and Z ◦ Y is A-independent.
(3) If Z is an A-basis of B, Y is a B-basis of V , then Z ◦ Y is an A-basis of V .
Special case Suppose that A, B are fields and dimAB, dimB V are finite. Then dimA Vis finite and dimA V = dimAB · dimB V .
Proof. (1) Let v ∈ V . By hypothesis, there exist elements y1, . . . , yn ∈ Y , b1, . . . , bn ∈ Bsuch that v =
∑j∈n bj◦yj, and, for all j ∈ n, elements z
(j)1 , . . . , z
(j)kj∈ Z, a(j)1 , . . . , a
(j)kj∈ A
such that bj =∑
i∈kj a(j)i • z(j)i . It follows that
v =∑
j∈n
(∑
i∈kj
a(j)i • z
(j)i
)◦ yj =
∑
j∈n
∑
i∈kj
aji ◦ (z(j)i ◦ yj) ∈ 〈Z ◦ Y 〉A .
(2) The mapping Z × Y → Z ◦ Y , (z, y) 7→ z ◦ y is surjective by definition of Z ◦ Y .If z, z′ ∈ Z, y, y′ ∈ Y such that z ◦ y = z′ ◦ y′, then z ◦ y − z′ ◦ y′ = 0V , hence y = y′,z = z′ as Y is B-independent and z, z′ 6= 0B. Now let U be a finite subset of Z ◦ Y . Foreach y ∈ Y set Zy := {z|z ∈ Z, z ◦ y ∈ U}, and for every pair (z, y) ∈ Z × Y such thatz ◦ y ∈ U let az,y ∈ A. Then
0V =∑
z∈Z,y∈Yz◦y∈U
az,y ◦ (z ◦ y) =∑
y∈Y
( ∑
z∈Zy
az,y • z)◦ y
implies that∑
z∈Zyaz,y • z = 0B for every y ∈ Y as Y is B-independent. Furthermore,
the A-independence of Z implies that for each y ∈ Y we have az,y = 0B for all z ∈ Zy.Hence Z ◦Y is A-independent. (3) is an immediate consequence of (1), (2), and triviallyimplies the special case.
2.7 Proposition (Generalized Independence Lemma (Dedekind)). Let R be an integraldomain, X a magma, H the set of nonzero homomorphisms of X into (R, ·). Then H isan R-independent subset of the R-module RX (see Example (4) on p. 33, A := K := R).
Proof. Otherwise let n ∈ N be minimal such that∑
i∈n cifi = 0RX for some ci ∈ R andmutually distinct f1, . . . , fn ∈ H. Then ci 6= 0R for all i ∈ n. Exploiting our R-linearrelation for the fi in two different ways we obtain, for arbitrary x, x′ ∈ X , the equations
∑
i∈nci(x
′fi)(xfi) =∑
i∈nci (x
′x)fi = 0R,
∑
i∈nci(x
′fn)(xfi) = x′fn∑
i∈nci (xfi) = 0R,
35
hence∑
i∈n−1 ci(x′fi−x′fn)(xfi) = 0R by subtraction. For arbitrary x′ ∈ X we conclude∑
i∈n−1
(ci x
′(fi − fn))fi = 0RX , implying ci x
′(fi−fn) = 0R for every i ∈ n, by the choiceof n. For all i ∈ n we have ci 6= 0R, hence fi−fn = 0RX , i. e., fi = fn, as R is an integraldomain. Since the functions fi are mutually distinct, we obtain n = 1. Hence c1f1 = 0RX
and finally f1 = 0RX as c1 6= 0R and R is an integral domain, a contradiction.
2.8 Proposition. Let K be a commutative unitary ring, A an associative unitary K-algebra, X a set.
(1) There exists a unital A-algebra module with A-basis X.
(2) Let β be a bijection of X onto a set X ′. Suppose that V , V ′ are unital A-algebramodules with A-bases X, X ′ resp. Then there exists a unique extension of β to anA-isomorphism of V onto V ′.In particular, a unital A-algebra module with basis X is unique up to A-isomorphism.
The proof of this 2nd application of the Extension Principle 2.1 follows the pattern of2.4:
(1) For every y ∈ X define
δy : X → A, x 7→{1A if x = y
0A otherwise
and Y := {δy|y ∈ X}. With reference to Example (4) on p. 33 we claim:
(∗) Y is an independent subset of the A-module AX
Clearly, the mapping X → Y , y 7→ δy, is a bijection. Let n ∈ N, y1, . . . , yn ∈ X bemutually distinct, a1, . . . , an ∈ A such that
∑i∈n ai ◦ δyi = 0AX . By applying the left
hand side to the elements yi we see that ai = 0A for all i ∈ n. This proves (∗).It follows that the A-submodule Y of AX generated by Y is free over Y . By 2.1 thereexists a set X containing X , an extension of β to a bijection β of X onto Y , andoperations that make X into an A-module which is isomorphic to Y via β. The claimin (1) follows. The universal property 2.5.2 allows us to show (2) in complete analogyto the proof of 2.4(2). �
We write AX for the free A-algebra module with basis X . For every element v of AXthere is a unique mapping of finite support X → A, x 7→ ax, such that v =
∑′x∈X ax ◦x.
In the following, we choose A = K and suppose that some operation · on the set Xis given. In other words, we take the underlying set of a magma (X, ·) as a basis of aK-space. The K-space KX is then equipped with a natural multiplicative operation •,given by ∑
x∈X
′cxx •
∑
x∈X
′dxx :=
∑
x,y∈X
′cxdy(x · y) (cx, dx ∈ K),
36
obviously an extension of the operation · given on X to KX . We distinguish herecarefully between different kinds of multiplications for the sake of clarity in the basicdefinitions. Nevertheless we will later adopt the convention to not write any multiplic-ation symbol at all in formulas if the nature of the factors (for example, “scalar timesvector”) allows a unique interpretation as to which type of product is being referred to.
The two distributive laws
∀u, v, w ∈ KX u • (v + w) = u • w + u • w, (v + w) • u = v • u+ w • u
are readily verified. Thus (AX,+, •) is a K-algebra.
2.8.1. Let K be a commutative unitary ring, X a magma, B a K-algebra, ϕ0 a multi-plicative homomorphism of X into B. Then
ϕ : KX → B,∑
x∈X
′cxx 7→
∑
x∈X
′cx(xϕ0)
is an extension of ϕ0 to an K-algebra homomorphism.
For a proof, it suffices to observe that ϕ is clearly K-linear and, for any choice of scalarscx, dx ∈ K (where only a finite number is different from 0K),
(∑′x∈X cxx•
∑′x∈X dxx
)ϕ =∑′
x,y∈X cxdy (x · y)ϕ0 =∑′
x,y∈X cxdy (xϕ0)(yϕ0) =(∑′
x∈X cxx)ϕ(∑′
x∈X dxx)ϕ. �
Clearly, • is associative (commutative resp.) if and only if · is associative (commutativeresp.). The algebra KX is called the semigroup algebra (or semigroup ring) of X overK if X is a semigroup, similarly the group algebra (or group ring) of X over K if Xis a group. The latter plays a fundamental role in the Theory of Representations andCharacters of Finite Groups.
If X is a monoid, the K-algebra KX is associative and unitary: 1X is its multiplicativelyneutral element. In the sequel we consider the case where X is a free commutative mon-oid over some set T (see p. 32). If |T | = 1, say T = {t}, thenX = T • = {1X, t, t2, t3, . . . }.In this case KX is a polynomial ring over K in the variable t (see 1.1).
2.9 Definition. Let K be a commutative unitary ring, (P,+, ·) an associative com-mutative unitary K-algebra. Suppose that there exists an independent subset T of thecommutative monoid (P, ·) the multiplicative closure of which is a K-basis of P . ThenP is called a polynomial ring over K in the set of variables (or indeterminates) T , and T iscalled a K-algebra basis of P .
Clearly, the independence of T means that the multiplicative closure of T is a free com-mutative monoid over T , hence the K-algebra isomorphism P ∼= KT • holds. Vice versa,we have
2.9.1. For every set T the K-algebra KT • is a polynomial ring over K in the set ofvariables T . �
A routine third application of the Extension Principle 2.1 now gives
37
2.9.2. If K ∩ T = ∅ there exists a polynomial ring over K in the set of variables Tcontaining K as a unital subring.
Proof. Define ϕ : K ∪ T → KT •, x 7→{x1A if x ∈ Kx if x ∈ T . Then ϕ is injective and ϕ|K is
a monomorphism. Applying 2.1 with M = KT •, B = K ∪ T , to both operations on K,M resp., we obtain the claim.
We write K[T ] for a polynomial ring over K in the set variables T containing K as aunital subring.
The elements of a polynomial ring P over T have the form
∑
j1,...,jk∈N0
′cj1,...,jkt
j11 · · · tjkk
where t1, . . . , tk are mutually distinct elements of T and Nk0 → K, (j1, . . . , jk) 7→ cj1,...,jk ,
is a mapping of finite support.
2.9.3 (The universal property). Let P be a polynomial ring over K in the set of variablesT , B be an arbitrary associative commutative K-algebra. Then every mapping of T intoB extends uniquely to a unital K-algebra homomorphisms of P into B.
Proof. Let F0 ∈ BT . Any extension F of F0 to a unital K-algebra homomorphism ofP into B must satisfy the following condition, for any choice of elements ti ∈ T andcj1,...,jk ∈ K:
(∗)( ∑
j1,...,jk∈N0
′cj1,...,jkt
j11 · · · tjkk
)F =
∑
j1,...,jk∈N0
′cj1,...,jk(t1F0)
j1 · · · (tkF0)jk
Hence there is at most one such extension. On the other hand, denoting by M themultiplicative closure of T in P , F0 extends to a unital multiplicative homomorphismF1 of M into B, by 2.3. We know that P ∼= KM so that 2.8.1 shows how to extend F1
to a K-algebra homomorphism from P into B. 38
As a consequence, we obtain by the standard argument as given in 2.4(2),
2.9.4. Let P, P ′ be polynomial rings over K in the set of variables T, T ′ resp. Let F0 be abijection of T onto T ′. Then the extension of F0 from 2.9.3 is a K-algebra isomorphismof P onto P ′.In particular, a polynomial ring over K in the set of variables T is unique up toisomorphism. �
38It turns out that the extension is indeed given by the rule (∗) which, with appropriate comments,could have been used as its definition. But then we would have had to prove the desired propertiesin detail which we obtained for free by applying 2.3 and 2.8.1.
38
Making further use of the notation introduced in the proof of 2.9.3, the unique extensionF of F0 as described in (∗) is called the replacement homomorphism39 with respect to F0:In every polynomial over T , each variable t ∈ T is replaced by tF0. If T is finite andconsists of the mutually distinct elements t1, . . . , tn, we write Fb1,...,bn for the replacementhomomorphism which maps ti to bi for all i ∈ n. Thus we obtain a function
P × Bn → B(f, (b1, . . . , bn)
)7→ fFb1,...,bn =: f(b1, . . . , bn).
Its left component functions ( . , (b1, . . . , bn)) : f 7→ f(b1, . . . , bn), are the replacementhomomorphisms of P into B. Its right component functions (f, . . . ) : (b1, . . . , bn) 7→f(b1, . . . , bn), are the polynomial functions of Bn into B which, of course, are no homo-morphisms in general. Thus each polynomial f ∈ P induces a polynomial function.Easy examples show that there may exist distinct polynomials which induce the samepolynomial function.40 Therefore one should bear in mind:
Polynomials and polynomial functions are distinct mathematical objects whichmay not be (in whatever sense) “identified”.41
Let P be a polynomial ring over K in n variables, b1, . . . , bn (not necessarily distinct)elements of a commutative associative unitary K-algebra B. We put
K[b1, . . . , bn] := PFb1,...,bn
which, by applying various parts of 1.3, is easily seen to be a unital K-subalgebra of B.Likewise it is clear that kerFb1,...,bn = {f |f ∈ P, f(b1, . . . , bn) = 0B} is a subalgebra42 ofP . The n-tuple (b1, . . . , bn) ∈ Bn is called algebraically independent if kerFb1,...,bn = {0P},i. e., if Fb1,...,bn is an isomorphism of P onto K[b1, . . . , bn]. Otherwise (b1, . . . , bn) iscalled algebraically dependent. For every f ∈ kerFb1,...,bn we have f(b1, . . . , bn) = 0B.
39This is the general form of the special case of one variable considered on p. 9.40For example, if K is the field with just two elements (p. 26), then all polynomials tj where j ∈ N
induce the same polynomial function on K: the identitity mapping, but they are clearly mutuallydistinct as polynomials (in one variable).
41The reader will have observed that we avoid throughout this by numerous authors frequently usedterm. The reason is simple: Either two objects are identical, making any “identifying process”superfluous, or they are not identical and “identifying” them means to make a mistake. The intentionin using the expression is usually to refer to some 1-1 correspondence between objects which isconsidered natural in the context, and to indicate that the smart author is not willing to meddlewith notational distinction. As a typical example – here mentioned only for convenience as we citedthis otherwise recommendable book anyway – see [St, p. 19]. Note that the author emphasizes thedistinction between polynomials and polynomial functions on p. 20. But be aware that, confusingly, asymmetric polynomial is later called a “symmetric function” (17.1), despite the author’s own warningin 1.30. Thus, in the terminology of [St], a symmetric function is not a function. Terminologicalinconsistencies of this kind often occur as a result of an attempted compromise between conceptualclearness and certain traditions of mathematical language.
42With respect to multiplication, kerFb1,...,bn is not only closed but has the much stronger property ofbeing an ideal of P (cf. 2.5). Clearly, in a commutative algebra there is no distinction between leftand right ideals.
39
Any n-tuple (b1, . . . , bn) ∈ Bn with this property is called a zero of f , generalizing thedefinition given on p. 9. The fundamental topic of Algebraic Geometry is to study the setof common zeros in Bn for a subset of P . For example, if n = 2, B = R, f = t21+ t22− 1,the set of zeros is {(b1, b2)|bi ∈ R, b21 + b22 = 1} – the unit circle (in the real plane).
If T consists of exactly n elements t1, . . . , tn we write K[t1, . . . , tn] instead of K[T ]. Wewill adopt the convention that, for a given commutative associative unitary ring K, thenotation K[t1, . . . , tn] – coming out of the blue – always stands for the polynomial ringover K in n variables (t1, . . . , tn) if the elements t1, . . . , tn have not been introducedwith a different meaning in the context. Then every π ∈ Sn gives rise to the specialreplacement homomorphism
π := Ft1π ,...,tnπ: K[t1, . . . , tn]→ K[t1, . . . , tn],
the canonical extension of the mapping F0 : T → K[t1, . . . , tn], ti 7→ tiπ for all i ∈ n.Clearly, π is an automorphism of K[t1, . . . , tn]. For example, if n = 3, π =
(1 2 33 2 1
),
then
(a) (1K + t1 + t2t3 + 2t21t2t3)π = 1K + t3 + t2t1 + 2t23t2t1 = 1K + t3 + t1t2 + 2t1t2t23,
(b) (t1t3 − t22)π = t3t1 − t22 = t1t3 − t22,
(c) (t41 + t42 + t43)π = t43 + t42 + t41 = t41 + t42 + t43.
Part (a) illustrates the typical behaviour of π: It modifies a polynomial, maintaining,however, certain characteristics of it. The action of π in the case of the polynomial in (b)turns out to be trivial. The intermediate feeling that “something is happening” vanisheswhen we compare the given polynomial with its image under π: t1t3 − t22 is a fixed pointfor π. While this observation clearly is due to the particular choice of π, the polynomialin (c) is not only a fixed point for π but in regard to any chosen permutation of {1, 2, 3}.A polynomial f ∈ K[t1, . . . , tn] is called symmetric if fπ = f for all π ∈ Sn. We writeKsym[t1, . . . , tn] for the set of all symmetric polynomials in K[t1, . . . , tn]. It is easily seento be a subalgebra of K[t1, . . . , tn].
43
2.9.5. Let n ∈ N, f ∈ Ksym[t1, . . . , tn], B a commutative associative unitary K-algebra.Then the polynomial function Bn → B, (b1, . . . , bn) 7→ f(b1, . . . , bn), is symmetric. �
The elementary symmetric functions, introduced in 1.12 in connection with Viete’s the-orem, arise (in the sense of 2.9.5) from the so-called elementary symmetric polynomials inK[t1, . . . , tn], defined by
sk :=∑
ij∈n
i1<i2<···<ik
ti1ti2 · · · tik for all k ∈ n,
43Later we will see a simple general reason for this fact.
40
s0 := 1K , sk = 0 for k > n.44 Their importance lies in the following result which willnot be proved here; for a proof see, e. g., [Bo, 4.4]:
Main Theorem on symmetric polynomials. Let K be a commutative unitary ring,n ∈ N. Then {s1, . . . , sn} is a K-algebra basis of Ksym[t1, . . . , tn].
This means that K[s1, . . . , sn] is a polynomial ring over K in the set of variables{s1, . . . , sn}. Thus the replacement homomorphism of K[t1, . . . , tn] into Ksym[t1, . . . , tn]given by ti 7→ si for all i ∈ n (in other words: f 7→ f(s1, . . . , sn)) is a K-algebraisomorphism:45
Ksym[t1, . . . , tn] = K[s1, . . . , sn] ∼= K[t1, . . . , tn].
A straightforward generalization of the above example (c) leads to the notion of a power-sum symmetric polynomial, defined for any k ∈ N0 by
tk1 + tk2 + · · ·+ tkn =: pk.
Applying Viete’s theorem 1.11 with bi = ti for all i ∈ n, we obtain (t− t1) · · · (t− tn) =∑nk=0(−1)ksktn−k =
∑nk=0(−1)n−ksn−ktk, hence 0K =
∑nk=0(−1)n−ksn−ktkj for all j ∈ n.
It follows that46n∑
k=0
(−1)n−ksn−kpk = 0K .
We will finish this chapter by a fourth application of the Extension Principle 2.1: Clearly,any unital subring of a field is an integral domain. Our aim is to show that all integraldomains arise as subrings of fields. Given an integral domain R, the main step consistsin the construction of a field which in a natural way emanates from the structure of R.We start with the observation that R×R is a commutative monoid with respect to bothof the operations ·, ⋆ defined as follows:
∀a, b, c, d ∈ R (a, c) · (b, d) := (ab, cd), (a, c) ⋆ (b, d) := (ad+ bc, cd).
The neutral elements are (1R, 1R) (with respect to ·), (0R, 1R) (with respect to ⋆). As Ris an integral domain, the subset
B := R× Rof R×R is closed with respect to both operations. As it contains both neutral elements,it is a twofold unital submonoid, with respect to · and with respect to ⋆.
44These supplementary definitions are consistent as s0 is the “sum of all empty products” over{t1, . . . , tn}, i. e., s0 =
∑1K = 1K while sk, for k > n, is the empty sum in K[t1, . . . , tn], namely
the sum of all products of k mutually distinct factors over {t1, . . . , tn}: there are none.45This is another example for the phenomenon that a structure is isomorphic to a proper substructure,
as in the familiar case of the non-trivial subgroups kZ (k ∈ N) of the group (Z,+).46Equivalently, nsn =
∑k∈n(−1)k−1sn−kpk. This is the n-th of the famous Newton-Girard formulae,
proved in the 17th century:
∀m ∈ N msm =∑
k∈m
(−1)k−1sm−kpk (in K[t1, . . . , tn])
41
Note that the mapping ι : R → B, a 7→ (a, 1R), is a monomorphism of the integraldomain (R,+, ·) into the double magma (B, ⋆, ·).The main idea is to define a partitionM on the set B and two operations onM whichmake it into a field that allows an embedding of R. Our procedure consists of six steps:
(I) The setMLet ∼ be the relation on B defined by
(a, b) ∼ (a′, b′) if and only if ab′ = a′b
for arbitrary a, a′ ∈ R, b, b′ ∈ R. Clearly, ∼ is reflexive and symmetric. Leta, a′, a′′ ∈ R, b, b′, b′′ ∈ R such that (a, b) ∼ (a′, b′), (a′, b′) ∼ (a′′, b′′). Thenab′ = a′b, a′b′′ = a′′b′, hence ab′b′′ = a′bb′′ = a′′b′b. It follows that b′(ab′′−a′′b) = 0Rwhich implies that ab′′ − a′′b = 0R as b′ 6= 0R and R is an integral domain. Thus(a, a′′) ∼ (b, b′′). This proves that ∼ is transitive, hence an equivalence relationon B.
For all a ∈ R, b ∈ R we write [a, b] for the equivalence class of (a, b) with respectto ∼ andM for the set of all equivalence classes of ∼. It is well known thatMis a partition of B. The mapping
κ : B →M, (a, b) 7→ [a, b],
is a surjection.
(II) The multiplication ⊡ onMWe know from 1.2 that the multiplication · in B induces a natural operation onP(B). For the product of two subsets U, V of B we write here simply UV (={u · v|u ∈ U, v ∈ V }). The problem is that, for U, V ∈ M, in general the setUV will not be an equivalence class with respect to ∼ so that the ordinary subsetproduct does not define an operation onM. The key observation, however, is thatthere exists a unique equivalence classW which contains UV : AsM is a partitionand UV 6= ∅, there exists trivially at most one element ofM such that UV ⊆ W .We have to prove existence: Let U, V ∈ M, a, c, a′, c′ ∈ R, b, d, b′, d′ ∈ R suchthat (a, b), (a′, b′) ∈ U , (c, d), (c′, d′) ∈ V . We claim that47
(a, b) · (c, d) ∼ (a′, b′) · (c′, d′).
As (a, b), (a′, b′) are elements of the same equivalence class we know that ab′ = a′b.Similarly, cd′ = c′d. It follows that ab′cd′ = a′bc′d, i. e., (ac, bd) ∼ (a′c′, b′d′).Hence any two elements of UV are equivalent, i. e., UV ⊆W for some W ∈M.
The above key observation now allows us to define the desired operation onM:
47A relation ∽ on a set X is called compatible with an operation · on X if x ∽ y, x′ ∽ y′ impliesx · x′ ∽ y · y′, for arbitrary x, x′, y, y′ ∈ X . Making use of this notion, the claim means that theequivalence relation ∼ on B is compatible with ·.
42
For all U, V ∈M define U ⊡ V to be the unique element ofM containing UV .
If (a, b) ∈ U , (c, d) ∈ V , then (ac, bd) ∈ UV ⊆ U ⊡ V , hence U ⊡ V is theequivalence class of (ac, bd). Hence we have the rule
∀a, c ∈ R ∀b, d ∈ R [a, b]⊡ [c, d] = [ac, bd].
In other words, κ is an epimorphism of (B, ·) onto (M,⊡). By 1.3, ⊡ is associative,commutative, and [1R, 1R] is neutral with respect to ⊡. As usual, we put 1M :=[1R, 1R]. We call ⊡ the multiplication onM.
(III) The addition ⊞ onMImitating the idea in (II), we show that for every U, V ∈M there exists a uniqueequivalence class which contains U ⋆ V . Uniqueness is trivial as in (II). As forexistence, we claim, under the same hypotheses as in (II), that
(a, b) ⋆ (c, d) ∼ (a′, b′) ⋆ (c′, d′).
From ab′ = a′b, cd′ = c′d we conclude that (ad + bc)b′d′ = (a′d′ + b′c′)bd, hence(a, b) ⋆ (c, d) = (ad + bc, bd) ∼ (a′d′ + b′c′, b′d′) = (a′, b′) ⋆ (c′, d′). Hence for everyU, V ∈M there exists a unique S ∈M such that U ⋆ V ⊆ S.
Again, this puts us in a position to define the desired operation on M: For allU, V ∈M define U⊞V to be the unique element ofM containing U ⋆V . Similarlyto the line of reasoning in II, we obtain
∀a, c ∈ R ∀b, d ∈ R [a, b]⊞ [c, d] = [ad+ bc, bd].
Thus κ is an epimorphism of (B,+) onto (M,⊞). By 1.3, ⊞ is associative, com-mutative, and [0R, 1R] is neutral with respect to ⊞. As usual, we put 0M :=[0R, 1R]. We call ⊞ the addition onM.
(IV) The group properties of (M,⊞) and (M,⊡)
We first describe the neutral elements of (M,⊞) and of (M,⊡):
0M = {(0R, b)|b ∈ R}, 1M = {(b, b)|b ∈ R},
as (a, b) ∈ 0M ⇔ (a, b) ∼ (0R, 1R) ⇔ a = 0R, (a, b) ∈ 1M ⇔ (a, b) ∼ (1R, 1R) ⇔a = b, for all a ∈ R, b ∈ R.It follows that [a, b] ⊞ [−a, b] = [0R, b
2] = 0M for all a ∈ R, b ∈ R. Hence thecommutative monoid (M,⊞) is a group. If [a, b] 6= 0M we know that a 6= 0R,hence (b, a) ∈ B and [a, b] ⊡ [b, a] = [ab, ab] = 1M. Hence the commutativemonoid (M,⊡) is a group. From our proofs we obtain the following formula forthe inverses: −[a, b] = [−a, b], [a, b]−1 = [b, a] if a 6= 0R.
43
(V) The distributive laws in (M,⊞,⊡)
Let a, c, e ∈ R, b, d, f ∈ R. Observing that (x, y) ∼ (xb, yb) for all x ∈ R, y ∈ R,we have
[a, b]⊡ ([c, d]⊞ [e, f ]) = [a, b]⊡ [cf + de, df ] = [a(cf + de), bdf ]
= [a(cf + de)b, bdfb] = [acbf + bdae, bdbf ]
= [ac, bd]⊞ [ae, bf ] = [a, b]⊡ [cd]⊞ [a, b]⊡ [e, f ].
Therefore (M,⊞,⊡) is a field.
(VI) The embedding of (R,+, ·) into (M,⊞,⊡)
Put ϕ := ικ, i. e., aϕ := [a, 1R] for all a ∈ R. As a composition of two unitalring homomorphisms, ϕ is a unital ring homomorphism. Let a ∈ R such thataϕ = 0M. Then a = 0R by the first equality in (IV). Hence ϕ is injective.
By 2.1, there exists a field Q containing R as a subring and an extension ϕ of ϕ to anisomorphism of Q ontoM. We observe that our field Q has the following property:
2.9.6. ∀x ∈ Q ∃a ∈ R ∃b ∈ R x = ab−1
Proof. As Q ∼=M via ϕ it suffices to prove that for all U ∈M there exist a ∈ R, b ∈ Rsuch that U = (aϕ)⊡ (bϕ)−1. Let U ∈M. Then there exist a ∈ R, b ∈ R such that
U = [a, b] = [a, 1R]⊡ [1R, b] = [a, 1R]⊡ [b, 1R]−1 = (aϕ)⊡ (bϕ)−1,
making use of the last equation in (IV).
2.10 Definition. Let R be an integral domain. A field Q is called a quotient field of Rif R is a subring of Q with the property 2.9.6.
2.10.1. Suppose R is a subring of a field L. Set Q := {ab−1|a ∈ R, b ∈ R}. Then Q isa subfield of L.
Proof. Clearly, R ⊆ Q as a = a1−1R ∈ Q for all a ∈ R. If a, c ∈ R, b, d ∈ R, we have
ab−1 − cd−1 = (ad− bc)(bd)−1 ∈ Q and, if b 6= 0R, ab−1(cd−1)−1 = (ad)(bc)−1 ∈ Q.
Obviously, Q in 2.10.1 is a quotient field of R, the uniquely determined quotient field ofR in L. Thus a quotient field of R is immediately found if a field is given containing Ras a subring. The non-trivial point and aim of our construction was the existence of aquotient field of R without having an enveloping field at our disposal from the beginning.
2.10.2. Let ψ be an isomorphism of R onto an integral domain R. Suppose that (Q,+, ·)is a quotient field of R, (Q, +, ·) is a quotient field of R. Then there is a unique extensionψ of ψ to an isomorphism of (Q,+, ·) onto (Q, +, ·).
44
Proof. Let a, a′ ∈ R, b, b′ ∈ R. Then
a · b−1 = a′ · b′−1 ⇔ ab′ = a′b⇔ (aψ)(b′ψ) = (a′ψ)(bψ)⇔ (aψ) · (bψ)−1 = (a′ψ) · (b′ψ)−1.
Therefore the mapping ψ : Q → Q, a · b−1 7→ (aψ) · (bψ)−1, is well-defined48, bijectiveand an extension of ψ. The homomorphism properties of ψ are immediate.
2.11 Theorem. Let R be an integral domain.
(1) There exists a quotient field of R.
(2) Any two quotient fields of R are isomorphic.
(3) Let Q be a quotient field of R. Then every automorphism of R extends uniquely toan automorphism of Q.
Proof. We obtain (1) by the above 6-step construction, (2) by choosing R = R, ψ = idR,(3) by choosing R = R, Q = Q, ψ ∈ AutR in 2.10.2.
2.11(2) is the justification of the frequently seen use of the definite article in the formu-lation “the quotient field of R” which does not express an absolute uniqueness, but auniqueness up to isomorphism. As a consequence of 2.11(3), we note
2.11.1. Let Q be a quotient field of an integral domain R. Then AutR isomorphic to asubgroup of AutQ.
As a proof, it suffices to observe that the extension of an automorphism of R to anautomorphism of Q given by 2.11(3) defines a monomorphism of AutR into AutQ. �
If Q is a quotient field of an integral domain R and a ∈ R, b ∈ R, the product ab−1 isalternatively also denoted by a
b, called fraction notation. This notation is motivated by
the classical case of R = Z in which a quotient field is called field of rational numbersand commonly denoted by Q. Traditionally, the elements of Q are written as fractions.The above steps (I) – (VI) may be viewed as a general version of a construction of Qfrom Z. The element a
bof Q is the element corresponding to the equivalence class [a, b]
defined at the end of step (I). If (a′, b′) ∈ R× R is any representative of the same class,the fraction a′
b′therefore denotes the same element of Q. Reducing (expanding resp.) the
fraction abmeans passing to a′
b′where |a′| < |a| (|a′| > |a| resp.), (a, b) ∼ (a′, b′).
From 1.1.2 it follows inductively that, for any integral domain R, the polynomial ringin n variables R[t1, . . . , tn] is an integral domain. If Q is a quotient field of R, we writeQ(t1, . . . , tn) for the quotient field of R[t1, . . . , tn]. Its elements are commonly written asfractions f
gwhere f, g ∈ R[t1, . . . , tn], g 6= 0R. In particular, if K is a field, K(t1, . . . , tn)
stands for the quotient field of K[t1, . . . , tn].49 We know that every π ∈ Sn determines
48The image of an element x ∈ Q is given only in dependence of the elements a ∈ R, b ∈ R chosen torepresent x in the form a · b−1. Although this representation is not unique, we have shown that theelement in Q assigned to x is unique: It is the same for any choice of a, b with x = a · b−1.
49More generally, if K is a subfield of a field L and b1, . . . , bn ∈ L, then K(b1, . . . , bn) is the quotientfield of K[b1, . . . , bn] in L (cf. 2.10.1).
45
an automorphism π of K[t1, . . . , tn] (see p. 40). By 2.11(3), π extends uniquely to anautomorphism of K(t1, . . . , tn). Assigning this automorphism to π ∈ Sn, we obtain amonomorphism of Sn into AutK(t1, . . . , tn). Every automorphism obtained in this wayinduces the identity automorphism on the subring Ksym[t1, . . . , tn], hence also on itsquotient field Ksym(t1, . . . , tn) in K(t1, . . . , tn). From the definition of Ksym[t1, . . . , tn]we obtain that Ksym(t1, . . . , tn) is the set of all fixed points with respect to the set ofautomorphisms induced by the symmetric group Sn. We conclude:
2.11.2. Let n ∈ N, K a field. Then there exists a monomorphism of Sn into the subgroup{α|α ∈ AutK(t1, . . . , tn), ∀q ∈ Ksym(t1, . . . , tn) qα = q} of AutK(t1, . . . , tn). �
Later (see p. 73) we will see that this monomorphism is in fact an isomorphism, inother words, that every automorphism of K(t1, . . . , tn) which fixes each element ofKsym(t1, . . . , tn) must necessarily permute the set of variables {t1, . . . , tn}.
46
3. Cosets and homomorphisms
Let A, B sets, ϕ ∈ BA. For all b ∈ Aϕ we set
bϕ− := {a|a ∈ A, aϕ = b},
called the complete pre-image of b. LetM be the set of all pre-images bϕ− where b ∈ Aϕ.The condition of having the same image with respect to ϕ defines an equivalence relationon A, andM is the set of its equivalence classes. It follows that
3.0.1. M is a partition of A, and ϕ− is a bijection of Aϕ ontoM,
because ϕ− is a surjection of Aϕ ontoM by the definition ofM, and bϕ− = b′ϕ− (whereb, b′ ∈ Aϕ) implies, for an arbitrary a ∈ bϕ−, that b = aϕ = b′, hence b = b′. �
3.1 Definition. Let (G, ◦) be a group, H ≤ G, and set ρ : G → P(G), g 7→ H ◦ g (:={h ◦ g|h ∈ H}). Let g ∈ G. The set H ◦ g is called the right coset of g with respect toH . We have, for any g′ ∈ G,
g′ ∈ (H ◦ g)ρ− ⇔ g′ρ = H ◦ g ⇔ H ◦ g′ = H ◦ g ⇔ g′ ◦ g−1 ∈ H ⇔ g′ ∈ H ◦ g,
hence (H ◦ g)ρ− = H ◦ g. We write G/.H for the set of all right cosets of H in G. Themapping H → H ◦ g, h 7→ h ◦ g, is a bijection. Applying 3.0.1 with A = G, B = P(G),ϕ = ρ,M = G/.H , we obtain:
3.1.1. G/.H is a partition of G which consists of subsets of G all of which are equipotentto H.
If the number of right cosets of H in G is finite, this number is called the index ofH in G and denoted by |G : H|. The following important statement is an immediateconsequence of 3.1.1:
3.1.2 (Lagrange’s theorem). Let G be finite. Then |G| = |G : H||H|. �
|G| is called the order of G. By 3.1.2, we have for every subgroup H of a finite group G
|H|∣∣ |G|, |G : H| = |G||H| .
Instead of ρ, we could have considered λ : G→ P(G), g 7→ g ◦H . The set g ◦H is calledthe left coset of g with respect to H . The set of all left cosets of H in G is a partition
47
of G, denoted by G./H . We observe that the bijection ι : G → G, g 7→ g−1, induces abijection of G/.H onto G./H as
(Hg)ι = {(hg)−1|h ∈ H} = {g−1h−1|h ∈ H} = g−1H for all g ∈ G.
Hence |G/.H| = |G./H|. But in general, g ◦ H 6= H ◦ g. For example, let G = S3,H =
{id3,
(1 2 32 1 3
)}, g =
(1 2 33 2 1
). Then
H � g ={(1 2 3
3 2 1
),
(1 2 32 3 1
)}6=
{(1 2 33 2 1
),
(1 2 33 1 2
)}= g � H.
A subgroup H of (G, ◦) is called normal if g ◦H = H ◦ g for all g ∈ G which is denotedby H E G. For a normal subgroup, the set of right cosets coincides with the set of leftcosets so that there is no need to use distinct notation. The set of all cosets of H in G isdenoted by G/H if H EG. Clearly, the subgroups {1G} and G are normal. The groupG is called simple if G 6= {1G} and {1G}, G are the only normal subgroups of G. Thefollowing remark is trivial:
3.1.3. If (G, ◦) is abelian, then every subgroup of G is normal. �
In S3, the subgroup of order 3 is a first not completely trivial example of a normalsubgroup. Still there is a simple general reason for this fact because a subgroup it is ofindex 2 in S3:
3.1.4. Every subgroup of G of index 2 in G is normal in G.
Proof. Suppose H < G, |G : H| = 2. Certainly, g ◦ H = H = H ◦ g for all g ∈ H .Now let g 6∈ H . Then H ◦ g 6= H 6= g ◦H , hence H ◦ g = G r H = g ◦H by 3.1.1 as|G/.H| = 2 = |G./H|.
3.1.5. Suppose H E G, and let ◦ be the operation on P(G) defined in 1.2. Then ρ is ahomomorphism of (G, ◦) into (P(G), ◦), ker ρ = H, Gρ = G/H.
Proof. Let g, g′ ∈ G. We have, making use of the normality of H in the third step,
(g ◦ g′)ρ = H ◦ g ◦ g′ = H ◦H ◦{g}◦{g′} = H ◦{g}◦H ◦{g′} = (H ◦ g)◦(H ◦ g′) = gρ ◦ g′ρ,
g ∈ ker ρ⇔ gρ = 1Gρ⇔ H ◦ g = H ⇔ g ∈ H . The last assertion is trivial.
Let H EG. By 1.3(4) and 3.1.5, G/H is a subgroup of (P(G), ◦). The group (G/H, ◦)is called the factor group of G over H . Its elements are the cosets H ◦ g (g ∈ G). Themapping ρ from 3.1.5 is called the canonical epimorphism of G onto G/H . We do notexaggerate when we say that the notion of a factor group is a fundamental element ofmathematics. We are now in a position to prove one of the most influential algebraicinstruments:
48
3.2 Theorem (Homomorphism theorem for groups). Let (G, ◦) be a group, ϕ a homo-morphism of (G, ◦) into a magma (B, ·), H := kerϕ.
(1) H EG, ∀g ∈ G (gϕ)ϕ− = H ◦ g,
(2) ϕ− is an isomorphism of (Gϕ, ·) onto (G/H, ◦).
Proof. By 1.3(4), Gϕ is a subgroup of (B, ·), kerϕ a subgroup of (G, ◦). Let g ∈ G,b := gϕ. We show
(∗) H ◦ g ⊆ bϕ− ⊆ g ◦H :
For all h ∈ H we have (h◦ g)ϕ = (hϕ) · (gϕ) = (1Gϕ) · (gϕ) = (1G ◦ g)ϕ = gϕ = b. HenceH ◦ g ⊆ bϕ−. Let x ∈ bϕ−. Then xϕ = b = gϕ, hence (g−1 ◦ x)ϕ = (gϕ)−1 · (xϕ) =b−1 · b = 1Gϕ by 1.3(3). It follows that g−1 ◦ x ∈ H , x ∈ g ◦H . Thus bϕ− ⊆ g ◦H . Thisproves (∗).For all g ∈ G it follows that H ◦ g ⊆ g ◦H , thus also H ◦ g−1 ⊆ g−1 ◦H . We concludeg ◦H = g ◦H ◦ g−1 ◦ g ⊆ g ◦ g−1 ◦H ◦ g = H ◦ g. Now H ◦ g = bϕ− = g ◦H , i. e., (1).
This implies, by 3.0.1, that ϕ− is a bijection of Gϕ onto G/H . Let b, b′ ∈ Gϕ andg, g′ ∈ G such that gϕ = b, g′ϕ = b′. Then (g ◦ g′)ϕ = (gϕ) · (g′ϕ) = b · b′, hence(b · b′)ϕ− = H ◦ g ◦ g′ = (H ◦ g)◦(H ◦ g′) = bϕ− ◦ b′ϕ−, making use of (1) and 3.1.5.
Combining 3.1.4 and 3.2(1), we obtain the following characterization of normal sub-groups:
3.2.1. The normal subgroups of a group G are exactly the kernels of the homomorphismsof G. �
The aim of 3.2 is the statement that every homomorphic image of a group is isomorphicto the factor group over the kernel (which automatically is a normal subgroup). Thus,to apply the homomorphism theorem, one determines the image Y of a group G undera homomorphism and its kernel H . Then 3.2(2) implies that
G/H ∼= Y.
Whenever an isomorphism between a factor group G/H of G and a group Y is to beproved, we have the method of 3.2 at our disposal: It suffices to find a homomorphismof G onto Y the kernel of which is H . Done that, there is no further step necessary : Thedesired isomorphism follows directly from 3.2(2). This strategy of proving isomorphismsis the main point of the homomorphism theorem and makes it into an indispensableinstrument with countless applications.
For example, in the proof of 1.3.2 and in 1.3.3 we have considered various homomorph-isms. Accordingly, we obtain the following group isomorphisms:
(a) (C/K(0, 1), ·) ∼= (R>0, ·). (Consider C→ R, z 7→ |z|.)
(b) (Q/Z,+) ∼= (R(1), ·). (Consider Q→ C, x 7→ exp(x · 2πi).)
49
(c) (C/Rn(1), ·) ∼= (C, ·) for all n ∈ N. (Consider C→ C, z 7→ zn.)In particular, (C, ·) is an example of a group which is isomorphic to a proper factorgroup.
(d) (Z/nZ,+) ∼= (Rn(1), ·) for all n ∈ N. (Consider Z→ C, k 7→ exp(k·2πin
).)
We add a few more examples:
Examples. (1) Let K be a field, n ∈ N. The determinant defines a multiplicativeepimorphism of the group GL(n,K) of invertible n× n matrices (called the generallinear group) over K onto the multiplicative group K of K. Its kernel is the setSL(n,K) of all n × n matrices over K of determinant 1K (called the special lineargroup). We obtain:
SL(n,K)EGL(n,K) and GL(n,K)/SL(n,K) ∼= K.
(2) Let n ∈ N>1. Mapping every permutation π ∈ Sn to its sign is an epimorphism ofthe symmetric group Sn onto the multiplicative group {1,−1}. Its kernel is the setAn of even permutations of n (called the alternating group on n). We obtain:
An E Sn and Sn/An ∼= {1,−1}.
(3) Let X be a (multiplicatively written) group, x ∈ X , ϕx : Z→ X, k 7→ xk. Then ϕxis a homomorphism of the group (Z,+) into the group X . We have
kerϕx = {k|k ∈ Z, xk = 1X} = nZ
for a unique n ∈ N0, by 1.2.1. The image Zϕx is the smallest subgroup of Xcontaining x, also called the subgroup generated by x and denoted by 〈x〉. By 3.2,
Z/nZ ∼= 〈x〉.
If n 6= 0, n is called the order of x and denoted by o(x). We then have
{1X , x, x2, . . . , xo(x)−1} = Zϕx ≤ X.
By 3.1.2, we obtain
3.2.2. Let X be a finite group. Then o(x) = min{k|k ∈ N, xk = 1X}∣∣ |X| for all
x ∈ X. �
If n = 0, x is called of infinite order.
For all T ⊆ X we put 〈T 〉 := ⋂H≤XT⊆H
H . As an intersection of a non-empty set
of subgroups of X , 〈T 〉 is a subgroup of X and obviously the smallest subgroupcontaining T , called the subgroup generated by T . If T = {x1, . . . , xk}, it is alsodenoted by 〈x1, . . . , xk〉.
50
3.3 Definition. A group X is called cyclic if there exists an element x ∈ X such that〈x〉 = X . Every x with this property ia called a generating element or generator of X .We have seen in the above Example (3) that every cyclic group is a homomorphic imageof (Z,+). More precisely, we have obtained:
3.3.1. Let (X, ·) be a cyclic group. Then
(X, ·) ∼={(Z,+) if X is infinite,
(Z/nZ,+) if |X| = n ∈ N.
In particular, (X, ·) is commutative. �
For example, the group (Rn(1), ·) (see 1.1.8, 1.3.2) is cyclic and Pn its set of generators, by1.4.1. The group (Z,+) is cyclic, generated by 1 and by −1, and these are the only gen-erating elements of (Z,+) (cf. 1.2.1 ). (Q,+), ({(1, 1), (1,−1), (−1, 1), (−1,−1)}, ·) areexamples of non-cyclic commutative groups. The isomorphism (Rn(1), ·) ∼= (Z/nZ,+)implies, by 3.3.1:
3.3.2. The number of generating elements of a finite cyclic group of order n is ϕ(n).50�
Now let G be an arbitrary finite group und C the set of its cyclic subgroups. As x ∈ 〈x〉for all x ∈ G, we have ⋃ C = G. We define a relation ≈ on G by putting, for all x, y ∈ G,
x ≈ y if and only if 〈x〉 = 〈y〉.
Then ≈ is an equivalence relation on G. The equivalence class of an element x ∈ Gconsists of the set of generating elements of 〈x〉, hence contains exactly ϕ(|〈x〉|) elements.As G is the disjoint union of all equivalenve classes with respect to ≈, we obtain:
3.3.3. Let G be a finite group, C the set of its cyclic subgroups. Then |G| =∑
X∈Cϕ(|X|).�
3.4 Theorem (Main theorem on finite cyclic groups.). Let n ∈ N, G a group of ordern, Tn := {d|d ∈ N, d|n}. The following are equivalent:
(i) G is cyclic
(ii) For every d ∈ Tn there exists exactly one subgroup of G of order d.
(iii) For every d ∈ Tn there exists at most one subgroup of G of order d.
(iv) For every d ∈ Tn there exists at most one cyclic subgroup of G of order d.
Proof. (i)⇒(ii) If G is cyclic, we have G ∼= Rn(1) by 3.3.1. Therefore it suffices to provethat (Rn(1), ·) has the property (ii). If d ∈ Tn, we have Rd(1) ≤ Rn(1) by the first partof 1.3.2, |Rd(1)| = d. Furthermore, if U ≤ Rn(1), |U | = d, then ud = 1 for all u ∈ U , by3.2.2. Hence U ≤ Rd(1). It follows that U = Rd(1).
50where ϕ is the Eulerian function, see 1.4.
51
The implications (ii⇒(iii)⇒(iv) are trivial.
(iv)⇒(i) Let C as in 3.3.3 and D := {d|d ∈ N, ∃X ∈ C |X| = d}. Clearly D ⊆ Tn, by3.1.2. It follows that
n =∑
X∈Cϕ(|X|) =
∑
d∈Dϕ(d) ≤
∑
d∈Tn
ϕ(d) = n,
by means of 3.3.3 (first equality), (iv) (second equality), (1.4.4) (last equality). Weconclude that D = Tn which, in particular, implies that n ∈ D. Hence G is cyclic.
3.5 Corollary. Let K be a field, G a finite subgroup of (K, ·). Then G is cyclic. Inparticular, if K is finite, then (K, ·) is cyclic.
Proof. Let n := |G|. For every d ∈ N, in particular for every d ∈ Tn, the polynomialtd− 1 has at most d zeros in K, by 1.6(3). By 3.2.2, the elements of a subgroup of orderd of (K, ·) are zeros of that polynomial. Hence there is at most one such subgroup. By3.4, this implies the claim.
3.6 Corollary. Let G be a cyclic group, H ≤ G. Then H and G/H are cyclic.
Proof. Let x ∈ G such that G = 〈x〉. Then 〈Hx〉 = G/H , hence G/H is cyclic. If G isinfinite we may w. l. o. g. assume that G = Z. By 1.2.1, H = nZ for some n ∈ N0, henceH = 〈n〉 so that H is cyclic. If G is finite, then T|H| ⊆ T|G|, by 3.1.2. Hence 3.4(iii) isinherited from G to H . By 3.4, H is cyclic.
Lagrange’s theorem 3.1.2 is a main reason for a strong link between finite group theoryand (multiplicative) arithmetic. Let G be a finite group, n := |G|, U(G) the set of allsubgroups of G. The image of the mapping
µ : U(G)→ N, H 7→ |H|,
is a subset of Tn, and U ≤ V ∈ U(G) implies that |U |∣∣ |V |, both by 3.1.2. The subgroup
relation (≤) defines a partial order on the set U(G), the relation of being a divisor (|)on the set Tn, and µ constitutes a link between both partially ordered sets (also called“posets”). We now consider this situation in general:
3.7 Definition. Let (M,�), (M ′,�′) be partially ordered sets51. A homomorphismof (M,�) into (M ′,�′) is a mapping χ of M into M ′ with the property that for allm,n ∈M
m � n⇒ mχ �′ nχ.
If χ is bijective and m � n ⇔ mχ �′ nχ for all m,n ∈ M , χ is called an isomorphism(of the partially orderd set (M,�) onto the partially ordered set (M ′,�′)).52 If χ is
51A partial order is a reflexive, transitive and antisymmetric relation. If � is a partial order on M ,then so is the relation �, defined by: m � n if and only if n � m (m,n ∈ M). We write m ≺ n ifm � n and m 6= n.
52Note that a bijective homomorphism need not be an isomorphism of partially ordered sets: Theidentity on T6 is a bijactive homomorphism of (T6, |) into (T6,≤), and 2 ∤ 3, 2 ≤ 3. Hence id6 is notan isomorphism.
52
bijective and m � n⇔ mχ �′ nχ for all m,n ∈M , χ is called an anti-isomorphism. Letm,n ∈M such that m � n. Then the set
[m,n]� := {x|x ∈M, m � x � n}
is called the interval between m and n. If the reference to � is clear from the context,we simply write [m,n]. Clearly, ([m,n],�) is then again a partially ordered set.
We shall mainly be interested in lattices, defined as posets (M,�) such that for allm,m′ ∈M there exists
(i) an element i ∈M such that i � m,m′ and i � k for all k � m,m′,
(ii) an element s ∈M such that m,m′ � s and s � l for all l � m,m′.
As a consequence of the antisymmetry, the element i in (i) (s in (ii) resp.) is uniquelydetermined. It is called the infimum (supremum resp.) of m,n, and denoted by inf(m,n)(sup(m,n) resp.).
For example, for every group G, (U(G),≤) is a lattice, called the subgroup lattice of G.For all U, V ≤ G we have inf(U, V ) = U ∩V , sup(U, V ) = 〈U ∪V 〉. Similarly, the normalsubgroups of G also form a lattice with respect to ≤. While sup(U, V ) may be difficultto calculate for arbitrary subgroups, it turns out to be just the product set (see 1.2) ifU or V is normal. More generally, we note
3.7.1. Let G be a group and U, V ≤ G such that UV = V U . Then UV ∈ U(G), hencesup(U, V ) = UV .
Proof. If UV = V U it follows that (UV )(UV ) = U(V U)V = U(UV )V = (UU)(V V ) =UV , i. e., UV is a submonoid. Let u ∈ U , v ∈ V . Then (uv)−1 = v−1u−1 ∈ V U = UV .Hence UV ∈ U(G), UV ⊆ 〈U ∪ V 〉 = sup(U, V ) ≤ UV . The claim follows.
Viceversa it is easily seen that, for subgroups U, V of G, the product set UV is a sub-group of G only if UV = V U . A glance at the definition of a normal subgroup (see 3.1)suffices to see that normality of U (or V ) implies that UV = V U . If both U and V arenormal in G, so is UV : For all g ∈ G we then have gUV = UgV = UV g.
If S, T, U are subgroups of G and U ≤ S, the hypotheses of 2.2.1 are satisfied so that“Dedekind’s law” holds. But the equations in 2.2.1 deal with equality of subsets, notof subgroups of G as long as it is not known if the involved subgroup products are sub-groups. Therefore, “Dedekind’s law” is a statement about the subgroup lattice U(G)only under appropriate additional hypotheses.
For every n ∈ N, (Tn, |) is a lattice and inf(m,m′) = gcd(m,m′), sup(m,m′) = lcm(m,m′)for all m,m′ ∈ Tn. If G is finite and n = |G|, the mapping µ introduced before 3.7 isa lattice homomorphism of (U(G),≤) into (Tn, |). The equivalence of the conditions3.4(i),(ii),(iii) may be reformulated as follows:
G is cyclic ⇔ µ is a lattice isomorphism of (U(G),≤) onto (Tn, |)⇔ µ is injective.
53
The mapping ı : Tn → Tn, d 7→ nd, is an antiautomorphism of the divisor lattice (Tn, |),
or, denoting by ג the relation “is divisible by”, an isomorphism of (Tn, |) onto (Tn, .(גA finite poset (M,�) determines an oriented graph with vertex setM and edges (m,m′)whenever m ≺ m′ and [m,m′] = {m,m′}. This graph is imaged by a system of dots andline segments as follows: Each dot represents an element of M , and dots representingm,m′ are joined by an ascending line segment from m to m′ if (m,m′) is an edge. Forexample, the subgroup lattice of the group (R12(1), ·), its image (T12, |) under µ, and theimage (T12, (ג of (T12, |) under ı are depicted as follows:
(U(R12(1)),≤) (T12, |) (T12, (ג
{1}
R2(1)
R4(1)
R3(1)
R6(1)
R12(1)
1
2
4
3
6
12
✲µ
12
6
3
4
2
1
✲ı
Passing from a lattice (M,�) to (M,�) means, in this representation, to “turn the graphupside down”. Divisor lattices have the property that the resulting image has the sameouter appearance as before; so the above behaviour of T12 under ı is typical. But forlattices in general, such an effect must not be expected.
For an arbitrary finite group G, the image of U(G) under µ consists of the divisors of|G| which occur as subgroup orders. If G is not cyclic, we know from 3.4 that µ is notinjective. The example of the group A4 shows that, moreover, µ needs not be surjectiveeither: The reader is strongly recommended to check that in A4 there are exactly threesubgroups of order 2, four subgroups of order 3, and unique subgroups of orders 1, 4, 12(which are exactly the normal subgroups of A4), but there is no subgroup of order 6.
The cornerstone of finite group theory, to be proved in part II of this course, is the resultthat, for every finite group G, U(G)µ contains all divisors of |G| which are prime powers :
Theorem (1st Theorem by Sylow (1872)) Let G be a finite group, d a divisor of |G|which is a power of a prime. Then there exists a subgroup of G of order d.
We will not digress further but take the opportunity and define the fundamental notionwhich arises from this result: Let p be a prime, G a finite group and m ∈ N0 maximalwith the property pm
∣∣ |G|. A subgroup of order pm of G is then called a Sylow p-subgroupof G.
We extend the definition of a complete pre-image from the beginning of this chapter,passing from elements to subsets of the target set of a mapping. Let A, B sets, ϕ ∈ BA.
54
For all T ⊆ B we set53
Tϕ− := {a|a ∈ A, aϕ ∈ T}.Then ϕ− is a mapping of P(B) to P(A). For all X ⊆ A we have Xϕϕ− ⊇ X .
3.7.2. Let ϕ be a surjective mapping of a set A onto a set B, S, T ⊆ B. Then Tϕ−ϕ = Tand S ⊆ T ⇔ Sϕ− ⊆ Tϕ−.
Proof. For all b ∈ B we have (bϕ−)ϕ = {b}, hence the first claim. We apply this in thelast of the following chain of implications:
S ⊆ T ⇒ Sϕ− ⊆ Tϕ− ⇒ Sϕ−ϕ ⊆ Tϕ−ϕ⇒ S ⊆ T.
The claimed equivalence follows.
3.8 Theorem (Correspondence theorem for groups). Let ϕ be an epimorphism of agroup G onto a group G. Then γ := ϕ−|U(G) is an isomorphism of (U(G),≤) onto([kerϕ,G],≤).
{1G}
{1G}
G G
✛
γ
kerϕ
(The ellipse sketches the subgroup lattice of G (the subgroup interval [kerϕ,G] resp.).The same assertion holds regarding the lattices of normal subgroups of G and of G(containing kerϕ).
We insert a few comments before we prove this result:
From 1.3(4) we know that Uϕ ≤ G if U ≤ G. The correspondence assertion of 3.8 meansthat each subgroup of G arises in this way, and that the respective subgroup U of Gis uniquely determined under the condition that kerϕ ≤ U . By 3.1.5, for any N E Gthe mapping ρ : G → G/N , g 7→ Ng, is an epimorphism. By 3.8, the subgroups ofG/N are exactly the quotients X/N where N ≤ X ≤ G. Applying 3.8 to the canonicalhomomorphism ρ, we thus obtain
([N,G],≤) ∼= (U(G/N),≤)For example, we may determine the subgroups of a factor group (Z/nZ,+) of (Z,+):Every subgroup X of (Z,+) has the form kZ for a unique k ∈ N0, by 1.2.1. We havenZ ≤ kZ if and only if k|n. Hence the subgroups of (Z/nZ,+) are the sets kZ/nZ wherek|n. (Z/nZ,+) has no nontrivial proper subgroup if and only if n = 1 or n is a prime.Given an arbitrary group G, a subgroup H of G is called maximal (denoted by H <
maxG)
if H < G and there is no subgroup X such that H < X < G. Thus
53We re-obtain our former definition if we consider a singleton T = {b} with b ∈ Aϕ. The drawbackthat {b} should be distinguished from b does less harm than introducing a further function name,different from ϕ−, for this reason.
55
3.8.1. nZ <max
Z if and only if n is a prime. �
Proof of 3.8. By 1.3(4), kerϕ = 1Gϕ− ⊆ Y γ for all Y ≤ G. Furthermore, for any
g, h ∈ Y γ we have (gh−1)ϕ = (gϕ)(hϕ)−1 ∈ Y , hence gh−1 ∈ Y γ. Thus γ is a mappingof U(G) into [kerϕ,G], and injective by the first assertion in 3.7.2. We prove that γ issurjective: Let X ∈ [kerϕ,G], Y := Xϕ. Then Y γ = Xϕϕ− ⊇ X . Let g ∈ Y γ. Thengϕ ∈ Y = Xϕ. Hence there exists an element x ∈ X such that gϕ = xϕ. It follows thatgx−1 ∈ kerϕ ≤ X , thus g ∈ Xx = X . This shows that Y γ = X . Thus γ is a bijection.Now it suffices to apply the second assertion in 3.7.2 to complete the proof.
If X ≤ G and Y := Xϕ, then (gX)ϕ = (gϕ)Y , (Xg)ϕ = Y (gϕ). Hence kerϕ ≤ X EGimplies that X = Y γ, Y E G. Conversely, if Y E G, there exists a homomorphism ψ of Ginto some other group W such that Y = kerψ, by 3.2.1. Then ϕψ is a homomorphismof G into W , and g ∈ kerϕψ if and only if gϕ ∈ kerψ = Y . Hence kerϕψ = Y γ, thuskerϕ ≤ Y γ E G by 3.2.1. This proves the additional assertion about the lattices ofnormal subgroups. �
3.9 Proposition. Let G be a group, N EG.
(1) Let U ≤ G. Then U ∩N E U , NU ≤ G and U/U ∩N ∼= NU/N .
(2) Let N ⊆M ≤ G, M/N EG/N . Then M EG and (G/N)/(M/N) ∼= G/M .
G
N
U
NU
U∩N
G
N
M
Proof. 54 (1) Let ϕ be the restriction of the canonical homomorphism ρ from 3.1.5 to U :
ϕ : U → G/N, u 7→ Nu (= Nnu for all n ∈ N).
54It is useful to sketch lattices not in every detail but to depict just the part which is of interest in thecontext. For example, in 3.9(1), the subgroups N,U,U ∩ N,NU are of interest. We do not knowanything about the subgroup intervals [U ∩N,U ], [U ∩N,N ], [N,NU ], [U,NU ], but we know thatU ∩ N = inf(U,N) and NU = sup(U,N) (by 3.7.1). Therefore we outline the situation as shownbeside the proof although we have no information about the mentioned subgroup intervals. (In acomplete representation of the subgroup lattice we would, for example, join U ∩ N and U by anascending line segment if and only if U ∩N <
maxU !) In such an illustration, a normal subgroup and
another subgroup will always give rise to a parallelogram if it is not known whether one of the twocontains the other. By 3.8, the line segment leading from U ∩N to U stands for the whole subgrouplattice of U/U ∩N . The analogous comment holds for NU/N . (So to speak, we have replaced theellipses in the sketch after the formulation of 3.8 by line segments.) Thus the parallelism of the twosides of the parallelogram which correspond to these two factor groups “signalizes” the isomorphismasserted in (1). – Furthermore, it is useful to distinguish (interesting) normal subgroups (of the“mother group” G) from non-normal subgroups by different kinds of dots: Our dots for N (and M ,G in (2)) are slightly thicker.
56
Then ϕ is a homomorphism of U onto NU/N , U∩N = {u|u ∈ U, Nu = N} = kerϕEU ,U/U ∩N = U/ kerϕ ∼= Uϕ = NU/N by 3.2.
(2) Let ϕ : G → (G/N)/(M/N), g 7→ (M/N)Ng. Then ϕ is the composition of twocanonical epimorphisms:
G→ G/N → (G/N)/(M/N), g 7→ Ng 7→ (M/N)Ng,
hence is an epimorphism. We have
g ∈ kerϕ⇔ Ng ∈M/N ⇔ ∃x ∈M Ng = Nx⇔ g ∈ NM ⇔ g ∈M.
By 3.2, it follows that G/M = G/ kerϕ ∼= Gϕ = (G/N)/(M/N).
We now turn to structures with more than one operation:
3.10 Definition. Let K be a commutative unitary ring, (A,+, ·) a K-algebra. An ideal(more precisely, “K-ideal”) of A is a K-subspace J of A such that J ·A ⊆ J , A · J ⊆ J ,for which we will use the notation J E
KA.55 As the group (A,+) is abelian, there is
no danger of confusion with the notation used for normal subgroups of groups: Everysubgroup of (A,+) is normal (3.1.3), so the use of the symbol E, regarding an algebra,should reasonably express something else: it is the ideal property. We even omit theletter K if the reference to K is clear, thus removing even the only distinguishing detailbetween the normal subgroup notation and the ideal notation. Clearly, there is a goodreason for this double usage of the same symbol. In short terms, it will soon becomeclear that, in the theory of algebras, ideals are the analogue of normal subgroups ingroup theory. A first (and rather vague) observation in the direction of this analogueis that ideals are special subalgebras, distinguished by a property which depends on thecontaining algebra A: Like the notion of a normal subgroup, that of an ideal is a relativeone, meaningful only with respect to A. The defining property of an ideal is obviouslymuch stronger than the ordinary (internal) multiplicative closure property. An idealbehaves like a “big zero” in an algebra: Multiplying any element by an element of anideal, the result is “swallowed” by the ideal, like 0A · x = 0A = x · 0A for any algebraelement x. Clearly, {0A} and A are ideals of A, the so-called trivial ideals. A unitaryalgebra which has only the trivial ideals is called simple. Every field is an example of asimple Z-algebra. More in general, we have:
3.10.1. If (A, ·) is a group, then A is unitary and associative, and {0A}, A are the onlyone-sided ideals of A. In particular, A is simple.
Proof. As A = A ∪ {0A}, the existence of a neutral element and the associative law in(A, ·) imply that A is unitary and associative. Let R be a right ideal of A, 0A 6= x ∈ R.Then 1A ∈ xA ⊆ R, hence R ⊇ xA = (xA)A = A, R = A. The proof for left ideals isanalogous.
55Thus an ideal J is both a left ideal and a right ideal of A.
57
A is called an (associative) division algebra if (A, ·) is a group. The starting point of thetheory of algebras (around 1840) was the discovery of a 4-dimensional non-commutativeassociative division algebra over R, the so-called quaternion algebra. The algebra of alln× n matrices over a field K is an example of a simple algebra which, for n > 1, is nota division algebra. In this course, however, we will mainly be interested in commutativealgebras; even more specifically, in fields.
In many respects, the familiar example of the integral domain (A = K =)Z is atypical asan algebra. It is of course an important example which should be completely understood.Clearly, an ideal (of any algebra) must be an additive subgroup. The additive subgroupsof Z have been determined in 1.2.1. As a trivial consequence we note:
3.10.2. The ideals of the Z-algebra (Z,+, ·) are exactly the subgroups of (Z,+). �
We write J ⊳max
A if J is a maximal ideal of A, i. e., if J is a proper ideal of A and there is
no ideal I of A such that J ⊂ I ⊂ A. For a simple algebra A – in particular for a field–, 3.10.1 implies that {0A} ⊳
maxA. By 3.10.2 and the discussion prior to the proof of 3.8,
nZ ⊳max
Z if and only if n is a prime.
Let A be an arbitraryK-algebra and JEA. Then (J+a)·(J+b) ⊆ J ·J+J ·b+a·J+ab ⊆J + ab for all a, b ∈ A, by the ideal property of J . It follows that
(∗) ∀U, V ∈ A/J ∃W ∈ A/J U · V ⊆W.
Recall that a very similar observation enabled us in steps (II), (III) of the constructionof a quotient field (preceding 2.10) to define the crucial operations (⊡, ⊞). Here againwe have a partition (the set A/J of all additive cosets of J in A) which is compatiblewith an operation (·), in the sense of (∗). Clearly, W in (∗) is uniquely determined byU , V . Therefore we define UV to be the additive coset of J in A in which the set U · Vis contained.56 As in step (II) on p. 42, we cannot expect U · V to be a complete coset.
56The general notion of a quotient is based on that of a partition M of the underlying set X of amagma (X, ◦) which is compatible (with ◦) in the following sense:
∀U, V ∈M ∃W ∈ M U ◦V ⊆W
As thenW is uniquely determined by the couple (U, V ) we may define UV to be thatW ∈ M whichcontains the product set U ◦V . This defines an operation onM, thus giving rise to a magma withunderlying setM. The mapping x 7→ [x] (where [x] denotes the unique element ofM containing x)is obviously an epimorphism of X onto our magmaM. A magma which is given in the describedway by a compatible partition of X is called a quotient of X . Without difficulty we obtain thefollowing
General homomorphism theorem. Let ϕ be a homomorphism of a magma (X, ◦) into a magma
(Y, ·), M := {bϕ−|b ∈ Xϕ}. Then Xϕ is a submagma of (Y, ·), M is a quotient of X, and ϕ− is
an isomorphism of Xϕ ontoM.
Note, however, that only for a group (X, ◦) we have the satisfytory description of the partitionelement [x] as the coset of x with respect to the kernel of ϕ (see 3.2(1)), a regularity which we maynot even dream of in the case of an arbitrary magma.
58
But passing to the uniquely determined coset that contains U · V (and calling it UV ),we obtain a (multiplicative) operation on A/J . In order to calculate the product UV ,it suffices to determine just one element of it (and form its coset with respect to J). Forany a ∈ U , b ∈ V , certainly ab ∈ U · V ⊆ UV , hence we have
∀a, b ∈ A (J + a)(J + b) = J + ab.
In other words, the canonical K-space epimorphism κ : A → A/J, a 7→ J + a, is alsoa multiplicative homomorphism, hence a K-algebra epimorphism. By the propertiesproved in 1.3, the K-double magma A/J , endowed with the usual coset addition andthe multiplication just defined, turns out to be a K-algebra, as an epimorphic image ofA. It is called the factor algebra of A by J . Clearly, ker κ = J . Now we are prepared toprove the analogue of 3.2 for algebras:
3.11 Theorem (Homomorphism theorem for algebras). Let K be a commutative unitaryring, A a K-algebra, ϕ a K-algebra homomorphism of A into a K-double magma B,J := kerϕ.
(1) J EKA,
(2) ϕ− is a K-algebra isomorphism of Aϕ onto A/J .
Proof. We shall write + for the first operation (in both A and B) and express the secondoperation simply by juxtaposition. In particular, ϕ is aK-linear mapping of theK-space(A,+) into the K-magma (B,+). The isomorphism ϕ− of (Aϕ,+) onto (A/J,+) (see3.2(2)) is therefore K-linear, hence an isomorphism of K-magmas. It remains to show
ad (1) ∀x ∈ J ∀a ∈ A xa, ax ∈ J : Let x ∈ J , a ∈ A. Then (xa)ϕ = (xϕ)(aϕ) =(0Aa)ϕ = 0Aϕ, similarly (ax)ϕ = 0Aϕ and (cx)ϕ = 0Aϕ for all c ∈ K.
ad (2) ϕ− is a multiplicative homomorphism: Let a, a′ ∈ A. Making use of 3.2(1),we obtain
((aϕ)(a′ϕ)
)ϕ− = (aa′)ϕϕ− = J + aa′ = (J + a)(J + a′) = (aϕ)ϕ−(a′ϕ)ϕ−.
Hence ϕ− is an isomorphism of the K-double magma Aϕ onto the K-algebra A/J . Thisimplies, in particular, that the K-double magma Aϕ is a K-algebra57.
We observed at the end of 3.10 that every ideal J of a K-algebra A is the kernel of thecanonical K-algebra homomorphism of A onto A/J . Combining this with 3.11(1), weobtain the following analogue of 3.2.1:
3.11.1. The ideals of a K-algebra A are exactly the kernels of the K-algebra homo-morphisms of A. �
The theorem includes the homomorphism theorem for K-spaces (by considering themas K-algebras in which the product of any two elements is defined to be zero) and forrings (being (associative) Z-algebras). Now let R be any commutative unitary ring.
57Thus, an – albeit trivial – separate verification of this fact is not necessary.
59
The mapping ϕ : Z → R, k 7→ k1R, is an additive homomorphism58. In particular,(−k)ϕ = −(kϕ) for all k ∈ Z. It is also a multiplicative homomorphism:
∀k, l ∈ Z (kl)ϕ = (kl)1R = (k1R)(l1R) = (kϕ)(lϕ),
the crucial mid equation being a consequence of the distributive law in R.59 As inExample (3) on p. 50, define n ∈ N0 by the equation kerϕ = nZ. By 3.11(2), we havethe isomorphism Z/nZ ∼= 〈1R〉 not only regarding the additive groups (as noted inExample (3) on p. 50) but also multiplicatively, hence with respect to the ring structuresof Z and R.
3.12 Definition. Let R be a commutative ring and ϕ as above. The number n ∈ N0
such that kerϕ = nZ is called the characteristic of R and denoted by charR. We haveZ/nZ ∼= Zϕ and distinguish the following two cases:
1st case: n = 0. Then Zϕ ∼= Z.
2nd case: n > 0. Then Zϕ = {0R, 1R, 1R + 1R, . . . , 1R + · · ·n−1
+ 1R}.
In the second case, R is called of positive characteristic. In both cases, we obtain anexplicit description of the subring generated by 1R which we call the prime ring of R.
3.12.1. Let R be an integral domain. Then charR equals 0 or is a prime.
Proof. Let n := charR and assume n 6= 0. Then n > 1 as R is unitary. Assume thatthere are k, l ∈ n− 1 such that kl = n. Then k1R, l1R 6= 0R and 0R = n1R = (k1R)(l1R),a contradiction as R has no zero divisors 6= 0. Hence n is a prime.
Given a K-algebra A, we write T (A) for the set of all subalgebras of A. As in the caseof subgroups of a group, these form a lattice with respect to set-theoretic inclusion. Thesame holds for the set of ideals of A. Writing 4
Kfor the relation of being a K-subalgebra,
we have inf(U, V ) = U ∩ V , sup(U, V ) = 〈U + V 〉 := ⋂{W |U + V ⊆ W 4KA} for all
U, V 4KA. Clearly, sup(U, V ) = U + V if and only if U + V is a subalgebra of A. In
particular, this is the case if U EA or V EA as follows from the following trivial remark:
3.12.2. Let U, V ∈ T (A) such that UV, V U ⊆ U + V . Then U + V ∈ T (A). �
If both U and V are ideals of A, so is U + V : For all a ∈ A we then have a(U + V ) =aU + aV ⊆ U + V , likewise (U + V )a ⊆ U + V . Applying 3.11 like 3.2 in the proofs of3.8, 3.9, we readily obtain the following important analogues for algebras:
58This is a special case of Example (3) on p. 50 with additive notation, x := 1R.59Applying the rule for additive inverses mentioned beforehand, we may assume w. l. o. g. that k ∈ N0
and proceed by a routine induction on k in this case.
60
3.13 Proposition. Let K be a commutative unitary ring, A a K-algebra, J E A.
(1) (Correspondence theorem for algebras). Let ϕ be an epimorphism of A onto aK-algebra A. Then ϕ−|T (A) is an isomorphism of (T (A),4
K) onto ([kerϕ,A],4
K).
The same assertion holds regarding the lattices of the ideals of A and those of Acontaining kerϕ.
(2) Let U 4KA. Then U ∩ J E U , J + U 4
KA and U/U ∩ J ∼= (J + U)/J .
(3) Let J ⊆ I 4KG, I/J E A/J . Then I EA and (A/J)/(I/J) ∼= A/I. �
The results 3.9, 3.13(2),(3) are commonly called the isomorphism theorems for groups,algebras resp. In the sequel, we will choose K = Z and consider a commutative ring Rin place of A. By 3.10.1, every field is simple. We now observe the converse:
3.13.1. Let R be a commutative simple ring. The R is a field.
Proof. Let x ∈ R. Then 0R 6= x ∈ xR E R, hence xR = R. In particular, xy = 1R forsome y ∈ R. Clearly, y 6= 0R. It follows that (R, ·) is an abelian group.
3.13.2. Let R be a commutative unitary ring, J ⊳max
R. Then R/J is a field.
Proof. By 3.13(1) (in the version for ideals), it follows that the commutative unitaryfactor ring R/J is simple. The claim follows from 3.13.1.
Conversely, if R/J is a field for some ideal J of R, then J must be maximal, by 3.13(1).Making use of 3.8.1 and 3.10.2, we conclude
3.13.3. Let n ∈ N. The factor ring Z/nZ is a field if and only if n is a prime. �
This remark and the description in the 2nd case of 3.12 imply
3.13.4. Let K be a field such that p := charK > 0. Then the prime ring of K is asubfield of K and isomorphic to Z/pZ. �
We could have considered an integral domain R instead of a field here but would nothave gained more generality: The version for R follows from 3.13.4 by choosing K as aquotient field of R.
If K is a field of characteristic 0, the prime ring of K is isomorphic to Z (see the 1st casein 3.12). Hence its quotient field in K (see 2.10.1) is isomorphic to Q. It is obviouslythe smallest subfield of K containing 1K . In any field K, the subfield generated by 1Kis called the prime field of K. We summarize:
3.13.5. Let K be a field, K0 its prime field. If charK = 0, then K0∼= Q. If charK =
p > 0, then K0∼= Z/pZ, in particular, |K0| = p. �
61
The theory of fields therefore has a natural subdivision into two chapters of ratherdifferent peculiarities: (1) Fields of characteristic 0, (2) fields of prime characteristic.An ordered field is necessarily of characteristic 0 because a sum of 1’s can never attainthe value 0 in an ordered field. Clearly, a finite field must be of prime characteristic.The theory of finite fields may be viewed as a natural subchapter in field theory, andwe will derive its fundamentals as applications of the results of the next chapter. Thereare quite a number of theorems on fields which hold for every characteristic but hithertorequire different proofs for case (1) and case (2). In the theory of representations overa field K it makes a fundamental difference if charK = 0 (the so-called classical case)or charK is a prime (the so-called modular case), the latter generally imposing muchmore intricate obstacles than the rather well understood first case.
3.14 Definition. Let R be a commutative unitary ring. A subset J of R is called aprincipal ideal of R if there exists an element a ∈ R such that J = aR.60 Any suchelement a is called an (ideal) generator of J .
3.14.1. ∀a, b ∈ R (aR ⊆ bR⇔ ∃x ∈ R bx = a),
as aR ⊆ bR if and only if a ∈ bR because R is associative and unitary. �
Recall 2.2 where we introduced the notation b |R
a for the condition on the right in 3.14.1.
3.14.2. Let K be a field, f, g ∈ K[t]rK such that g |K[t]
f and g of minimal degree with
this property. Then g is irreducible in K[t].
Proof. Let r, s ∈ K[t] such that rs = g and assume that r 6∈ K. Then r |K[t]
f and
deg r ≤ deg g. It follows that deg r = deg g, hence s ∈ K by 1.1.2.
R is called a principal ideal domain if the following holds:
(i) R is an integral domain,
(ii) Every ideal of R is a principal ideal.
By 3.10.2 and 1.2.1, Z is a principal ideal domain. The polynomial ring Z[t] is not aprincipal ideal domain: The ideal generated by 2 and t is not a principal ideal. Moregenerally, it is an easy exercise to prove that R[t] is a principal ideal domain only if Ris a field. Conversely, this necessary condition is sufficient:61
60Clearly, any principal ideal of R is an ideal of R, thus deserves its name.61Polynomial rings in one variable over a field belong to the larger class of integral domains called
Euclidean domains. Essentially, these are integral domains for which there exists a function into N0
with properties that are similar to those of the degree function for K[t], in the sense that they allowthe line of reasoning as in the proof of 3.14.3. It is then not really surprising that, by the sameproof in general form, every Euclidean domain turns out to be a principal ideal domain. The generalnotion of a Euclidean domain can play a role in directions which are not pursued in this course.
62
3.14.3. Let K be field, S := K[t]. Then S is a principal ideal domain. If {0K} 6= JES,g ∈ J of minimal degree, then J = gS.
Proof. Condition (i) holds by 1.1.2. (ii) Let J E S. If J = {0S}, then J = 0SS, henceprincipal. Now let J 6= {0S}, g ∈ J of minimal degree. Clearly, gS ⊆ J as g ∈ J E S.We claim that gS ⊇ J . We have gS = cgS for all c ∈ K. Therefore we may choosec with the property that cg is normed, then consider cg instead of g. In other words,we may assume w. l. o. g, that g is normed. Let f ∈ S. By 1.6(1), there exist h, r ∈ Ssuch that f = gh + r, r = 0K or deg r < deg g. Then r = f − gh ∈ J so that, by theminimality of the degree of g, we conclude r = 0K . It follows that f = gh ∈ gS.
3.15 Proposition. Let K be a field, 0K 6= f ∈ K[t], J := fK[t]. Then
(1) dimK K[t]/J = deg f ,
(2) dimK K[t]/J is a field if and only if f is irreducible.
Proof. (1) Let n := deg f . By 1.6(1), for every f ∈ K[t] there exists a polynomialr ∈ K[t] of degree < n such that J + f = J + r. Hence {J +1K , J + t, . . . , J + tn−1} is aset of generators of theK-spaceK[t]/J . As J does not contain any (nonzero) polynomialof degree < n, it is a K-linearly independent set of n elements. The claim follows.
(2) We apply 3.13.2 and its converse to conclude that K[t]/J is a field if and only ifJ ⊳
maxK[t]. From 3.14.1 and 3.14.3 it follows that the maximality of the ideal J is
equivalent to the property that the ideal generator f is irreducible.
Let R be a principal ideal domain, M := R. Then (M, ·) is a commutative monoid.Let q ∈ M be indecomposable. From 3.14.1 it follows that qR ⊳
maxR. Suppose that
q ∤M
a, b where a, b ∈M . Then the ideals aR + qR, bR + qR properly contain qR, hence
equal R. Thus both of them contain 1R. It follows that 1R ∈ (aR + qR)(bR + qR) ⊆abR + qR. Hence ab /∈ qR, i. e., q ∤
M
ab. Thus we have shown, by contraposition, that
the indecomposable elements q of the multiplicative monoid M of nonzero elements ofany principal ideal domain have the property (∗) on p. 29. From this it follows easilythat the same holds with respect to the submonoid N of (Z, ·) and the submonoid N of(V , ·) in the examples (1), (2) on p. 28.
63
4. Field extensions
4.1 Definition. A field extension is a pair (K,M) of fields K, M such that K is asubfield of M . Then M is called an extension field of K. A basic idea of everything thatfollows is to view M as a K-vector space (cf. Example (2) on p. 33). A field extension(K,M) is called finite if dimKM is finite. In particular, this is the case if M is finite.The interpretation of M as a vector space over any subfield K has the following strikingconsequence:
4.1.1. If M is finite, there exists an integer n such that |M | = |K|n. In particular, |M |is a power of the prime charM .
Proof. As M is a finite dimensional vector space over K we have the K-space isomorph-ism M∼=Kn for some n ∈ N. Hence |M | = |K|n. The last assertion is the special casewhere K is the prime field of M (see 3.13.5).
Let b be an element of an extension field M of the field K, Fb the replacement ho-momorphism K[t] → M , f 7→ f(b). The element b is called algebraic over K ifkerFb 6= {0K}, i, e., if there exists a nonzero polynomial f ∈ K[t] such that f(b) = 0K .
62
For example, 3√5 is algebraic over Q as it is a zero of t3 − 5. Every root of unity is
algebraic over Q, being a zero of tn − 1 for some n ∈ N. Trivially, every element b ∈ Kis algebraic over K as it is a zero of t − b. If kerFb = {0K}, b is called transcendentalover K. Equivalently, this means that the subring K[b] of M generated by K and b is apolynomial ring over K in the variable b (see 1.1 or, more generally, 2.9). The numberse and π are famous examples of real numbers which are transcental over Q.63 We set
AK(M) := {b|b ∈M, b is algebraic over K}.
The extension (K,M) is called algebraic if AK(M) = M . For example, Q[√2] is a field
and (Q,Q[√2]) is algebraic as will be recognized as a very small special case of the
general remark 4.1.5. (Q,R) is not algebraic: We could refer to the transcendency of eor π or, simpler, to the argument mentioned in footnote 63.
4.1.2. If (K,M) is finite, then (K,M) is algebraic.
62Resuming the notion introduced on p. 39, this means that the 1-tuple (b) is algebraically dependent.The transcendency of b means the contrary, i. e., that (b) is algebraically independent.
63The first proofs for the transcendency of e, π resp., were discovered by Ch.Hermite (1873, fore), C. L. F. v. Lindemann (1882, for π) resp. The set of all transcendental real numbers is non-denumerable because R is non-denumerable while the set of all algebraic real numbers is denu-merable, as is not very difficult to prove. By contrast and hitherto without exception, a proof oftranscendency for a specific real number requires non-trivial analytical methods. Therefore questionsof this kind are usually treated within the area of Analytic number theory.
65
Proof. Let n := dimKM , b ∈M . The (n+1)-tuple (1K , b, . . . , bn) is K-linearly depend-
ent, i. e., there exist c0, . . . , cn ∈ K, not all of them zero, such that c0+ c1b+ · · ·+ cnbn =0K . Thus b is a zero of the nonzero polynomial
∑nj=0 cjt
j ∈ K[t].
For every b ∈ AK(M) there exists a unique polynomial with zero b which is of minimaldegree and normed. This polynomial is called the minimal polynomial of b over K and isdenoted by minb,K . From 3.14.3 we obtain
4.1.3. Let b ∈ AK(M). Then kerFb = minb,K K[t]. �
4.1.4. Let b ∈ AK(M), f ∈ K[t]. Then f = minb,K if and only if f is irreducible andnormed and f(b) = 0K.
Proof. Suppose f = minb,K and let g, h ∈ K[t] such that f = gh. Then f is normedand g(b)h(b) = 0K . As K has no zero divisors 6= 0K , it follows that g(b) = 0K orh(b) = 0K . Hence, by 4.1.3, f |
K[t]
g or f |K[t]
h. As f = gh, this implies h ∈ K or g ∈ K.
Thus f is irreducible. Conversely, let f be irreducible, normed, and f(b) = 0K . Thenf ∈ kerFb, hence minb,K |
K[t]
f by 4.1.3. As f is irreducible in K[t] and minb,K /∈ K, it
follows that f = cminb,K for some c ∈ K. But f and minb,K are normed, thus c = 1K ,f = minb,K .
Our next remark will have most important applications in the sequel:
4.1.5. Let b ∈ AK(M), J := minb,KK[t]. Then there exists an isomorphism ψ of K[b]onto K[t]/J with the properties bψ = J + t, cψ = J + c for all c ∈ K. In particular,K[b] is a field, and dimK K[b] = degminb,K.
Proof. We haveK[b] = K[t]Fb and, by 4.1.3, J = kerFb. Thus the isomorphism assertionnow follows from 3.11(2). The final assertions follow both from 3.15 as minb,K generatesJ by 4.1.3 and is irreducible by 4.1.4.
For example, Q[√2] is a field and of dimension 2 over Q as min√
2,Q = t2 − 2. If b ∈ Mis transcendental over K, K[b] is a polynomial ring over K, not a field, hence properlycontained in its quotient field K(b) in M . If b ∈ AK(M), then A[b] = A(b) by 4.1.5.Hence, for an arbitrary b ∈M , we have
4.1.6. b ∈M is algebraic over K if and only if K[b] = K(b). �
4.1.7. AK(M) is a subfield of M .
Proof. 64 Clearly, 0K , 1K ∈ AK(M). Let b, b′ ∈ AK(M). We have K[b, b′] = (K[b])[b′],
64The reader should be warned that an approach to the assertion by considering minimal polynomialsin K[t] of elements a, b ∈M is no good idea. It is generally hard to derive information about minab,Kor mina+b,K from mina,K , minb,K . We mention the following result:
Theorem. (Isaacs 1970) Let charK = 0 and dimK K[a+b] = dimK K[a] dimK K[b]. Then K[a, b] =K[a+ b].
The hypothesis is satisfied if dimK K[a] and dimK K[b] are relatively prime so that in this casewe have degmina+b,K = degmina,K degminb,K , by 4.1.5. The proof in [Is] should give a sufficientimpression about the intricacies of the topic.
66
and the trivial inclusion AK(M) ⊆ AK[b](M) implies that the extension (K[b], K[b, b′])is finite. By the special case of 2.6, dimK K[b, b′] = dimK K[b] dimK[b]K[b, b′] is finite,hence (K,K[b, b′]) is algebraic by 4.1.2. Thus b′− b and (if b′ 6= 0K) bb
′−1 are elements ofan algebraic extension field of K and hence algebraic over K. This shows that AK(M)is additively, AK(M)r {0K} multiplicatively a group.
Generalizing inductively the argument in this proof, we observe that for a finite num-ber of algebraic elements a0, a1, . . . , an of an extension field of K, the extension fieldK[a0, . . . , an] = K[a0][a1] . . . [an] is of dimension
∏j∈n dimK[aj−1]K[aj ] over K. In par-
ticular, its dimension is finite, hence it is an algebraic extension of K by 4.1.2. Now thefollowing is a simple consequence:
4.1.8. Let (K,L), (L,M) be algebraic field extensions. Then (K,M) is algebraic.
Proof. Let b ∈ M . As (L,M) is algebraic, there exists an m ∈ N0 and elementsa0, . . . , am ∈ L, am 6= 0K , such that
∑mj=0 ajb
j = 0K . It follows that b is algebraicover the field K[a0, . . . , am]. By the above argument, K[a0, . . . , am, b] is an algebraicextension of K. In particular, b is algebraic over K.
We are going to pursue now questions about the zeros of polynomials as presented inour first chapter. To this end we introduce some useful notation: For any f ∈ K[t], theset of all zeros of f in M will be denoted by ZM(f). Given an isomorphism ϕ of Konto a field K, we write ϕ for the extension of ϕ to the isomorphism of K[t] onto thepolynomial ring K[t] over K with the properties tϕ = t. Explicitly,
ϕ : K[t]→ K[t],∑
j
cjtj 7→
∑
j
(cjϕ)tj
4.2 Lemma. Let K be a field, g ∈ K[t] irreducible.
(1) There exists an extension field L of K such that ZL(g) 6= ∅.
(2) Let L be as in (1), b ∈ ZL(g), ϕ a monomorphism of K into a field X, K := Kϕ,g := gϕ. Then for each b ∈ ZX(g) there exists a unique monomorphism ϕb of K[b]into X such that ϕb|K = ϕ, bϕb = b.
Before we embark on the proof we insert a simple remark,under the same hypotheses as in (2):
4.2.1. Let ψ be any extension of ϕ to a monomorphism ofK[b] into X. Then bψ ∈ ZX(g),
because if g =∑n
j=0 cjtj (cj ∈ K) then
K K
X
K[b]L ≥ K[b]
✲
✲
ϕ
ϕb
�
0X = 0Kψ = g(b)ψ =( n∑
j=0
cjbj)ψ =
n∑
j=0
(cjϕ)(bψ)j = g(bψ)
Therefore we have the following consequence of (2):
67
4.2.2. The assignment b 7→ ϕb is a bijection of ZX(g) onto the set of extensions of ϕ toa monomorphism of K[b] into X. Hence their number equals |ZX(g)|. �
Proof of 4.2. We assume w. l. o. g. that g is normed and put J := gK[t]. By 3.15(2),K[t]/J is a field.
(1) (Kronecker’s construction) Let σ : K → K[t]/J , c 7→ J + c. This is a field mono-morphism, being the restriction to K of the canonical epimorphism of K[t] onto K[t]/J .
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L
K
K[t]/J
σ
σClearly, J ∩ K = {0K}. By 2.1, there exists an extension field L ofK and an extension σ of σ to an isomorphism of L onto K[t]/J . Letb := (J + t)σ−1, c0, . . . , cn ∈ K such that g =
∑nj=0 cjt
j . Then
(g(b)
)σ =
( n∑
j=0
cjbj)σ =
n∑
j=0
(J+cj)(J+t)j = J+
n∑
j=0
cjtj = J+g = J,
hence g(b) = 0K .
(2) Let b ∈ ZX(g). Since K[b] is generated by K and b as a ring, there is at most onefield monomorphism of K[b] into X which maps b to b and c to cϕ for all c ∈ K. Itremains to prove that there is at least one such monomorphism.
As g is irreducible in K[t], g is irreducible in K[t]. We have g = ming,K , g = minb,K , by
4.1.4. Let J := Jϕ = gK[t]. Then ϕ induces an isomorphism ¯ϕ of K[t]/J onto K[t]/Jsuch that (J + t) ¯ϕ = J + t, (J + c) ¯ϕ = J + cϕ for all c ∈ K.
Let ψ be the isomorphism of K[b] onto K[t]/J , in the same way ψ the isomorphism ofK[b] onto K[t]/J which is given by 4.1.5. Put ϕb := ψ ¯ϕψ−1:
K[b] →ψ
K[t]/J →ϕ
K[t]/J →ψ−1
K[b]
Then ϕb is a composition of isomorphisms, hence an isomorphism of K[b] onto K[b] andmaps b to b, c to cϕ for all c ∈ K. �
4.3 Theorem (Splitting field theorem). Let K be a field, f ∈ K[t], f 6= 0K .
(1) There exists a splitting field of f over K.
(2) Let M be a splitting field of f over K, ϕ a monomorphism of K into a field Xwhich is a splitting field of fϕ over Kϕ. Then there exists an extension ψ of ϕ toan isomorphism of M onto X. In particular, for any two splitting fields M,X of fover K there exists an isomorphism ψ of M onto X such that cψ = c for all c ∈ K.
Proof by induction on deg f .65 If deg f = 0, then f ∈ K and both (1), (2) are trivial. Forthe inductive step, let deg f > 0 and assume (1), (2) to be true for nonzero polynomials
65This is the common short way of announcing a proof by induction on n for the assertion: “For alln ∈ N0 the following holds: If K is a field, f ∈ K[t], f 6= 0K , deg f = n, then the assertions (1), (2)follow.” Note that the assertion is not about one given field K but the universal quantifier regardingK is part of the claimed proposition. The proof starts with a field K, but the inductive hypothesiswill be applied to a different field!
68
of smaller degree over arbitrary fields. By 3.14.2, there exists an irreducible polynomialdivisor g of f , and we may assume g to be normed.
(1) By 4.2(1), there exists an extension field L of K such that ZL(g) 6= ∅. Let b ∈ ZL(g).
K Kϕ
XM
K[b]
✲ϕ
Then f(b) = 0K as g |K[t]
f . By 1.6(2), there exists a polynomial
h ∈ (K[b])[t] such that f = (t−b)h. By 1.1.2, deg h = deg f−1.By our inductive hypothesis (for (1)), there exists a splittingfield M of h over K[b]. Since b ∈ ZM(f), M contains a splittingfield of f over K. But the latter contains K[b] and all zeros ofh in M , hence equals M .
(2) Put f := fϕ, K := Kϕ, g := gϕ. Then ZM(g) ⊆ ZM(f), ZX(g) ⊆ ZX(f). Letb ∈ ZM(g), b ∈ ZX(g). Let ϕg as given by 4.2(2). By 1.6(2), there exist h ∈ (K[b])[t],h ∈ (K[b])[t] such that f = (t− b)h, f = (t− b)h. Then M is a splitting field of h overK[b], X is a splitting field of h over K[b]. Inductively there exists an isomorphism ψ ofM onto X such that ψ|K[b] = ϕb. A fortiori, ψ|K = ϕ.
The final assertion is the special case where ϕ = idK . �
Let X be a field. If (K,L) is a field extension, ϕ a monomorphism of K into X , wewrite ϕL for the set of all monomorphisms of L into X which are extensions of ϕ.
4.3.1. Let (K,L), (L,M) be field extensions, ϕ a monomorphism of K into X. Then
ϕM =⋃
ψ∈ϕL
ψM ,
because the non-empty sets in this union are exactly the equivalence classes for theequivalence ∼ on ϕM defined by: ω ∼ ω′ if and only if ω|L = ω′|L (ω, ω′ ∈ ϕM).66 �
4.4 Proposition. Let X be a field, (K,M) a finite field extension, ϕ a monomorphismof K into X, K := Kϕ. Let b ∈M , g := minb,K , g := gϕ. Then
|ϕM | =∑
b∈ZX(g)
|ϕMb| ≤ |ZX(g)| dimK[b]M ≤ deg g dimK[b]M = dimKM.
Proof. By means of 4.3.1 and 4.2.2, we have |ϕM | =∑
ψ∈ϕK[b]
|ψM | =∑
b∈ZX(g)
|ϕMb|. This
proves the first equality. The last equality follows from 4.1.5 and the special case of 2.6which imply deg g dimK[b]M = dimK K[b] dimK[b]M = dimKM . We have deg g = deg g,hence |ZX(g)| ≤ deg g by 1.6(3). It remains to prove the first inequality in our claim.This will be a consequence of the following intermediate result:
(∗) |ϕM | ≤ dimKM
which we prove by induction on dimKM . If dimKM = 1, then K =M , hence ϕM = {ϕ}and the claim is trivial. For the inductive step, let dimKM > 1 and assume that (∗) holds66This proof deals with absolute nothingness but is nevertheless correct if ϕM = ∅.
69
for all field extensions of smaller dimension. Let b ∈MrK. Then dimK[b]M < dimKM .Making use of the parts we have already proved, we obtain from our inductive hypothesis
|ϕM | =∑
b∈ZX(g)
|ϕMb| ≤
∑
b∈ZX(g)
dimK[b]M = |ZX(g)| dimK[b]M ≤ dimKM.
This proves (∗) which we will now apply to the monomorphisms ϕb (in place of ϕ) ofK[b] (in place of K) into X . We obtain
∑
b∈ZX(g)
|ϕMb| ≤
∑
b∈ZX(g)
dimK[b]M = |ZX(g)| dimK[b]M.
4.4.1. Let (K,M) be a finite field extension, X =M , ϕ ∈ AutK. Then ϕM ⊆ AutM ,
because for all ψ ∈ ϕM we have dimKM = dimKϕMψ = dimKMψ and M ∼=Mψ ≤Mwhich implies, by the finiteness of (K,M), that M =Mψ. �
In the following we consider the special case where X =M , ϕ = idK :
4.5 Definition. Let (K,M) be a field extension. Then
AutKM := {α|α ∈ AutM, ∀c ∈ K cα = c}
is a subgroup67 of AutM and is called the Galois group of (K,M).
4.5.1. ∀β ∈ AutM AutKβM = β−1(AutKM)β.
Proof. Let α, β ∈ AutM . Then
α ∈ AutKβM ⇔ ∀c ∈ K (cβ)α =cβ ⇔ ∀c ∈ K cβαβ−1 = c
⇔ βαβ−1 ∈ AutKM ⇔ α ∈ β−1(AutKM)β.
In an arbitrary group G, put ug := g−1ug for all u, g ∈ G, called the conjugate of u byg.68 Set Ug := g−1Ug for all U ⊆ G, g ∈ G. A subgroup U is normal if and only if
67It contains idM , is closed with respect to composition, and cα = c if and only if c = cα−1, for allc ∈ K.
68For every g ∈ G, the mapping G→ G, u 7→ ug, is an automorphism of G. The mapping
G→ AutG, g 7→[G → Gu 7→ ug
],
is a homomorphism. Its kernel {z|z ∈ G, ∀g ∈ G gz = zg} is called the centre of G and denotedby Z(G). Its image is a normal subgroup of AutG, called the group of inner automorphisms of Gand denoted by InG. Thus, by 3.2,
G/Z(G) ∼= InGEAutG.
These assertions are recommended as routine exercises. They belong to the basics in group theory.
70
Ug = U for all g ∈ G as a glance at the definition (3.1) shows. We may express 4.5.1 inthe following form:
4.5.1’ ∀β ∈ AutM AutLβM = (AutLM)β. �
4.5.2. Let M be a splitting field of a nonzero polynomial f ∈ K[t]. Then AutKM isisomorphic to a subgroup of SZM (f). In particular, |AutKM |
∣∣ k! where k := |ZM(f)|.
Proof. For all α ∈ AutKM the restriction α|ZM (f) is a permutation of ZM(f) as followsfrom 4.2.1 in the special case X =M , ϕ = idK . The restriction mapping ρ : AutKM →SZM (f), α 7→ α|ZM(f), is a homomorphism. Let α, β ∈ AutKM such that αρ = βρ. Thenα = β becauseM is generated as a field by K∪ZM(f). Hence ρ is a monomorphism.
As an immediate consequence of 4.4 (for X =M , ϕ = idK (hence g = g)) we note:
4.5.3. If (K,M) is finite, b ∈M , then |AutKM | ≤ |ZM(minb,K)| dimK[b]M ≤ dimKM .�
The question as to when these inequalities become, in fact, equalities will now lead toground-laying notions of Galois theory:
4.5.4. Let (K,M) be finite and |AutKM | = dimKM . Then |ZM(minb,K)| = degminb,Kfor all b ∈M , hence
(i) For every b ∈M , minb,K splits in M [t] into linear factors.
(ii) There is no b ∈M such that minb,K has a multiple root in M .
The first claim follows by specializing 4.4 as before. This means, by 1.6(3), that Mcontains a splitting field of minb,K , hence (i), and that the number of zeros of minb,K inM attains the maximal possible value, hence (ii). �
(K,M) is called normal if it is algebraic and satisfies condition (i). It is called separableif it is algebraic and satisfies condition (ii). A finite, normal, separable field extension iscalled Galois or a Galois extension.69
4.5.5. Let (K,L), (L,M) be algebraic field extensions.
(1) If (K,M) is normal, so is (L,M).
(2) If (K,M) is separable, so is (L,M).
(3) If (K,M) is Galois, so is (L,M).
Proof. The first two claims follow from the fact that, by 4.1.3 (with L in place of K),minb,L |
L[t]
minb,K for all b ∈M . Clearly, (3) is implied by (1) and (2).
69Note that there are authors who do not require a Galois extension to be finite.
71
A nonzero polynomial f ∈ K[t] is called separable over K if no irreducible factor of f hasa multiple zero in any extension field of K. Clearly, this is equivalent to the conditionthat no irreducible factor of f has a multiple zero in a splitting field of f over K. Thecondition that f itself has this property is certainly stronger70 but sometimes easy tocontrol. The following remark presents an important example:
4.5.6. Let p := charK > 0 and k ∈ N such that p|k. Then tk− t is separable over K. IfK is finite, k := |K|, then K = ZK(t
k − t). In particular, K is a splitting field of tk − tover its prime field.
Proof. We have (tk−t)′ = −1 as p|k, hence tk−t is separable by 1.8: Its formal derivativeis of degree 0, thus has no zero. The hypotheses on K imply that xk−1 = 1K for allx ∈ K, by 3.2.2. Therefore xk = x for all x ∈ K, hence K = ZK(t
k − t), and tk − t is apolynomial over the prime field of K.
By 4.1.1, k = |K| implies that k is a power of p. Of course, it is a typically finitephenomenon that a field consists of the zeros of some polynomial. Returning to thegeneral case, we observe the following criterion of separability:
4.5.7. A nonzero polynomial f ∈ K[t] is separable if and only if g′ 6= 0K for everyirreducible polynomial divisor g of f in K[t].
Proof. Applying 4.3(1), we may consider a splitting field M of f over K. Clearly, Mcontains a splitting field over K for every polynomial divisor of f in K[t]. If g is sucha divisor, irreducible and g′ = 0K , then every zero of g in M is a multiple zero by 1.8.Hence f is not separable. Conversely, if f is not separable, there exists an irreduciblepolynomial divisor g of f in K[t], w. l. o. g. normed, which has a multiple zero b ∈ M .By 4.1.4, g = minb,K . But 1.8 implies that g′(b) = 0K which forces g′ = 0K .
If charK = 0, then g′ 6= 0K for every g ∈ K[t]rK. Hence 4.5.7 implies
4.5.8. If charK = 0, every nonzero polynomial over K is separable. �
4.6 Proposition. Suppose the hypotheses of 4.3(2) and let f be separable, n := dimKM .There there exist exactly n extensions of ϕ to an isomorphism of M onto X.
Proof by induction on n. The case n = 1 is trivial. For the inductive step, let n > 1 andassume the claim to be true for smaller values of n. Let b ∈ ZM(f) rK, g := minb,K ,g := gϕ, b ∈ ZX(g) and ϕb according to 4.2(2).
70For example, f = t4 + 2t2 + 1 = (t2 + 1)2 ∈ Q[t] is separable but has multiple zeros in C. Note,however, that there are authors who define a polynomial to be separable if it has no multiple zerosin any extension field, thus excluding cases like our example f . Clearly, for irreducible polynomialsthe two definitions coincide. An algebraic element b ∈ M is called separable over K if minb,K isseparable. From 4.1.4 it is clear that this definition is independent of the choice of the definition ofseparability for arbitrary polynomials.
72
K K
XM
K[b] K[b]
✲
✲
✲
ϕ
ϕb
Then M is a splitting field of f over K[b], X is a splitting fieldof f over K[b], and dimK[b]M < n. Clearly, deg g = deg g, and
f (f resp.) is separable over K[b] (K[b] resp.). By our inductivehypothesis, there exist exactly dimK[b]M extensions of ϕb to anisomorphism of M onto X . Hence, by 4.3.1 and 4.2.2,
|ϕM | = |ZX(g)| dimK[b]M = deg g dimK[b]M = dimK K[b] dimK[b]M = dimKM.
where the last two equations make use of 4.1.5 and the special case of 2.6. �
4.7 Theorem. Let (K,M) be a field extension. The following are equivalent:
(i) (K,M) is Galois,
(ii) There exists a nonzero separable polynomial over K with splitting field M ,
(iii) |AutKM | = dimKM <∞.
Proof. (ii)⇒(iii) follows from the special case of X = M , ϕ = idX in 4.6, and (iii)⇒(i)is the contents of 4.5.4. Suppose (i). Let B be a K-basis of M , f :=
∏b∈Bminb,K .
By 4.1.4 and our hypothesis, f is separable over K. Furthermore, f splits over M intolinear factors as B ⊆ M and (K,M) is normal. As ZM(f) ⊇ B, K ∪ ZM(f) cannot becontained in any proper subfield of M . Hence M is a splitting field of f over K. Theproof is complete.
Example. Let K be a field, n ∈ N. Then(Ksym(t1, . . . , tn), K(t1, . . . , tn)
)is a Galois
extension, and its Galois group G is isomorphic to Sn: Viete’s theorem 1.11 showsthat K(t1, . . . , tn) is a splitting field of the separable polynomial
∑nk=0(−1)ksktn−k ∈
Ksym(t1, . . . , tn)[t]. By 4.7, this proves the first assertion. By 2.11.2, Sn is isomorphicto a subgroup of G. Vice versa, by 4.5.2, G is isomorphic to a subgroup of Sn. HenceG ∼= Sn.The definition of the Galois group of a field extension is a special case of the followinggeneral notion:
4.8 Definition. Let (M, ◦) be a magma, G a unital submonoid of End (M, ◦)71. For ally ∈M , the set
CG(y) := {α|α ∈ G, yα = y}
is called the centralizer of y in G. For all α ∈ G,
FixM(α) := {y|y ∈M, yα = y}
is called the set of fixed elements of α. We observe
4.8.1. Let y ∈ M , α ∈ G. Then CG(y) is a unital submonoid of G, FixM(α) is asubmagma of (M, ◦). �
71Recall 1.3.1 and its comments.
73
For every U ⊆ M let CG(U) :=⋂y∈U CG(y), for every H ⊆ G similarly FixM(H) :=⋂
α∈H FixM(α). It follows from 4.8.1 (and 2.2.2) that CG(U) is a unital submonoid ofG, FixM(H) a submagma of (M, ◦). The following remark is also immediate:
4.8.2. Let U, U ⊆M , H, H ⊆ G. Then
U ⊆ U ⇒ CG(U) ⊆ CG(U), H ⊆ H ⇒ FixM(H) ⊆ FixM(H).
�
Let T (G) be the set of all unital submonoids of G, V(M) the set of all submagmasof (M, ◦). Then (T (G),⊆), (V(M),⊆) are lattices, and the foregoing remarks may bereformulated as follows:
4.8.3. The mappings
ζ : V(M)→ T (G), U 7→ CG(U), ϕ : T (G)→ V(M), H 7→ FixM(H)
are lattice anti-homomorphisms. �
4.8.4. (1) ∀U ∈ V(M) Uζϕ ⊇ U, Uζϕζ = Uζ,
(2) ∀H ∈ T (G) Hϕζ ⊇ H, Hϕζϕ = Hϕ,
(3) Mζ = {idM}, {idM}ϕ =M .
G
{idM}M
U
FixM(CG(U)
)
CG(U)✶
✐
✲
ϕ
ζ
ζ
G
{idM}M ϕ
FixM (H)
CG
(FixM (H)
)
Hϕ
ζ q
✾
✛
Proof. (1) Let U ∈ V(M). Every element of U is fixed by every element of Uζ , henceU ⊆ Uζϕ, Uζ ⊆ Uζϕζ . Let α ∈ Uζϕζ . Then α centralizes Uζϕ, a fortiori U . Henceα ∈ CG(U) = Uζ .
(2) Let H ∈ T (G). Every element of Hϕ is centralized by every element of H , henceH ⊆ Hϕζ , Hϕ ⊆ Hϕζϕ. Let y ∈ Hϕζϕ. Then y is a fixed element of Hϕζ , a fortioriof H . Hence y ∈ FixM(H) = Hϕ.
(3) is trivial.
Now let M be a field, K a subfield of M and G := AutKM . For every subgroup H ofAutM , FixM(H) is a subfield ofM : By 4.8.1 it is additively and multiplicatively closed.By the homomorphism rule for inverses (1.3(3)), it is also closed with respect to takinginverses. Clearly, it contains 0M and 1M . The prime field ofM must hence be containedin FixM(H), in other words:
4.8.5. If K is the prime field of a field M , then cα = c for all c ∈ K, α ∈ AutM . �
If H ≤ G, clearly K ≤ FixM(H), i. e., FixM(H) is an element of the interval [K,M ] ofthe lattice of all subfields of M . As in 3.7, we write U(G) for the subgroup lattice ofG. For all L ∈ [K,M ], CG(L) is a subgroup of G; it is just a different notation for the
74
Galois group AutLM of the extension (L,M). Restricting our above ζ to [K,M ], ϕ toU(G), we obtain the following lattice anti-homomorphisms:
Γ : [K,M ]→ U(G), L 7→ CG(L)(= AutLM),
Φ : U(G)→ [K,M ], H 7→ FixM(H).
4.9 Proposition. Let (K,M) be a Galois extension. Then (AutKM)Φ = K.
In other words: The fixed field of the Galois group of a Galois extension equals the groundfield of the extension: It is not larger!
Proof. Clearly, KΓΦ ⊇ K (4.8.4(1)). To prove the reverse inclusion, put G := KΓ,L := GΦ and observe that K ≤ L, AutLM = LΓ = KΓΦΓ = KΓ, by 4.8.4(1). It followsthat dimK L dimLM = dimKM = |G| = |AutLM | ≤ dimLM , where we applied firstthe special case of 2.6, then 4.7, and finally 4.5.3. Hence dimK L = 1, i. e., K = L.
4.10 Corollary. Let (K,M) be a Galois extension. Then ΓΦ = id[K,M ]. In particular,Γ is injective, Φ is surjective, [K,M ] is finite.
Proof. Let L ∈ [K,M ]. By 4.5.5(3), (L,M) is Galois. Hence, by 4.9, LΓΦ = L. ThusΓΦ = id[K,M ]. From 4.5.3 we know that, in particular, the number of subgroups ofAutKM is finite. As Φ is surjective it follows that [K,M ] is finite.
On our way to the main theorem of Galois theory, 4.10 may be viewed as the goal ofthe first part. The second part will consist in the proof of the analogous result for thereverse composition ΦΓ. Before we embark on this aim, let us consider an illustratingexample:
Example. Let M be the splitting field of t3 − 2 over Q in C, w a primitive 3rd rootof unity. Then M = Q[ 3
√2, w], by 1.5. As M 6⊆ R, we have dimQQ[ 3
√2] dimQ[ 3
√2]M =
3 ·2 = 6. From 4.1.5 it follows that min 3√2,Q = t3−2, minw,Q = t2+t+1 (see Example (4)
on p. 10). As M is generated as a field by Q, 3√2 and w, 4.2.1 implies that at most the
six assignments listed in the first two lines of the following table can have an extensionto an automorphism of M :
3√2 7→ 3
√2 3√2w 3
√2w−1 3
√2 3√2w 3
√2w−1
w 7→ w w w w−1 w−1 w−1
α1 α2 α3 α4 α5 α6
By 4.5.8 and 4.7, M is a Galois extension of Q. By 4.8.5 and again by 4.7, |AutM | =|AutQM | = 6. More precisely, 4.5.2 implies that AutM ∼= S3. Hence all six assignmentsgiven by the table extend to an automorphism of M . We write αi (i ∈ 6) for theautomorphism which extends the i-th assignment in our table (3rd line).
There exist at least the following subfields of M :
M, Q[3√2], Q[
3√2w], Q[
3√2w−1], Q[w], Q.
75
In the order given, these have the dimensions 6, 3, 3, 3, 2, 1 over Q. The three subfieldsof dimension 3 are mutually distinct (as the union of any two of them generates M asa field). By 4.10, Γ is injective, hence |[Q,M ]| ≤ |U(S3)| = 6. It follows that the abovelist contains every subfield of M . We conclude that Γ, Φ are lattice anti-isomorphisms:
Q
Q[ 3√2w] Q[ 3
√2w−1] Q[ 3
√2]
Q[w]
M
2
3 3 3
3 2 2 2
AutM
〈α6〉 〈α5〉 〈α4〉〈α2〉
= 〈α3〉
{idM}
2
3 3 3
3 2 2 2
Γ−→Φ←−
(The small numbers are the relative extension degrees (subgroup indices, resp.). The subfields
which are normal extensions of Q (the normal subgroups, resp.) have been marked by bigger
nodes than the others.)
Our aim is to show that the exact correspondence observed in this example holds for everyGalois extension. As a preparation, we resume and slightly generalize the terminologyintroduced in 4.8: Let (M, ◦) be a magma, H ⊆ End (M, ◦), k ∈ N. We put
FixMk(H) := {(y1, . . . , yk)|yi ∈ M, ∀α ∈ H yiα = yi}.
Clearly, FixMk(H) =(FixM(H)
)k. A subset T of Mk is called H-invariant if for all
y1, . . . , yk ∈M , α ∈ H ,
(y1, . . . , yk) ∈ T ⇒ (y1α, . . . , ykα) ∈ T,
which is denoted in short by Tα ⊆ T . A trivial example of an H-invariant subset isFixMk(H).
4.11 Proposition. Let M be a field, H ≤ AutM , k ∈ N.
(1) Let V be an H-invariant subspace of Mk, V 6= {(0M , . . . , 0M)}. Then
V ∩ FixMk(H) 6= {(0M , . . . , 0M)}.
(2) Let H be finite, α1, . . . , αm the elements of H (mutually distinct). Let b1, . . . , bk ∈ M .Then the kernel of the M-linear mapping defined by the matrix multiplication
µ :Mk → Mm, (x1, . . . , xk) 7→ (x1, . . . , xk) ·(biαj
)i∈kj∈m
is H-invariant.
Consequently, ker µ ∩ FixMk(H) 6= {(0M , . . . , 0M)} if k > m.
76
Proof. (1) Let x1, . . . , xk ∈ M , not all = 0M , such that (x1, . . . , xk) ∈ V , chosen withthe property that the number n of subscripts j ∈ k with xj 6= 0M is minimal. W. l. o. g.assume x1, . . . , xn 6= 0M , xn+1 = · · · = xk = 0M . As V is an M-subspace of Mk weobtain
(x−1n x1, . . . , x
−1n xn−1, 1M , 0M , . . . , 0M) = x−1
n (x1, . . . xk) ∈ V.Hence there exist y1, . . . , yn−1 ∈ M such that (y1, . . . , yn−1, 1M , 0M , . . . , 0M) ∈ V . Forall α ∈ H we have 1Mα = 1M , 0Mα = 0M , V α ⊆ V . Hence
(y1α−y1, . . . , yn−1α− yn−1, 0M , . . . , 0M)
=(y1α, . . . , yn−1α, 1M , 0M , . . . , 0M)− (y1, . . . , yn−1, 1M , 0M , . . . , 0M) ∈ V.
By the choice of n this implies that yjα−yj = 0M , hence (y1, . . . , yn−1, 1M , 0M , . . . , 0M) ∈V ∩ FixMk(H).
(2) Let x1, . . . , xk ∈ M such that (x1, . . . , xk) ∈ ker µ, α ∈ H , αj′ := αjα for all j ∈ m.Let j ∈ m. Then
∑i∈k xi(biαj) = 0M , hence
∑i∈k(xiα)(biαj′) = 0Mα = 0M . It follows
that (x1α, . . . , xkα) ∈ ker µ.
If k > m, then kerµ 6= {(0M , . . . , 0M)} by the dimension formula for linear mappings.Applying (1) to V := kerµ we obtain the claim.
4.12 Proposition (Artin). LetM be a field, H a finite subgroup of AutM , L := FixMH.Then
(1) dimLM = |H|,
(2) H = AutLM ,
(3) (L,M) is Galois.
Proof. (1) Our main step will be to show that dimLM ≤ |H|: Let αi, µ be as in 4.11(2),k ∈ N and b1, . . . , bk ∈ M such that (b1, . . . , bk) is L-linearly independent. We want toapply the last part of 4.11 to derive the desired inequality k ≤ |H|. All we have to do isto prove that kerµ ∩ Lk = {(0M , . . . , 0M)}, because Lk = FixMk(H).
Let (x1, . . . , xk) ∈ ker µ ∩ Lk. Assuming w. l. o. g. that α1 = idM , we consider the firstcomponent of (x1, . . . , xk)µ and obtain
0M = (x1, . . . , xk)
b1...bk
= x1b1 + · · ·xkbk,
hence x1 = · · · = xk = 0M as (b1, . . . , bk) is L-linearly independent.
Thus dimLM ≤ |H| ≤ |AutLM |. By 4.5.3, this proves (1) and, moreover, (2) asH ≤ AutLM . Furthermore, it now suffices to apply 4.7 to obtain (3).
4.13 Corollary. Let (K,M) be a Galois extension. Then ΦΓ = idU(G) (where G =AutKM). In particular, Γ is injective, Φ is surjective.
77
Proof. By 4.5.3, G is finite. Let H ≤ G, L := FixM(H). Then HΦΓ =(FixM(H)
)Γ =
LΓ = AutLM = H , by 4.12(2).
As a consequence, we obtain the following further characterization of Galois extensions:
4.14 Corollary. Let (K,M) be a field extension. The following are equivalent:
(i) (K,M) is Galois,
(ii) There exists a finite subgroup G of AutM such that FixMG = K.
Proof. If (i) holds, let G := AutKM . Then FixMG = GΦ = KΓΦ = K by 4.9. If (ii)holds, we apply 4.12(3) to obtain (i).
4.15 Proposition. Let (K,M) be a Galois extension, L ∈ [K,M ], G := AutKM . Then(K,L) is normal if and only if L is G-invariant (i. e., Lα ⊆ L for all α ∈ G)72.
Proof. For all b ∈ M , α ∈ G we have minb,K = minbα,K , by 4.2.1 (with ϕ := idK ,X := M , ψ := α|K[b]). Hence, if (K,L) is normal, b ∈ L, then bα ∈ L. Conversely, letL be G-invariant, b ∈ L. As (K,M) is normal, [K,M ] contains a splitting field Y ofminb,K over K. Let b ∈ ZY (minb,K). By 4.2(2), there exists an isomorphism ψ of K[b]onto K[b] which maps b onto b and c onto c for all c ∈ K. By 4.7 and 4.3(2), ψ extendsto an automorphism α of M . Clearly α ∈ G, hence b = bα ∈ Lα ⊆ L.
4.16 Theorem (Main theorem of Galois theory). Let (K,M) be a Galois extension andG := AutKM . Let
Γ : [K,M ] → U(G), L 7→ AutLM,Φ : U(G) → [K,M ], H 7→ FixM(H).
(1) Γ,Φ are mutually inverse lattice anti-isomorphisms. If L, L∗ ∈ [K,M ] such thatL ⊆ L∗, then dimL L
∗ = |LΓ : L∗Γ|.
(2) Let L ∈ [K,M ]. Then (K,L) is normal73 if and only if LΓEG.
(3) Let L ∈ [K,M ], (K,L) normal. Then AutKL ∼= G/LΓ.
G
{idM}M
HΦ = FixMH = L H = AutLM = LΓ
K
L∗ L∗Γ
✛✲
Γ
Φ
72Equivalently, Lα = L for all α ∈ G. For the non-obvious implication in this equivalence consider anyα ∈ G. Thanks to the universal quantifier in the apparently weaker condition we may conclude thatLα−1 ⊆ L and therefore also L ⊆ Lα, hence L = Lα as both inclusions hold.
73As (K,L) is clearly separable and finite, this means that (G,L) is Galois.
78
Proof. (1) By 4.10 and 4.13, ΓΦ = id[K,M ], ΦΓ = idU(G). By 4.8.2, this proves the firstassertion. Let L, L∗ ∈ [K,M ] such that L ⊆ L∗. We apply the special case of 2.6, 4.7and 3.1.2 to conclude
dimL L∗ =
dimLM
dimL∗ M=|LΓ||L∗Γ| = |LΓ : L∗Γ|.
(2) By 4.15 (including footnote 72) and (1), we obtain
(K,L) normal ⇔ ∀α ∈ G Lα = L⇔ ∀α ∈ G (Lα)Γ = LΓ
⇔ ∀α ∈ G (AutLM)α = AutLM ⇔ AutLM EG,
where we make use of 4.5.1 for the 3rd equivalence.
(3) Let L ∈ [K,M ], (K,L) normal. By 4.15, α|L ∈ AutKL for all α ∈ G. Let
ρ : G→ AutKL, α 7→ α|L .
Then ρ is a homomorphism, ker ρ = {α|α ∈ G, α|L = idL} = AutLM = LΓ. Everyβ ∈ AutKL is, in particular, a K-monomorphism of L into M , hence extends to anautomorphism α ∈ AutKM , by 4.7 and 4.3(2). As β = α|L ∈ Gρ, ρ is surjective. By3.2(2), G/LΓ = G/ ker ρ ∼= Gρ = AutKL.
We will see in part 2 of this course that 4.16 has crucial consequences for the problem ofsolubility of polynomial equations by radicals. We finish this chapter with applicationsof 4.16 in a different direction, deriving important structural properties of finite fields.
4.17 Definition. Let p be a prime,M a commutative unitary ring such that charM = p.We set
ϕp :M →M, x 7→ xp.
4.17.1. ϕp is a ring endomorphism.
Proof. Let x, y ∈M . Then (xy)ϕp = (xy)p = xpyp = (xϕp)(yϕp), and
(x+ y)ϕp = (x+ y)p =
p∑
j=0
(p
j
)xp−jyj = xp + yp = xϕp + yϕp
as p|(pj
)for all j ∈ p− 1.
The mapping ϕp is called the Frobenius endomorphism of M .
4.17.2. If M is a field 74, ϕp is a monomorphism. For a finite field M , ϕp is anautomorphism.
74As the proof shows, it is sufficient that there is no nonzero element x ∈ M such that xp = 0M . (Inan arbitrary ring R, an element x is called nilpotent if there exists a positive integer n such thatxn = 0R.)
79
Proof. With respect to addition, let x ∈ kerϕp. Then 0M = xϕp = xp, hence x = 0M .It follows that ϕp is injective. If M is finite, this implies that ϕp is surjective.
If M is a finite field, ϕp is called the Frobenius automorphism of M .
4.18 Theorem (1st main theorem on finite fields). Let p be a prime, n ∈ N.
(1) There exists a field of order pn.
(2) Any two fields of order pn are isomorphic.
Proof. Recall footnote 34: (2) Let M be a field of order pn. By 4.5.6, M is a splittingfield of the polynomial tp
n − t over the prime field of M . By 3.13.5, the prime fieldsof any two fields of characteristic p are isomorphic. Applying 4.3(2) to the polynomialf = tp
n − t we obtain the claim.
(1) Making use of 4.3(1), we consider a splitting field X of f := tpn − t over a field
K of order p. By 4.5.6, f is separable over K, hence |ZX(f)| = pn. We show thatZX(f) is a subfield of X : Clearly, 0X , 1X ∈ ZX(f). Let a, b ∈ ZX(f). By 4.17.1,(a− b)pn − (a− b) = (a− b)ϕnp − (a− b) = aϕnp − bϕn − a+ b = 0M , i. e., a− b ∈ ZX(f).If b 6= 0M , (ab−1)p
n
= apn
(bpn
)−1 = ab−1, i. e., ab−1 ∈ ZX(f). By the definition of asplitting field, it follows that X = ZX(f). Hence X is a field of order pn.
The proof shows that a finite fieldM (the order of which is, by 4.1.1, necessarily a powerof a prime) is a splitting field of a separable polynomial over its prime field K, hencea Galois extension field of K, by 4.7. We give a second line of reasoning for this fact,considering the Frobenius automorphism ϕp of M .
4.18.1. Let p be a prime, n ∈ N. With respect to a field M of order pn, o(ϕp) = n.
Proof. By 3.5 there exists an element x ∈ M such that, multiplicatively, M = 〈x〉. Letk ∈ N. then
ϕkp = idM ⇔ xϕkp = x⇔ xpk
= x⇔ xpk−1 = 1M ⇔ o(x)|pk − 1,
(cf. Example (3) on p. 50).) Hence o(ϕp) = min{k | pn − 1|pk − 1} = n.
We have AutM = AutKM and, by 3.2.2 and 4.5.3,
dimKM = n = o(ϕp)∣∣ |AutM | ≤ dimKM,
hence |ϕp〉| = o(ϕp) = |AutM |. It follows that 〈ϕp〉 = AutM . As dimKM = n =|AutKM |, 4.7 implies that (K,M) is Galois. We derived this fact before, but now wehave even determined the Galois group of (K,M): AutM = 〈ϕp〉 is cyclic. Now 3.4implies (see 3.7) that
µ : U(AutM)→ N, H 7→ |H|,is an isomorphism of (U(AutM),≤) onto (Tn, |).
80
4.19 Theorem (2nd main theorem on finite fields). Let M be a finite field, K its primefield, p := |K|. Then there exists a positive integer n such that |M | = pn, AutM = 〈ϕp〉,and the mapping
δ : [K,M ]→ N, L 7→ dimK L,
is a lattice isomorphism of ([K,M ],≤) onto (Tn, |).
Proof. The first two assertions have already been obtained by our foregoing remarks. Itremains to prove the claim about δ. Let ı : Tn → Tn, k 7→ n
k(see p. 54). Then Γµ ı is
a composition of two anti-isomorphisms and an isomorphism of lattices, hence a latticeisomorphism of ([K,M ],≤) onto (Tn, |).
K
M
L
✲
✲
✲
Γ
AutM
{idM}
AutLM
✲
✲
✲
µ
n
1
|AutLM |
✲
✲
✲
ı
1
n
n|AutLM|
For all L ∈ [K,M ] we have, by 4.16(1), LΓµı =n
|AutLM |=
dimKM
dimLM= dimK L, thus
Γµ ı = δ.
81
A. Appendix
On irreducibility of polynomials over Q
A polynomial f =∑n
j=0 cjtj ∈ Z[t] is called primitive if gcd(c0, . . . , cn) = 1. Clearly,
this conditions holds if one of the coefficients ci equals 1 or −1. In particular, normedpolynomials in Z[t] are primitive. A further trivial remark is the following:
A.0.1. Let f ∈ Z[t]r Z be irreducible in Z[t]. Then f is primitive,
as a common divisor d 6= 1,−1 of c0, . . . , cn allows the decomposition f = d ·∑nj=0
cjdtj
into non-units of Z[t]. �
The set of primitive polynomials is a unital submonoid of (Z[t], ·). This is the contentsof the following, known as Gauss’s lemma:
A.0.2. Let g, h ∈ Z[t] be primitive. Then gh is primitive.
Proof. Let k, l ∈ N0, ai, bj ∈ Z for all i, j ∈ N0 such that g =∑k
i=0 aiti, h =
∑lj=0 bjt
j ,
ak, bl 6= 0 where we put ai := 0 =: bj for i > k, j > l. Then gh =∑k+l
m=0
(∑mi=0 aibm−i
)tm.
Assume there is a prime p such that p|∑mi=0 aibm−i for all m. As g, h are primitive, there
exists a minimal r ≤ k and a minimal s ≤ l such that p ∤ ar, p ∤ bs. We have
p|r+s∑
i=0
aibr+s−i = a0br+s + · · ·+ ar−1bs+1 + arbs + ar+1bs−1 + · · ·+ ar+sb0.
Each of the first r and the last s summands in this sum is divisible by p as p dividesa0, . . . , ar−1, b0, . . . , bs−1. Hence p|arbs which implies, as p is a prime, that p|ar or p|bs(Recall that p has the property stated for q in (∗) on p. 29.) This contradiction showsthat gh is primitive.
A.0.3. Let g ∈ Q[t], g 6= 0, m := min{n|n ∈ N, ng ∈ Z[t]}. Then g = zmg for some
z ∈ Z and a primitive polynomial g ∈ Z[t]. If g is normed, z = 1 may be chosen.
Proof. Set g∗ := mg. Let z be the greatest common divisor of the coefficients of g∗ andg ∈ Z[t] such that g∗ = zg. Then g is primitive and g = z
mg. Now suppose that g is
normed and let d ∈ N be a common divisor of all coefficients of g∗. Then d|m as g isnormed, and m
dg = 1
dg∗ ∈ Z[t]. Hence d = 1 by the minimality of m.
A.0.4. Let h ∈ Z[t] be primitive and g ∈ Q[t]. If gh ∈ Z[t], then g ∈ Z[t].
83
Proof. According to A.0.3, let z ∈ Z, m ∈ N, g ∈ Z[t] be primitive such that gcd(z,m) =1, g = z
mg. As gh ∈ Z[t], a prime divisor of m would have to divide each coefficient
of the polynomial ugh, hence of gh by (∗) on p. 29. But by A.0.2, gh is primitive. Itfollows that m has no prime divisor, i. e., v = 1 as v ∈ N. Thus g = zg ∈ Z[t].
A.0.5. Let g, h ∈ Q[t], h normed. If gh ∈ Z[t], then g ∈ Z[t].
Proof. Choosing m as in A.0.3 (with h in place of g), we obtain that mh is primitiveand g(mh) = mgh ∈ Z[t]. By A.0.4, g ∈ Z[t].
Consequently, a product of two normed polynomials over Q has integer coefficients ifand only if both factors have integer coefficients.
A.0.6. Let b ∈ C be a zero of a normed polynomial f ∈ Z[t].75 Then minb,Q ∈ Z[t].
Proof. By 4.1.3, f = minb,Q h for some h ∈ Q[t]. As f is normed, also h is normed. Theclaim follows from A.0.5.
A.0.7. Let f ∈ Z[t]r Z be irreducible in Z[t]. Then f is irreducible in Q[t].
Proof. Let f ∈ Z[t], g, h ∈ Q[t] such that f = gh, deg g, deg h > 0. According to A.0.3,let q ∈ Q such that h = qh, for some primitive polynomial h ∈ Z[t]. Then qg ∈ Q[t] and(qg)h = f ∈ Z[t]. By A.0.4, it follows that qg ∈ Z[t]. As deg qg = deg g, deg h = deg h,f is not irreducible in Z[t].
A.1 Theorem (Eisenstein 1850). 76 Let n ∈ N, f =∑n
j=0 cjtj ∈ Z[t]. Suppose that
there exists a prime p such that p|c0, . . . , cn−1, p ∤ cn, p2 ∤ c0. Then f is irreducible in
Q[t].
Proof. We may factor out the greatest common divisor d of the coefficients cj and,writing f = df , observe that the primitive polynomial f satisfies the same hypothesesas f . As d is a unit in Q[t], it suffices to prove the assertion for f in place of f . In otherwords, we assume in the sequel, w. l. o. g., that f is primitive. By A.0.7, it then sufficesto show that f is irreducible in Z[t]:
Let g, h ∈ Z[t] such that f = gh. Let k ∈ N0, a0, . . . , ak, b0, . . . , bn−k ∈ Z such thatg =
∑ki=0 ait
i, h =∑n−k
j=0 bjtj, ak 6= 0 6= bn−k. We have p|c0 = a0b0. As p is a prime, it
follows that p|a0 or p|b0 (by (∗) on p. 29). W. l. o. g. assume that p|a0. Then p ∤ b0 asp2 ∤ c0. Since f is primitive, p ∤ cn = akbn−k, hence p ∤ ak.
Let m ∈ k be minimal with the property that p ∤ am. Then p|am−j for all j ∈ m. Theassumption k < n would imply m < n so that, by hypothesis, p|cm =
∑mj=0 am−jbj ,
hence p|(cm −
∑j∈m am−jbj
)= amb0. But p ∤ am, b0, a contradiction as p is a prime. It
follows that k = n, hence h ∈ Z. This implies that h|c0, . . . , cn so that h ∈ {1,−1} as fis primitive, i. e., h is a unit of Z[t].
75Numbers b with this property are called algebraic integers. They are the classical object of study inAlgebraic Number Theory.
76Frequently, a result is erroneously not called after its discoverer. Eisenstein’s theorem is no exception:In substance, it was anticipated by Schonemann in 1846.
84
By A.1, every polynomial tn − p where n ∈ N, p a prime, is irreducible in Q[t]; moregenerally, every polynomial tn − pk where k ∈ Z and p ∤ k. The polynomials of degree 5which lead to the equations mentioned on p. 22 are of “Eisenstein type”. While it is atrivial fun to produce countless examples of irreducible polynomials in Q[t] by choosingsuitable coefficients in accordance with Eisenstein’s theorem, there are other applicationsof A.1 which are less obvious. As an example, we consider the p-th cyclotomic polynomial
φp = tp−1 + · · ·+ t+ 1 =tp − 1
t− 1
(where p is a prime). In Example (4) on p. 10, we considered the case p = 3. As p isa prime, the zeros of φp in C are exactly the primitive p-roots of unity, i. e., the p-throots of unity 6= 1. At first glance surprisingly, we may apply A.1 to prove that φp isirreducible in Q[t]:
To this end we note first that the substitution endomorphism Ft+1 of Q[t] is an auto-morphism: Clearly, f and fFt+1 have the same degree, for any f ∈ Q[t] r {0}. HencekerFt+1 is trivial and Ft−1 is injective. Furthermore, t = (t − 1)Ft+1 so that Q[t]Ft+1
contains t. It follows that Ft+1 is surjective. A polynomial f ∈ Q[t] is irreducible if andonly if its image with respect to any automorphism of Q[t] is irreducible. In particu-lar, the claim that φp is irreducible will follow if we succeed in showing that φpFt+1 isirreducible. As we have
φpFt+1 =(t+ 1)p − 1
(t+ 1)− 1=
∑pj=0
(pj
)tj − 1
t=
∑
j∈p
(p
j
)tj−1 ,
we see that φpFt+1 satisfies the hypotheses of A.1, hence is indeed irreducible.
Recall that for arbitrary n ∈ N, the n-th cyclotomic polynomial is given by
φn :=∏
w∈Pn
(t− w)
(cf. p. 24). The field Qn := Q[w] where w ∈ Pn is called the n-th cyclotomic field. From1.4.4 we conclude
tn − 1 =∏
d|nφd for all n ∈ N
as both polynomials are normed of degree n and have the same n zeros in C: the n-throots of unity.
A.1.1. φn ∈ Z[t], for every n ∈ N.
Proof by induction on n. The claim is trivial for n = 1. Let n > 1 and assume thatφd ∈ Z[t] for all d < n. In C[t] we have the equation
tn − 1 = φnh where h :=∏
n 6=d|nφd.
85
Thanks to our induction hypothesis, h ∈ Z[t] and φn = tn−1h∈ Q(t) ∩ C[t] = Q[t].77 As
h is obviously normed, we conclude that g ∈ Z[t] by A.0.4. �
Finally we show that the n-th cyclotomic polynomial is irreducible in Q[t], for everyn ∈ N. If n is a prime, Eisenstein’s theorem A.1 provides a very simple proof as wehave seen above. For arbitrary n, the proof is more complicated. It is a nice examplefor a line of reasoning which makes use of rings of prime characteristic to obtain anassertion for characteristic 0. We will make use of the standard notation introduced onp. 67 in the more general framework of a commutative unitary ring K and a unital ringepimorphism ϕ of K onto a unitary ring K. Obviously ϕ extends to a ring epimorphismϕ of K[t] onto the polynomial ring K[t] with the property tϕ = t (cf. 2.5.1).
A.2 Lemma. Let p be a prime, g, h ∈ Z[t] normed and irreducible, b ∈ C such thatg(b) = 0 = h(bp). Put Zp := Z/pZ and consider the standard extension Z[t] → Zp[t],f 7→ f , of the canonical epimorphism of Z onto Zp. Suppose that g, g′ have no commonzeros. Then g = h.
Proof. As h(bp) = 0, b is a zero of the polynomial h(tp). It follows that h(tp) = fg forsome f ∈ Z[t], by 4.1.3, 4.1.4 and A.0.5. Applying 4.17.1, we obtain
hp = h(tp) = f g.
Let X be a splitting field of h over Fp (4.3(1)). Then ZX(g) ⊆ ZX(h). Our hypothesison g implies, by 1.8, that |ZX(g)| = deg g. As bp ∈ Q[b], we conclude, by 4.1.5 and 4.1.4,that
deg g = |ZX(g)| ≤ |ZX(h)| ≤ deg h = deg h = dimQQ[bp] ≤ dimQ Q[b] = deg g = deg g.
It follows that deg g = deg h, ZX(g) = ZX(h), hence g = h as both polynomials arenormed.
A.3 Theorem. φn is irreducible in Q[t], for every n ∈ N.
Proof. Let n ∈ N, w ∈ Pn, g := minw,Q. For every prime p such that p ∤ n we show:g(wp) = 0.
To this end put h := minwp,Q. As p ∤ n we have bp ∈ Pn, hence g, h ∈ Z[t] and g, h |Z[t]
φn
77There is a much more general reason for this statement, independent of any induction and a typicalapplication of Galois theory:
Proposition. Let (K,M) be a Galois extension, G its Galois group. Let Q be a finite G-invariantsubset of M . Then
∏b∈Q(t− b) ∈ K[t].
Proof. Let n := |Q| and b1, . . . , bn be the elements of Q. Then sk(b1, . . . , bn) = sk(b1α, . . . , bnα) =sk(b1, . . . , bn)α for all k ∈ n ∪ {0}, α ∈ G. By 1.11, the coefficients of the polynomial
∏b∈Q(t − b)
are contained in FixM (G), hence in K, by 4.9.
The step in the proof of A.1.1 deals with the special case where K = Q, M is a splitting field of tn−1over Q, Q = Pn.
86
by A.0.6, 4.1.3. We have to show that g = h and make the assumption that g 6= h.Then gh |
Z[t]
φn, hence
gh |Zp[t]
φn |Zp[t]
tn − 1Zp.
The polynomials tn − 1Zpand (tn − 1Zp
)′ = ntn−1 have no common zero, hence tn − 1Zp
has no multiple zeros in any extension field of Zp, by 1.8. The same must hold for its
divisor g so that we may apply A.2 and conclude that g = h. It follows that g2 |Zp[t]
φn,
a contradiction because tn − 1Zphas no multiple zeros in any extension field of Zp. It
follows that g = h, thus g(wp) = 0.
An easy induction now shows that g(wk) = 0 for every k ∈ N such that gcd(k, n) = 1:The claim is trivial for k = 1. For the inductive step, let k > 1. Let p be the smallestdivisor > 1 of k. Then p is a prime for which we have shown above that g(wp) = 0.Now wp ∈ Pn, g = minwp,Q by 4.1.4, and k
pis prime to n and < k. Hence, applying the
inductive hypothesis to the primitive root of unity wp, we conclude that 0 = g((wp)kp ) =
g(wk).
By 1.4.2 and its converse, the set of all wk where gcd(k, n) = 1 equals Pn. HenceZC(g) = Pn so that g = φn. The claim follows.
Consequently, dimQ Qn = ϕ(n) for every n ∈ N. Now it is easy to determine the Galoisgroup AutQn. Proceeding as in the Example on p. 75, we know that every automorphismof Qn is uniquely determined by the image of w where w is a fixed element of Pn. Theonly candidates for this image are the elements of Pn, and these are the powers wk withgcd(k,m) = 1. It suffices to consider k ∈ n which shows that every such assignmentw 7→ wk extends to an automorphism αk of Qn as their number equals dimQ Qn, henceϕ(n). Furthermore, the simple rule (wk)l = wkl shows that the mapping
ϕ : {k|k ∈ Z, gcd(k, n) = 1} → AutQn, k 7→ αk
is a multiplicative epimorphism. For all k with gcd(k, n) = 1, the anti-image αkϕ−
equals the coset nZ+k. The set of these cosets is the unit group of the factor ring Z/nZand isomorphic (via ϕ−) to AutQn (see footnote 56). Thus AutQn is an abelian groupof order ϕ(n).
Within the universe of all Galois extensions of Q with abelian Galois group, the cyclo-tomic fields play a dominant role because of the following famous result:
Theorem (Kronecker-Weber 1886) Let (Q,M) be a Galois extension with abelian Galoisgroup. Then there exists a positive integer n such that M is a subfield of Qn.
Theorems A.1 and A.3 deal with important classes of irreducible polynomials over Q butthey do not give an answer to the question how to decide whether a given polynomialf over Q is irreducible or not. Due to Kronecker, there is an algorithmic answer to thisquestion, thanks to A.0.7 which allows us to assume that f has integer coefficients. Let
87
f =∑n
j=0 cjtj ∈ Z[t]. If f = gh where g, h ∈ Z[t], we have deg g ≤ n
2or deg h ≤ n
2.
Thus it suffices to decide if f has a divisor g of degree k ≤ n2in Z[t]. Let k ≤ n
2and con-
sider arbitrary mutually distinct integers x1, . . . , xk+1. The assumption that g|f clearlyimplies that g(xi)|f(xi) for all i ∈ k + 1. Now each of the values f(xi) has only finitelymany divisors di ∈ Z. Therefore there exist only finitely many (k + 1)-tuples
(d1, . . . , dk+1) such that di|f(xi) for all i ∈ k + 1.
Given one of these (k + 1)-tuples, put qi :=di∏
j∈k+1j 6=i
(xi−xj) for all ∈ k + 1. Then
gd1,...,dk+1:=
∑
i∈k+1
qi(t− x1) · · · (t− xi−1)(t− xi+1) · · · (t− xk+1)
has the property that substitution of xi gives the value di for all i ∈ k + 1. It is theonly polynomial in Q[t] with this property because the difference of any two distinctsuch polynomials would be nonzero, of degree ≤ k and would have at least the k + 1zeros x1, . . . , xk+1 which is impossible by 1.6(3). Now it suffices to check if gd1,...,dk+1
hasinteger coefficients and divides f . Either this happens for a suitable choice of the divisortuple (d1, . . . , dk+1), or f is irreducible. Clearly it helps technically to assume that fis primitive and to choose the elements xi such that f(xi) has only very few integerdivisors.
88
Bibliography
[Bo] Bosch, Siegfried, Algebra, Springer 20066
[HW] Hardy, G.H., Wright, E.M., An introduction to the theory of numbers, OxfordUniversity Press 20086
[Is] Isaacs, I.M., Degrees of sums in a separable field extension,Proc.Amer.Math. Soc. 25, no. 3, (1970), 638-641
[Pi] Pierce, Richard S., Associative algebras, Springer 1982
[St] Stroth, Gernot, Algebra, de Gruyter 1998
89
Index
action of a K-space, 32algebra
K-, 7associative, 7commutative, 7homomorphism, 7simple, 57
algebraic-ally closed, 22-ally dependent, 39-ally independent, 39element, 65field extension, 65integer, 84
alternating group, 50
basisK-, 6of a commutative monoid, 28of a module, 34of an algebra, 37
canonical epimorphism, 48centralizer, 73centre of a group, 70characteristic, 60compatible
relation, 42with an operation, 58
conjugate, 70coset
left, 47right, 47
cyclic group, 51cyclotomic field, 85
Dedekind’s law, 27degree, 8division algebra, 58divisor in a commutative ring, 29
elementary symmetric function, 24Euclidean domain, 62Euler’s formula, 11Eulerian function, 16even permutation, 50exponent for a radical, 9extension field, 65
factor algebra, 59factor group, 48field extension, 65finite field extension, 65fixed elements, 73fixed point, 40free commutative monoid, 28freely generated by a subset, 28Frobenius endomorphism, 79
Galois extension, 71Galois group, 70general linear group, 50generating element, 51generating system
K-, 6of a module, 34of a monoid, 27
generator of a cyclic group, 51generator of an ideal, 62group algebra, 37group ring, 37
91
homomorphism of posets, 52
ideal, 57ideal
right, 32left, 32minimal one-sided, 33one-sided, 32
idempotent, 27indecomposable element, 29independent subset
K-linearly, 6of a module, 34of a monoid, 28
infimum, 53infinite order, 50inner automorphism, 70integral domain, 9interval in a poset, 53invariant subset, 76invariant subspace, 32irreducible element, 29isomorphism of posets, 52
kernel, 11Kronecker’s construction, 68
Lagrange’s theorem, 47lattice, 53
magma, 13maximal ideal, 58module
K-space left-, 32K-algebra left-, 32K-algebra right-, 32K-space right-, 32homomorphism, 34unital, 32
Moivre’s formula, 11monoid, 13monomial, 7monomorphism, 7multiplicative closure in a monoid, 27multiplicity of a zero, 18
nilpotent element, 79normal field extension, 71
order of a group, 47order of a group element, 50
polynomial, 7polynomial
elementary symmetric, 40function, 39minimal, 66normed, 8power-sum symmetric, 41primitive, 83pure, 9ring, 7ring over a set, 37separable, 72symmetric, 40
power series, 8power series algebra, 8prime field, 61prime ring, 60primitive root of unity, 15principal ideal, 62principal ideal domain, 62product set, 13
quotient field, 44quotient of a magma, 58
radical, 9replacement homomorphism, 39root
of a real number, 10of unity, 10
semigroup, 13semigroup algebra, 37semigroup ring, 37separable
element, 72field extension, 71
simple group, 48skew field, 17
92
space, K-, 6special linear group, 50splitting field, 22subgroup
lattice, 53maximal , 55normal, 48
submagma, 13submonoid
generated unital, 27unital, 27
subsemigroup, 13substitution, 9supremum, 53Sylow subgroup, 54symmetric group, 14symmetric mapping, 24
transcendental, 65
unit in a ring, 29unital, 13unitary, 5
Viete’s theorem, 23
zeromultiple, 18of a polynomial, 9, 40
zero divisor, 8
93