Introduction to Ring Theory

Sachi HashimotoProof School

Block 2, 2016

1 Basic Definitions and Examples

1.1 What are we talking about?

Broadly speaking, a ring is a set of objects which we can do two things with: add andmultiply. In many ways it will look like our familiar notions of addition and multiplication,but sometimes it won’t. We have to decide what properties of addition and multiplicationwe are going to require to be true, and which ones we can do without. Before we give theseproperties, let’s go through a few familiar examples.

Example 1. Our basic example of a ring will be the integers, which we will write as Z. Asa set, this is just the numbers {0, 1,−1, 2,−2, . . . } and addition and multiplication work“as usual”.

Example 2. Another key example that will come up again and again are the polynomials,which I will denote as Z[x]. Here, my objects are polynomials anx

n + an−1xn−1 + · · · +

a1x + a0, where the coefficients ai are integers. We could also take polynomials whosecoefficients are in R or Q or C, in which case we would write our ring as R[x] or Q[x] orC[x]. Polynomials come equipped with a usual notion of how to multiply them and howto add them.

These two basic examples in some sense represent all of the examples we will talk aboutin this class. Roughly speaking, we usually think of rings as either ‘functions’ or ‘numbers’.(Actually, we could generalize even further, and think of numbers as constant functions,but this is quite silly, and it’s better to just think of numbers like Z or Q.) Let’s look atthree more examples.

Example 3. The functions f : [0, 1]→ R form a ring which we will (nonstandardly) nameFun(I,R). When we add two functions f and g, we get a new function (f + g) : [0, 1]→ Rwhich takes r ∈ [0, 1] to f(r)+g(r). When we multiply two functions, f and g, we producea new function (f · g) : [0, 1] → R where (f · g)(r) = f(r)g(r). We call this pointwiseaddition and pointwise multiplication, because the rule to add or multiply two functionsis to add or multiply them at each point.

1

Example 4. The Gaussian integers are the subset of the complex numbers consisting ofall complex numbers C with integer coefficients {a + bi|a, b ∈ Z}. We write them as Z[i].The usual addition and multiplication of complex numbers apply.

Example 5. Modular arithmetic, Z/nZ, is a ring with the usual addition and multiplica-tion.

Example 6. The two by two matrices with entries in the real numbers R form a ring,under matrix multiplication and addition.

1.2 Laying Down the Rules

Now that we’ve seen a bunch of examples, let’s nail down a definition of a ring.

Definition. A ring, R, is a set of objects along with two binary operations, · multiplicationand + addition, with the following properties:

1. Closure: if a, b ∈ R then a+ b and a · b are in R.

2. Associativity: if a, b, c ∈ R then (a+ b) + c = a+ (b+ c) and (a · b) · c = a · (b · c).

3. Zero: there is an element 0 ∈ R such that for all a ∈ R, a+ 0 = 0 + a = a.

4. One: there is an element 1 ∈ R such that for all a ∈ R, 1 · a = a · 1 = a.

5. Commutativity of Addition: if a, b ∈ R then a+ b = b+ a.

6. Additive Inverses: for every a ∈ R, there exists an element −a ∈ R such thata+ (−a) = 0.

7. Distributivity: for any a, b, c ∈ R we have that a(b+c) = ab+ac and (b+c)a = ba+ca.

One thing that might seem funny about these examples is that while we require additionto be commutative and have inverses, we don’t require multiplication to be commutativeor have inverses. We want to be able to work with rings that have a looser multiplicativestructure: for example, we want our rings to model sets of functions, and one thing wenotice about functions is that they don’t always commute. The two by two matrices over Rare good simple example of a noncommutative ring. Also, in the case of rings of numbers,like our example Z, many numbers don’t have multiplicative inverses. The only integerswhich have multiplicative inverses are 1 and −1.

Remark. In this class, we will generally work with commutative rings (rings where mul-tiplication is commutative), but for the purpose of proving things in more generality, all ofthe proofs we do today will work whether or not we require our rings to be commutative.

Definition. A subring S of a ring R is a subset S ⊂ R that is also a ring.

It suffices to check that S contains the element 1 and is closed under subtraction andmultiplication.

2

1.3 Basic Properties of Rings

From the definition, we can deduce a few basic propositions. The goal in proving these isto say some facts about rings, but also to give you an example for how basic proofs in ringtheory go.

Proposition 1. The multiplicative identity 1 is unique.

Proof. Suppose R is a ring with two multiplicative identities, 1 and 1′. Then both of themsatisfy the property that for all r ∈ R, 1r = r1 = r and 1′r = r1′ = r.

In particular, we can let r = 1′ in the first equation and then we get the identity11′ = 1′1 = 1′ but in the second equation letting r = 1 we get 1′1 = 11′ = 1. This provesthat 1 = 1′.

Proposition 2. For any r ∈ R, 0r = r0 = 0.

Proof. We know that 0 + 0 = 0 since 0 is the additive identity. So r0 = r(0 + 0) = r0 + r0.Adding −(r0) to both sides (the additive inverse of r0, whatever it is!) we get that 0 = r0.To get 0r = 0 do the same thing with the other distributive equation.

Proposition 3. For any a, b ∈ R we have (−a)b = a(−b) = −(ab).

Proof. First consider (−a)b. Then we know that −a is the element such that a+(−a) = 0.So using distributivity we get that (a+ (−a))b = ab+ (−a)b and also (a+ (−a))b = 0b = 0by the previous proposition. Therefore ab + (−a)b = 0. Let’s add −(ab) to both sides:we get ab + (−a)b + (−(ab)) = −(ab) and we can commute the things on the left side toget 0 + (−a)b = −(ab) and so (−a)b = −(ab). You will show on the problem set thata(−b) = −(ab) to complete the proof.

We’re not going to be this explicit about associativity, zero, commutativity, and so onin the future, because it does get a little tedious. However, it’s good to have a groundingin the basics.

3

2 Problems Relating to Basic Definitions and Examples

2.1 The Basics

1. Prove that the additive identity 0 is unique. That is, if 0 and 0′ are two elements ofa ring R such that for all r ∈ R, r+ 0 = 0 + r = r and 0′+ r = r+ 0′ = r then 0 = 0′.

2. Emulating the proof in class, show that a(−b) = −(ab).

3. Convince yourself that each of the examples we talked about in class are actuallyrings. That is, go through the definition of ring, and verify any property that youaren’t sure you believe. (For example: what is the identity in the ring of functionsfrom [0, 1] to R?)

4. Show that additive inverses are unique: that is, show that if a+ b = 0 and a+ b′ = 0then b = b′.

5. Show that (−1)(−1) = 1.

6. The center of a ring R is defined as {z ∈ R|zr = rz for all r ∈ R} (that is, the set ofall elements which commute with every element of the ring.) Prove that the centerof a ring is a subring of that ring.

2.2 Getting your hands dirty

1. Convince yourself that Q( 3√

2) = {a + b 3√

2 + c 3√

4|a, b, c ∈ Q} is a ring. Most ofthe properties should be easy, but the closure under multiplication will require somejustification.

2. An element of a ring is called a unit if it has a multiplicative inverse. In other words,a is a unit in the ring R if there is some b ∈ R such that ab = 1. Which elements inthe ring Z[x] have multiplicative inverses? What about Q[x]? What about Fun(I,R)?

3. An nonzero element a of a ring is called a zero divisor if there is another nonzeroelement b such that either ab = 0 or ba = 0. For example, in the integers modulo 6,2 and 3 are zero divisors. Which elements in Fun(I,R) are zero divisors?

4. Determine which of the following are subrings of Q:

(a) the set of all rational numbers with odd denominators (when reduced)

(b) the set of all rational numbers with even denominators (when reduced)

(c) the set of all squares of rational numbers

4

5. Let X be a set and let P (X) be the power set of X, that is, the set of all subsets of X.(For example, ifX = {a, b, c} then P (X) = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}).Show that we can make P (X) into a ring where if Y,Z ∈ P (X) then Y + Z =(Y − Z) ∪ (Z − Y ) and Y · Z = Y ∩ Z. What is the identity? What is Y · Y for anyY ∈ P (X)?

2.3 To ponder

1. Consider the power series ring Q[[x]] that consists of elements of the form∑∞

n=0 anxn.

What elements have multiplicative inverses? What is the inverse of the power series1 + x? Note that Q[x], the polynomials in x with rational coefficients, sits insideQ[[x]] and 1 + x ∈ Q[x] but 1 + x is not a unit in Q[x]. Deduce that if S is a subringof R then all of the units of S are units of R but not all of the units of R containedin S are also units of S.

2. A commutative ring is called an integral domain if it does not have any zero divisors(see problem three from “Getting your hands dirty.”) Prove that in any finite integraldomain, every nonzero element is a unit. That is, every finite integral domain is afinite field.

3. Let R be a ring. A nonzero element a is called a left zero divisor in R if there isa nonzero element x such that ax = 0. Symmetrically, b 6= 0 is called a right zerodivisor if we can find a nonzero y such that yb = 0. An element u has a left inversein R if there is some s ∈ R such that su = 1 and an element v ∈ R has right inverseif there is some t such that vt = 1.

(a) Prove that if u has a right and a left inverse, then those inverses are equal.

(b) Prove that if u has a right inverse then u is not a right zero divisor.

(c) Find a ring R and an element r ∈ R such that r is a right inverse and a left zerodivisor.

5

3 Ring Homomorphisms, Kernels, and Ideals

Consider the rings Z[x] = {anxn + · · · + a1x + a0|ai ∈ Z}, polynomials in x with integercoefficients, and Z[y2] = {b2ny2n + b2n−2y

2n−2 + · · · + b2y2 + b0|b2i ∈ Z}, polynomials in

y with integer coefficients and only even powers of y. These two rings act and seem verymuch the same. In fact, we can make a one to one correspondence of objects in Z[x] withobjects in Z[y2] just by taking replacing each x with y2, or vice verse. How do we quantifythis notion of sameness? How can we in general talk about rings being the same?

Definition. We say two rings R and S are isomorphic if there is a bijective functionϕ : R→ S such that

• ϕ(a+R b) = ϕ(a) +S ϕ(b), (ϕ respects addition)

• ϕ(a ·R b) = ϕ(a) ·S ϕ(b), (ϕ respects multiplication)

• and ϕ(1R) = 1S . (ϕ preserves the multiplicative identity)

More generally, we can make analogies between rings by considering other functionsfrom one ring to another.

Definition. A ring homomorphism from R to S is a map ϕ : R→ S such that

• ϕ(a+R b) = ϕ(a) +S ϕ(b), (ϕ respects addition)

• ϕ(a ·R b) = ϕ(a) ·S ϕ(b), (ϕ respects multiplication)

• and ϕ(1R) = 1S . (ϕ preserves the multiplicative identity)

Here, we’re just relaxing the idea that it has to be a bijection. If our function is injective,then ϕ gives us a way to situate R in S as a smaller ring, and if our function is surjective,we will see that ϕ gives a way of grouping elements of R into “symmetry classes” so thatS is R mod some symmetries. In math, we often look at homomorphisms to tell us moreabout the structure of a ring, rather than looking at the individual ring itself.

Example 7. Consider the map ϕ : Z[x] → Z given by p(x) 7→ p(0). This just plucks outthe constant term of the polynomial p(x). This map is surjective but not injective. All thepolynomials with no constant term get sent to 0.

Notice that if p and q are polynomials that get sent to 0 under ϕ then p+ q and rp getsent to zero for any r ∈ Z[x].

Example 8. More generally, if we have a ring of functions like Fun(I,R), we can create aring homomorphism which is evaluation of the functions at a point. For example, the ring offunctions from [0, 1]→ R has a homomorphism ϕp : Fun(I,R)→ R for each point p ∈ [0, 1],sending f : [0, 1] → R to f(p). We can check that if f : [0, 1] → R and g : [0, 1] → R are

6

elements of Fun(I,R), then ϕp(f + g) = (f + g)(p) = f(p) + g(p) = ϕp(f) + ϕp(g). Alsoϕp(f · g) = (f · g)(p) = f(p)g(p) = ϕp(f)ϕp(g). Finally since the constant all ones functionc1 : [0, 1]→ R is the multiplicative identity in Fun(I,R) we check ϕp(c1) = c1(p) = 1.

Example 9. The map Z→ Z/nZ given by a 7→ (a mod n).

Example 10. The map Q(√

2) → Q(√

2) given by (a + b√

2) 7→ (a − b√

2) is an isomor-phism.

Example 11. There are injective maps Z→ Z[x] and Z→ Q realizing Z as a smaller ringinside these rings.

Definition. The image of a ring homomorphism ϕ : R → S is the set of elements s ∈ Ssuch that there exists r ∈ R with ϕ(r) = s.

Definition. The kernel of a ring homomorphism ϕ : R → S is the set of elements r ∈ Rsuch that ϕ(r) = 0.

Notice that we can characterize being injective as having the kernel equal to 0, andsurjective as having image equal to the target.

Proposition 4. The kernel of a ring homomorphism ϕ : R→ S has the following proper-ties:

1. If r ∈ ker(ϕ) then for all c ∈ R, cr ∈ ker(ϕ).

2. If r, s ∈ ker(ϕ) then so is r + s.

3. 0 ∈ ker(ϕ).

Proof. We’ll start with 0 ∈ ker(ϕ). Note that ϕ(0) = ϕ(0+0) = ϕ(0)+ϕ(0), so subtractingϕ(0) from both sides in S we see that 0 = ϕ(0). Then, consider r ∈ ker(ϕ) and suppose cis any element of R. Then ϕ(cr) = ϕ(c)ϕ(r) because ϕ is a homomorphism, and ϕ(r) = 0because r is in the kernel of ϕ. Thus ϕ(cr) = ϕ(c)ϕ(r) = ϕ(c)0 = 0. If r, s ∈ ker(ϕ) thenϕ(r + s) = ϕ(r) + ϕ(s) = 0 + 0 = 0 so r + s ∈ ker(ϕ).

Definition. We call any subset I of a ring R satisfying:

1. If r ∈ I then for all c ∈ R, cr ∈ I.

2. If r, s ∈ I then so is r + s.

3. 0 ∈ I.

an ideal of R.

7

Example 12. The subset I = 2Z = {2n|n ∈ Z} is an ideal of Z. More generally, aZ ={an|n ∈ Z} is an ideal of Z. In fact, these are the kernels of Z→ Z/nZ. We write these as(n), where this notation denotes the ideal generated by n.

Example 13. The subset {0} in R is always an ideal. So is R.

Example 14. Let p ∈ R and consider Mp = {f ∈ Fun(I,R)|f(p) = 0}. Then Mp is anideal of R and there are no other ideals other than Fun(I,R) containing Mp. (You willprove this on the problem set.)

Definition. We say M ⊂ R is a maximal ideal of R if it is not contained in any properideals of R.

8

4 Problems Relating to Ring Homomorphisms, Kernels, andIdeals

4.1 The Basics

1. Show that if I, J ⊂ R are ideals then so is I ∩ J .

2. Show that if ϕ : R → S is a ring homomorphism and J is an ideal of S thenϕ−1(J) = {r ∈ R|ϕ(r) ∈ J} is an ideal. If I is an ideal of R, is ϕ(I) necessarily anideal of S?

3. Let R be a commutative ring. A field is a commutative ring where every nonzeroelement has a multiplicative inverse. Show that {0} is a maximal ideal of R if andonly if R is a field. Hint: An ideal is a proper ideal if and only if it does not contain1. Consider the ideals (a) for a ∈ R, defined to be the multiples of the element (a):(a) = {ar|r ∈ R}.

4. Let I be an ideal of R. Suppose r ∈ R. Define r + I = {r + a|a ∈ I}. Show thatr + I = s + I as sets if and only if r − s ∈ I. (Hint, for one direction show thatr ∈ r + I and therefore r ∈ s + I and use this to write an equation r = s + a fora ∈ I.) Let R = Z and I = 5Z. What is the set m+ 5Z for m = 1, 2, 5, 7?

5. Which of the following are ideals of Z[x]:

(a) The set of all polynomials whose constant term is a multiple of 3.

(b) The set of all polynomials whose x2 term is a multiple of 3.

(c) The set of all polynomials with no constant, x or x2 term.

(d) The set of polynomials whose coefficients sum to zero.

4.2 Getting your hands dirty

1. Find all ring homomorphisms from Z to Z/30Z. Describe the kernel and image.

2. Show that Mp ⊂ Fun(I,R) defined as Mp = {f : I → R|f(p) = 0} is a maximal idealof Fun(I,R) for each p ∈ [0, 1].

3. What is the ideal (4, 6) generated by the elements 4 and 6 in Z? What about (2, 3)?What about (m,n)?

9

4.3 To ponder

1. Show that if I1 ⊂ I2 ⊂ . . . are ideals of a ring R then so is ∪∞n=1In. Come up withan example of an inclusion of ideals I1 ⊂ I2 ⊂ I3 ⊂ . . . such that In 6= In+1 for all n.(You will need to think of rings that we haven’t talked about yet, but only slightlyoff the beaten track.)

2. Let R be a commutative ring. Then we denote N(R) to be the nilradical of R, whereN(R) = {r ∈ R|rn = 0, for some n ∈ N} is the set of all nilpotent elements. Showthat N(R) is an ideal.

10

5 Tying Up Loose Ends

Some points I want to emphasize, having looked at your recent problem sets:

• Homomorhpisms are just as (or more) important than rings!

• A kernel is not a free-floating object: something is the kernel of a homomorphism(similarly for an image.)

• An ideal is not free-floating. A set is not just an ideal, it is an idea of a ring.

• To prove something is an ideal, try introducing a homomorphism.

Example 15. Consider inside of Z[x], the set of all polynomials whose coefficients sum tozero.

Notice: by plugging in 1 to p(x), we get the coefficient sum. However, we can’t saythis ideal is all multiples of (x − 1). Why? This is true in Z[x] but assumes a theoremabout polynomial division: namely that we can divide polynomials. Unfortunately, we canonly divide polynomials when our coefficients come from a field, which is why you haveonly proved the existence of polynomial division in your previous classes for polynomials inrings like R[x] and Q[x]. Now, it is true that if a polynomial has a root r ∈ Z then (x− r)is a factor, but we need a lemma to extend this factor theorem to Z[x], which is not a fieldand therefore doesn’t have division.

Instead, try constructing a homomorphism which has this set of polynomials as thekernel:

ϕ1 : Z[x]→ Zp(x) 7→ p(1)

ker(ϕ1) = {p(x) ∈ Z[x]|p(1) = 0.}

Example 16. Consider inside Z[x], the set of all polynomials whose constant term is amultiple of 3.

Construct the homomorphism

ϕ0 : Z[x]→ Z

p(x) 7→ p(0)

This takes a polynomial and plucks out the constant term.Then we have the ideal 3Z ⊂ Z consisting of all multiples of 3. The preimage ϕ−10 (3Z)

of this ideal under ϕ0 is an ideal and consists of all polynomials whose constant term is amultiple of 3. So the set described is an ideal of Z[x].

Note that we could also do this problem also by constructing ϕ : Z[x] → Z/3Z andtake p(x) 7→ p(0) and consider the kernel, but it is worth understanding the more flexiblemethod of taking preimages of ideals under homomorphisms. In some sense, the kernel ismerely the preimage of the zero ideal.

11

Try both methods here out for yourself:

Exercise 1. Prove that Mp, the set of all functions in Fun(I,R) which are zero at a fixedpoint p ∈ I, is an ideal.

Exercise 2. Prove that the set of polynomials in Z[x] whose coefficients sum to zeromodulo 5 is an ideal of Z[x].

12

6 Quotient Rings and First Isomorphism Theorem

Last week, we defined ideals and talked about how they arose from the kernel of ringhomomorphisms. This week, we will talk about how, given an ideal I of R we can finda ring S and a ring homomorphism ϕ : R → S where I = ker(ϕ). To do this, we willconstruct S as a quotient ring R/I.

On the homework, for r ∈ R and I an ideal of R, we constructed the set r + I ={r+a|a ∈ I}. For example, if R = Z and I = 5Z we computed that we have 0+5Z, 1+5Z,2 + 5Z, 3 + 5Z, and 4 + 5Z. If n is any other number then n+ 5Z = (n mod 5) + 5Z. Forexample, 7 + 5Z = 2 + 5Z.

Recall from the homework that we have the following proposition:

Proposition 5. The sets r + I and s+ I are equal if and only if r − s ∈ I.

Proof. We showed this by noting that if r−s ∈ I then r+I = r−s+s+I = {r−s+s+a|a ∈I} = {s + a|a ∈ I} since r − s + a ∈ I because I is closed under addition. Conversely, ifthe two sets are equal then 0 ∈ I implies r ∈ r+ I so r ∈ s+ I so r = s+ a for some a ∈ Iand so r − s = a ∈ I.

We would like to construct S = R/I to be the set of r + I for r ∈ R. In our example,S = {0 + 5Z, 1 + 5Z, 2 + 5Z, 3 + 5Z, 4 + 5Z}. For ease of notation, we often write r + Ias [r], for example S = {[0], [1], [2], [3], [4]} but we could just as easily say that S ={[5], [6], [7], [8], [9]}.

There is a natural way to define addition and multiplication on R/I. If [r] and [s] arein R/I then [r] + [s] = [r + s] and [r][s] = [rs]. One thing we have to do when workingwith quotient rings is to check that if we defined something in terms of r and s that thisoperation is well defined, meaning that no matter which name we give to [r] and [s] we getthe same answer for [r] + [s].

For example, is [0] + [6] the same as [0] + [1]? In our definition the first would be [6]whereas the second would be [1], but we know that [6] = [1].

Proposition 6. Addition and multiplication defined in this way on R/I are well defined.

Proof. Suppose r, r′ are two elements of [r] = [r′] and s, s′ are both elements of [s] = [s′].Then we want to show that [r+ s] = [r′+ s′] On the homework, we showed that r− r′ ∈ Iand s− s′ ∈ I. Let’s call these a and b. Then (r+ s)− (r′+ s′) = a+ b ∈ I since I is closedunder addition. Therefore by the same homework problem, we see that [r + s] = [r′ + s′].

Now consider [rs] and [r′s′]. We would like to show that rs− r′s′ ∈ I. Let’s replace r′

with r− a and s′ with s− b. Then rs− r′s′ = rs− (r− a)(s− b) = rs− rs+ as+ rb+ ab =as+ rb+ ab ∈ I, since anything times a or b is in I, as a, b ∈ I. Therefore [rs] = [r′s′].

13

This proposition showed us that it doesn’t matter what name we give for these setsr + I, the operations on R/I are well defined. In fact, these operations turn R/I into aring (if you don’t believe it, take some time on the homework to test out each of the ringaxioms.)

Note that there is a nice way to take elements of R to elements of R/I, by taking r tor + I.

Proposition 7. The map ϕ : R→ R/I given by r 7→ r + I is a surjective ring homomor-phism with kernel I.

Proof. Clearly this map is surjective. Note that we actually defined R/I to make this intoa ring homomorphism: suppose r, s ∈ R. Then ϕ(r + s) = [r + s] = [r] + [s] = ϕ(r) + ϕ(s)and similarly ϕ(rs) = [rs] = [r][s] = ϕ(r)ϕ(s). Check for yourself that ϕ(1) = [1] is theidentity in R/I.

What’s the kernel of ϕ? It’s the set of r ∈ R such that [r] = [0], i.e. the set of r ∈ Rsuch that r − 0 ∈ I or r ∈ I.

What’s neat is that there is a sort of converse to this idea:

Proposition 8 (First Isomorphism Theorem). Suppose ϕ : R→ S is any ring homomor-phism. Then R/ ker(ϕ) is isomorphic to ϕ(R) as rings.

Proof. You!

Thus, every ideal is the kernel of a homomorphism and every kernel is an ideal, and weknow exactly how that correspondence works.

Example 17. Let’s use the quotient ring construction in one particular case, Q[x]/(x2−2).Suppose p(x) is any polynomial in Q[x]. We can use polynomial long division (since x2− 2has leading coefficient 1) to write p(x) = r(x)+(x2−2)q(x) for some polynomials q(x), r(x)in Q[x] with the degree of r(x) less than 2. Then we can see that [p(x)] = [r(x)] sincep(x)− r(x) ∈ (x2 − 2) is a multiple of (x2 − 2). Conversely, for any two polynomials p(x)and r(x) with degree less than 2, then [p(x)] 6= [r(x)] because p(x) − r(x) has degree lessthan 2, and cannot be in the ideal (x2− 2) unless p(x)− r(x) = 0, as 0 is the only elementof (x2 − 2) of degree less than 2. Therefore p(x) = r(x).

So we can find equivalence class representatives by taking all of the elements of Q[x]less than degree 2: Q[x]/(x2 − 2) = {[a+ bx]|a, b ∈ Q}.

Let’s do a few computations: [x][x] = [x2] and we can write x2 = 2 + x2 − 2 usingdivision. Therefore [x]2 = [2].

This ring is actually the same as the ring Q[√

2] because x acts as a root of the poly-nomial x2− 2. In general, taking polynomial rings mod an ideal is a nice way to constructnew rings when we want to force x to act a certain way.

14

7 Problems Relating to Quotient Rings and First Isomor-phism Theorem

7.1 The Basics

1. Show that if I ⊂ R is an ideal of R and 1 ∈ R is the multiplicative identity then [1]is the identity in R/I and [0] is the additive identity. Find the additive inverse of [a]for a ∈ R and verify that this works for the toy example Z/5Z.

2. Prove the proposition from class: if ϕ : R → S is a homomorphism of rings thenR/ ker(ϕ) is isomorphic to ϕ(R).

3. Describe the elements of Q[x]/(x). What ring homomorphism that we have discusseddoes this correspond to? What about Q[x]/(x− 3)?

4. Suppose R is a commutative ring and M ⊂ R is a maximal ideal. Show that R/M isa field. Hint: on an earlier homework, you showed that {0} is a maximal ideal of S ifand only if S is a field. Combine this with another result from a previous homeworkthat if I ⊂ S is an ideal and ϕ : R → S a ring homomorphism then ϕ−1(I) is anideal of R.

5. A prime ideal P ⊂ R is an ideal of R such that if a · b ∈ P then a ∈ P or b ∈ P (orboth!). Show that nZ is a prime ideal of Z if and only if n is prime.

6. A domain is a ring S where for any a, b,∈ S, ab = 0 implies a = 0 or b = 0. Showthat R/P is a domain if and only if P is prime.

7.2 Getting your hands dirty

1. The ring Z[x]/(x2 + 1) is a familiar ring. What ring is that?

2. Write out the multiplication table for (Z/3Z)[x]/(x2 + 1). How many elements doesthis have? Note that this satisfies the field axioms, and is an example of a “finitefield”.

3. Suppose I want to construct a ring such that there is some element which is not zerobut whose square is zero. What quotient of Z[x] would I take?

7.3 To ponder

1. What is the map Fun(I,R) → Fun(I,R)/Mp where Mp is defined as before as themaximal ideal of functions f : I → R which are zero at p?

15

8 The Isomorphism Theorems

Theorem 1. Let R be a ring.

1. (The Second Isomorphism Theorem) Let A be a subring and let B be an ideal of R.Then A + B = {a + b|a ∈ A, b ∈ B} is a subring of R, A ∩ B is an ideal of A and(A+B)/B ∼= A/(A ∩B).

2. (The Third Isomorphism Theorem) Let I and J be ideals of R with I ⊆ J . ThenJ/I (the image of J under the canonical homomorphism) is an ideal of R/I andR/I/(J/I) ∼= R/J .

3. (The Fourth Isomorphism Theorem or Lattice Isomorphism Theorem) Let I be anideal of R. The correspondence J ↔ J/I is an inclusion preserving bijection betweenthe set of ideals J of R that contain I and the set of ideals of R/I. Furthermore, J(an ideal containing I) is an ideal of R if and only if J/I is an ideal of R/I.

Proof: You. The second isomorphism theorem is an extra practice problem, we won’tsee it come up a lot in the context of rings, but the fourth one is an especially good theorem.You can use it to take a second crack at the proof that M is a maximal ideal if and onlyif R/M is a field. It should provide a quicker proof.

Example 18. Consider the ring R = Z and the ideal I = 72Z. Then we can draw apicture of the ideal lattice in Z of ideals containing the fixed ideal I = 72Z on the righthand side. On the left hand side, we can draw a picture of the ideals of Z/I.

16

9 Euclidean Domains

We’ve been skirting around some very important issues of factorization for some time now.For example, from the beginning, I have asked you to hold off on using results aboutfactorization as much as possible. However, we have found it useful on occasion to citethe division algorithm for polynomials, to show that certain properties about polynomialideals are true.

Multiplication is an interesting topic of study in rings because, unlike addition, it isnot fully prescribed. In the following sections we will discuss the various ways in whichfactorization can work, or fail. We will begin with the very nicest possible factorization,which we see happening in the integers and the polynomial rings. In each of these rings,we have a division algorithm, the Euclidean algorithm, and a notion of norm.

Let R be an integral domain. (We will from hereon out restrict our attention to ringswhich are commutative and have no zero divisors, called integral domains, because thetopics here are useful in number theory.)

Definition. A function N : R→ N with N(0) = 0 is called a norm.

Definition. We say R is a Euclidean Domain (or has a division algorithm) if there is anorm N on R such that for any pair of elements a and b in R, with b 6= 0, there exist qand r in R satisfying

a = qb+ r with N(r) < N(b) or r = 0.

The element q is called the quotient and the element r is called the remainder. Notethat a division algorithm (which is what the above process of finding a q and r to makethe equation a = qb+ r is called) is not an algorithm at all.

As you can see, the division algorithm allows you to execute the Euclidean Algorithmby repeating it successively. In this way, we will be able to find a greatest common divisorof a and b.

a = q1b+ r1

b = q2r1 + r2

r1 = q3r2 + r3

...

rn−1 = qn+1rn

Note that there is no guarantee of uniqueness of remainders or quotients. We will see thatin executing the Euclidean Algorithm, we are able to take an ideal I = (a, b) and write itas a singly generated ideal, and more generally, an ideal with any number of generatorscan be pared down to a single generator.

17

Example 19. A field is a Euclidean Domain. Any norm will work, (the zero norm is fine),and for every pair of elements in the field, a = qb+ 0 where q = ab−1.

Example 20. The integers Z are a Euclidean Domain with the norm N(a) = |a|. Thelong division algorithm is a division algorithm.

Example 21. If F is a field, F [x] is a Euclidean Domain with norm N(p(x)) = deg(p(x)).The division algorithm is the usual polynomial division algorithm.

Definition. A principal ideal is an ideal I ⊆ R that can be generated by one elementI = (r), r ∈ R.

Proposition 9. Every ideal of a Euclidean Domain is principal.

Proof. Let R be a Euclidean Domain. Then let I be an ideal of R, and consider the setof norms N(a) such that a ∈ I. Pick some d ∈ I of minimum norm. The ideal (d) ⊆ I.Conversely, for any a ∈ I, we can divide a = qd+ r where N(r) < N(d) or r = 0. However,r = a − qd is a difference of two things in the ideal I and therefore r itself must be in I.But d is of minimum norm in I, so r = 0. Thus a = qd and so a ∈ (d). We can concludeI = (d).

Example 22. The ring Z[x] is not a Euclidean Domain, since (x, 2) is not principal.

18

10 Problems Relating to Euclidean Domains

The Basics

1. Let R be a commutative ring and a, b ∈ R with b 6= 0. A greatest common divisor ofa and b is a nonzero element d such that d|a and d|b and if d′|a and d′|b then d′|d.

(a) Show that d|a if an only if a ∈ (d) if and only if (a) ⊂ (d).

(b) Conclude that in a Euclidean Domain if d is the greatest common divisor of aand b then (a, b) = (d). (This might explain why the notation (a,b) is commonlyused for gcd.)

(c) If (d) = (a, b), is d necessarily the gcd of a and b?

2. Let R be an integral domain. If two elements d and d′ of R generate the sameprincipal ideal, that is, (d) = (d′) then d = ud′ for some unit u in R. In particular,if d and d′ are both greatest common divisors for some elements a and b, then theydiffer by multiplication by a unit.

3. Let R be a Euclidean Domain and let a and b be nonzero elements of R. Remindyourself that greatest common divisors can be computed algorithmically, and convinceyourself that this is true in a ring, not just a field.

4. Let R be a commutative ring with 1 and let a and b be nonzero elements of R. Aleast common multiple of a and b is an element e of R such that a|e and b|e and ifa|e′ and b|e′ then e|e′.

(a) Prove that the least common multiple of a and b (if it exists) is a generator forthe ideal (a) ∩ (b).

(b) Deduce that any two nonzero elements in a Euclidean Domain have a least com-mon multiple which is unique up to multiplication by a unit.

(c) Prove that in a Euclidean Domain the least common multiple of a and b is ab(a,b)

where (a, b) is the greatest common divisor of a and b.

Hint: Show that ( ab(a,b)) ⊆ (a) ∩ (b) (easy). Then show the opposite containment

(hard). If we call e the LCM of a and b , then a hint for the hard direction is toshow that ab/e divides a and b and therefore (a, b).

Getting your hands dirty

1. Find a generator for the ideal (x4 − 4x3 + 7x2 − 4x, x3 − 6x2 + 11x− 6) ⊆ Q[x].

2. Find a generator for the ideal (11391, 5673) ⊂ Z.

19

3. Prove the multiplicative property of norms in Z[√n], if it is defined as N(a+ b

√n) =

a2 − nb2.

4. Consider R = Z[√−5] the quadratic integer ring. This is an important ring in

number theory, and it has an associated field norm N(a+ b√−5) = a2 + 5b2, where

a, b ∈ Z. Consider the ideal I = (3, 2 +√−5). Suppose that I were principal, and

generated by some element c+d√−5. Then we must have that 3 = α(c+d

√−5) and

2 +√−5 = β(c + d

√−5) for some α, β ∈ R. For each of these two equations, take

norms, and consider the possible values of c2 + 5d2, and α. Derive a contradictionand show that R is not a Euclidean Domain.

5. Find a generator for the ideal (85, 1 + 13i) in Z[i]. Hint: Graph the multiples of1 + 13i in the complex plane, and find the one closest to 85.

To Ponder

1. Prove that the quotient Z[i]/I is finite for any nonzero ideal I. (Use an argumentabout norms.)

20

11 Factorization

This section should probably have come before the previous section, but it’s better latethan never, right? Some of these definitions will be familiar and some new.

Definition. Let R be an integral domain and a, b ∈ R. We say that a|b or a divides b ora is a factor of b if there exists c ∈ R such that b = ac.

We say that a nonzero element r ∈ R is irreducible if whenever a|r either a is a unit,or there exists a unit u ∈ R such that a = ur. Equivalently, whenever we can write r = ab,for a, b ∈ R one of a and b is a unit.

Note that any factor of a unit is also a unit, if u = ab then abu−1 = 1 so bu−1 is aninverse for a. This shows that even if we were to continue trying to break an irreducibleelement up into more pieces, it would remain in the form a unit times itself: factoring theunit u would only lead a product of more units.

Irreducible captures the idea of unbreakable—there is also a related notion called prime-ness that we have talked about, but it turns out that being irreducible is what we want forthe pieces of our factorizations. The notion of primeness, that if p|ab then p|a or p|b, thatwe discussed earlier in class, will line up with irreducibility in some rings but not othersand we will study these implications.

Definition. A domain R has unique factorization (or simply is a unique factorizationdomain UFD ) if every nonzero a ∈ R can be written uniquely as a product

a = u∏

pidi

where u is a unit and the pi are irreducible. Here, uniquely means up to reordering theirreducibles and multiplying them by units.

Example 23. The integers Z are a ring with unique factorization. The units are ±1 andthe irreducibles are the primes and their negatives.

Example 24. In the Gaussian integers, Z[i] the units are ±1,±i, which we can prove (andwe will do so using the norm). We have the norm N(a + bi) = a2 + b2 which has theproperty that N(xy) = N(x)N(y), that is, it is multiplicative. In particular, this impliesthat for any unit u,

1 = N(1) = N(uu−1) = N(u)N(u−1)

so the norm of u must be 1. Therefore ±1 and ±i are the only units in Z[i].Now we will show that if p is a prime number which factors in the Gaussian integers, it

must be the sum of two squares, and conversely, if it is the sum of two squares it factors. Wewill also see that if p cannot be written as the sum of two squares it must be irreducible, bynorm considerations, as the following proof will show. This will provide a full classificationof what prime numbers factor in Z[i].

21

Let p be a prime number and suppose that in Z[i] we can factor p = xy where x andy are not units. Then N(p) = p2 = N(x)N(y) and since x and y are not units, we havethat N(x) = N(y) = p so p must be a sum of two squares, say by writing x = a + bi weget p = a2 + b2. Furthermore, x and y much each be irreducible since p is a prime and thenorm is multiplicative. On the other hand, if p cannot be written as a sum of two squares,it is irreducible in Z[i].

In general, the irreducible elements of Z[i] are the elements of prime norm.

Example 25. Suppose we want to factor 14+6i in Z[i]. First we calculate its norm, whichis 142 + 62 = 232. Then we factor that into primes, 232 = 23 · 29. Note that 2 = 1 + 1 is asum of two squares while 29 = 4 + 25. So we can try (1 + i)3 = −2 + 2i to start, and thennote that multiplying this by (2 + 5i) gives −14 − 6i. So multiplying the whole thing by(−1) gives 14 + 6i.

12 Problems on Unique Factorization

12.1 The Basics

1. Factor 12 + 10i in Z[i].

2. Let R be a UFD and let p ∈ R be irreducible. Show that if p|ab for some a, b ∈ Rthen either p|a or p|b. (That is, every irreducible element is prime.)

3. Is a field a UFD? Explain.

4. Suppose R is a UFD. Show that every pair of elements has a gcd, and that it isunique up to multiplication by a unit.

12.2 Getting Your Hands Dirty

1. Show that Z[2i] = {a + bi|a, b ∈ Z and b is even} is not a UFD. Note: the norm inthis ring is a2 + 4b2. Think about small numbers and find a number that factors intwo ways, prove that these factorizations are each irreducibles in Z[2i].

2. Show that Z[√−6] is not a UFD.

22

13 Principal Ideal Domains

We saw that Euclidean Domains were one example of a more general ring called a PrincipalIdeal Domain.

Definition. A principal ideal domain or PID is an integral domain such that every idealis principal.

We will see that PIDs are a general class of rings with unique factorization. We firstbegin by studying a model proof of unique factorization, in the integers Z.

Theorem 2. The integers Z have unique factorization.

Proof. Let n ∈ Z be a nonzero integer. By multiplying n by a unit (negative one) we canassume that n is positive and greater than or equal to one. We first show that n has afactorization (existence) by induction on n. The base case, n = 1, is trivial, because n is aunit and has no irreducible factors.

Suppose that for every k < n we can factor k into a product of irreducibles. If nis a prime number, it clearly is its own factorization, since the prime numbers are theirreducibles of Z. Otherwise, n is not irreducible and we can write n = ab where a and bare not units and we may assume both a and b are positive. It follows that a, b < n soby induction a and b have factorizations, and by multiplying these factorizations we get afactorization for n.

Note that this part of the proof, existence of factorizations, doesn’t seem to generalizemuch. We will need to be a little more creative if we are going to prove the existence offactorizations in general PIDs.

Now we must prove uniqueness: consider the following lemma.

Lemma 1. Let p ∈ Z be a prime number and suppose p|ab then p|a or p|b. To translateinto more general terms: let p ∈ Z be an irreducible element of Z, then p is a prime elementof Z.

Assuming the lemma, then uniqueness of factorizations in Z follows. Suppose we havetwo different factorizations for n,

n = u∏

pdii = v∏

qejj .

Then, going through each pi in succession, we note that since pi|v∏qejj and pi is not a

unit, pi must divide some qj and so we can pair up the pi and qj as being the same upto multiplication up to a unit. There must be the same number, or else we run into acontradiction, as we are left over with some qj and only a unit on the pi side. Thus thetwo factorizations are the same.

23

This motivates the following definition.

Definition. Let R be an integral domain and p ∈ R be a nonzero element which is not aunit. Then p is called prime if whenever p|ab, we have that p|a or p|b.

Every prime is irreducible: if p = ab then p|ab so, without loss of generality, p|a anda = pc for some c ∈ R. Then we can write p = ab = pbc and therefore bc = 1 proving thatb is a unit.

Furthermore, in a UFD irreducibles are prime: we can see this by factoring ab, if p|aband p irreducible then p|a or p|b by unique factorization, as you showed on your problemset. In fact, this is equivalent to unique factorization.

Theorem 3. Let R be an integral domain such that every nonzero a ∈ R has a factorizationinto irreducible elements. Then if every irreducible element of R is prime, R is a UFD.

Proof. Use the lemma, and use the proof in the case of R = Z. This generalizes completelywithout modification.

Theorem 4. In a PID all irreducibles are prime.

Proof. Let R be a PID and let p ∈ R be an irreducible. Suppose p|ab and p does not divideb. We want to show p divides a. Consider the ideal (p, b). This is principal, (p, b) = (c).Then c|p so since p is irreducible either c is a unit or c is the same as p, up to a unit. If cis the same as p then (p, b) = (p) so b ∈ (p) so p|b contradicting our assumption. thus c isa unit.

We can write c = rp+sb for some r and s. Multiply by a and c−1, a = rpac−1 +sbac−1

but p|ab so p divides the right hand side, so p|a.

So, to prove unique factorization in Z, we just need to prove the lemma, which comesdown to proving that Z is a PID.

Theorem 5. Z is a PID.

Proof. We’ve already proved this, but it’s good to remind ourselves of the proof. Let I ⊂ Zbe an ideal. If I = {0}, then I = (0) is principal. Otherwise, pick the smallest positiveelement n ∈ I. Then clearly (n) ⊆ I. Furthermore, for any m ∈ I, by division withremainder, we can write m = qn + r where 0 ≤ r < n. Since m ∈ I and qn ∈ I it followsthat r ∈ I. But r < n so by minimality of n we have r = 0 and so m = qn is divisible byn so I = (n).

24

14 Problems Relating to PIDs

14.1 The Basics

1. Let R be a domain and a ∈ R be irreducible. Show that a is also irreducible as anelement of the polynomial ring R[x].

2. Show that a ∈ R (an integral domain) is a prime element if and only if (a) is a primeideal.

14.2 Getting your hands dirty

1. From the lecture, we proved that if we have factorization into irreducibles, then we getuniqueness. In this pair of problems, you will prove factorization into irreducibles.First, start by showing we get a single factor. Let R be a PID and a ∈ R be anon-unit. Prove that a has an irreducible factor.

Hint: Suppose a = a0 has no irreducible factor. We can pick a nontrivial factora1|a0 and a1 cannot have an irreducible factor either. We get an infinite sequence. . . a2|a1|a0 which gives a sequence of ideals

(a0) ( (a1) ( (a2) ( . . .

Consider the ideal I = (a0, a1, a2, . . . ) generated by all the ai. Using the fact that Ris a PID derive a contradiction.

2. Show that if R is a PID then every nonzero element a ∈ R has a factorization intoirreducible factors (and a unit.) As a corollary, deduce that R is a UFD using thetheorems from class.

25

15 References

1. Dummit and Foote, Abstract Algebra, Third Edition

2. Waffle Wofsey, Notes on Unique Factorization, http://math.harvard.edu/ waffle/ufds2.pdf,from Canada/USA Mathcamp

26