trial solutions to introduction to commutative algebra · pdf file · 2017-03-041...
TRANSCRIPT
1
Trial solutions
to
Introduction to Commutative Algebra
(M.F.Atiyah &&&& I.G.MacDonald)
by M. Y.
URL www.geocities.jp/mathlife1/
This document is the translation of my learning notebook described in Japanese for the famous text
book ″Introduction to Commutative Algebra (M.F.Atiyah & I.G.MacDonald)″. Not all of exercises
are solved. Especially, if the hint attached to the exercise is regarded as a solution, I do not try to
solve it. Errors may be included in this document because of my own solutions.
Translated 2017 may 1
Chapter 1
Ex.1.1
xn=0 ⇒x∈r(0)⊆J(A)⇒1+ax is a unit for any a⇒1+x is a unit.
Let x be a nilpotent and u be a unit. u+x=u(1+u−1
x). u−1
x is nilpotent⇒1+u−1
x is a unit
by the above. u+x is a unit.
Ex.1.2
i) ⇐:By induction, f is a unit (ex.1.1).
⇒: By induction on degree n of f.
The case that n is 0, 0a is a unit.
The case that degree of f is less than n, suppose the result is true.
Let g=b0+b1x+…+bmxm be the inverse of f=a0+a1x+…+anx
n. a0b0=1 and anbm =0.
fang=an⇒ an2bm−1 is 0. Consequently, an
rbm-(r-1)=0 for r=2,…, m+1. an
m+1b0=0. b0 is a
unit⇒anm+1
=0. f−anxn is a unit of degree under n (ex.1.1). By the assumption of
induction, a0 is a unit and a1,…,an-1 are nilpotent.
ii)⇒: f is nilpotent ⇒1+f is a unit. Then 1+a0 is a unit and a1,…,an are nilpotent by (i).
a0∈(a1,…,an, f). a1,…,an and f are nilpotent⇒a0 is nilpotent.
⇐:clear.
iii) :⇒ Let f=a0+a1x+…+anxn. Let g=b0+b1x+…+bmx
m (bm≠0) be the polynomial of the
least degree such that fg=0. anbm=0. (ang)f=0. an g=0 because the degree of ang is less
than that of g. 0=fg=(a0+a1x+…+an−1xn−1
+anxn)g=(a0+a1x+…+an−1x
n−1)g. an−1bm=0.
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Then (an−1g)f=0. an−1g=0 because degree of an−1g is less than that of g. Consequently,
an-rg=0 (r=0,…,n). bm (≠0) satisfies bm f=0.
⇐:trivial.
iv) I(h) denotes the ideal which all of coefficients of h∈A[x] generate.
⇒:Clearly, I(fg) ⊆I(f) and I(fg)⊆ I(g). fg is primitive⇒ I(fg)=(1) ⇒I(f)=I(g)=(1).
⇐:Let I(f)=I(g)=(1). Suppose I(fg)≠(1). a maximal ideal ∃m⊇I(fg). Let as be the
coefficents of highest degree of f such that ai∉m. Let bt be the same of g. The
coefficient of degree s+t of fg, i.e. Σi+j=s+t aibj ∉m, because asbt∉m and other aibj∈m. It
contradicts to I(fg) ⊆m.
Ex.1.3
i) ”f is a unit⇔ a0 is a unit and a1,…,an are nilpotent.”
ii) ”f is nilpotent ⇔ all of coefficients of f are nilpotent.”
iii) ”f is a zero divisor ⇔ ∃a (≠0)∈A, af=0.”
iv) ”fg is primitive ⇔ both of f and g are primitive.”
Proof:
i): By induction on number of indeterminates.
The case of one indeterminate is proved in (ex. 1.2.).
When the number of indeterminates is less than n, assume the result is true. Let f be a
polynomial of n indeterminates, say x1,…,xn. Regard f as a polynomial of xn having
coefficients fi ∈A[x1, …,xn-1], i.e. f = f0+ f1 xn+…+ fkxnk. f is a unit ⇔ f0 is a unit and fi
is nilpotent for i=1,...,k by (ex. 1.2). f0 = a0+ f0’ (where f0’ has not a constant) is a unit⇔
a0 is a unit and other terms are nilpotent by assumption of induction. fi (i=1,...,k) is
nilpotent ⇔ all of coefficients are nilpotent by following (ii). Then, f is a unit ⇔
constant is a unit and other coefficients are nilpotent by assumption of induction.
ii):By induction on number of indeterminates.
The case of one indeterminate is proved in (ex. 1.2).
When the number of indeterminates is under n, assume the result is true. Let f be a
polynomial of n indeterminates, x1,…,xn. Regard f as a polynomial of xn having
coefficients fi ∈ A[x1, …,xn-1], i.e. f = f0+ f1xn+…+ fkxnk. f is nilpotent ⇔ fi is nilpotent
for i=0,...,k (ex.1.2). Each fi is nilpotent⇔ all of coefficients are nilpotent by assumption
of induction.
iii): ⇒: Let n
n
m
n
mm
m...mm x...xxaf 21
21 21Σ= be a polynomial of n indeterminates. Introduce
the lexicographical order (m1,…,mn)>(m’1,…,m’n) ⇔ mn >m’n. If mn=m’n, mn-1>m’n-1.
3
so on. into monomials. Let n
n
l
n
ll
l...ll x...xxbg 21
21 21Σ= be the least degree of polynomial
such that fg=0. Let nM...MMa
21be the highest coefficient of f. Let
nL...LLb21
be the highest
coefficient of g. fg=0⇒nn L...LLM...MM ba
2121=0.
nM...MMfa21
g=0⇒nM...MMa
21g=0, because the
order of ganM...MM 21
is lower than that of g. fg=( n
n
M
n
M
M...MM x...xaf 1
21 1− )g=0. The
highest order coefficient of n
n
M
n
M
M...MM x...xaf 1
21 1− is nM...MMa
21 1− . Since
nn L...LLM...MM ba2121 1− =0, gfa
nM...MM 21 1− =0. Because the order of ganM...MM 21 1− is lower
than that of g, ganM...MM 21 1− =0. Consequently,
nn L...LLm...mm ba2121
=0.ThennL...LLb
21f=0.
⇐: trivial.
iv):Let f,g∈A[x1,…,xn]. Let I(f) be the ideal which all of coefficients of f generate.
⇒:I(fg) ⊆I(f) and I(fg)⊆ I(g). fg is primitive⇒ I(fg)=(1). ⇒I(f)=I(g)=(1).
⇐:Let I(f)=I(g)=(1). Suppose I(fg)≠(1). A maximal ideal ∃m⊇I(fg). Introduce
lexicographic order into monomials by degrees of variable x1,…,xn. Let nm...mma
21be the
coefficient of the highest degree of f that does not belong to m. Let nl...llb
21 be the same
of g. The coefficient of fg of degree nn lm,...,lm,lm +++ 2211 ,i.e.
∉Σ +=++=++=+ nj...jjni...iinlmji,...,lmji,lmji bannn 212122221111
m. It contradicts to I(fg) ⊆m.
Ex.1.4
J(A)⊇r(0) always holds. Let f∈J(A[x]). 1+gf is a unit for any g∈A[x](1.9). Take x as g.
xf is nilpotent (ex.1.2i). Every coefficient of xf are nilpotent by (ex.1.2.ii). Every
coefficient of f is nilpotent. Then f is nilpotent. f∈ r(0).
Ex.1.5
i):⇒: let n
n
n xaf ∑∞
=
=0
andn
n
n xbg ∑∞
=
=0
. If fg=1, then a0b0=1. a0 is a unit.
⇐:If a0 is a unit, redefine f= a0(1− xg) where g∈A[[x]]. f−1
= a0−1
(1+ xg+ x2g
2 +…) is
definable, because coefficient of x n
is determined by first n+1 terms.
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ii):Let f=a0+a1x+…+anxn+…∈A[[x]] and f
m=0 for some m(>0). a0
m=0. Let
f1=f−a0=x(a1+a2x+…+anxn-1
+…). f1 is nilpotent because f1 is a finite sum of nilpotents.
Then a1 is nilpotent because a1+a2x+…+anxn-1
+… is nilpotent. Consequently, all of
coefficients of f are nilpotent.
A counter example :
Let A=k[x1,…,xn,…]/(x1,x22,…,xn
n,…) and f=x1y+x2y
2+…+xny
n+…∈ A[[y]]. For any
m(>0) , f m≠0.
iii) Let f=a0+a1x+…+anxn+…∈A[[x]].
⇒: f ∈ J(A[[x]])⇒1+bf is a unit for ∀b∈A. Then 1+ba0 is a unit by (i). Then a0∈J(A).
⇐: Let a0∈J(A). For ∀g=b0+b1x+…∈A[[x]], 1+gf=1+b0a0+(a1b0+a0b1)x+… is an unit
because 1+b0a0 is a unit by (i). Then f ∈J(A[[x]]).
iv):(mc,x)=m for maximal ideal ∀m of A[[x]]:
x ∉m⇒m⊂(m,x)=(1). ∃g=b0+b1x+…∈m and ∃h∈A[[x]], g+hx=1. b0 is a unit by (i). g
is a unit by (i). Contradictory. Then x∈m. ⇒g=b0+b1x+…∈m⇒b0∈mc.⇒ g∈(m
c,x)
⇒m⊆(mc,x). m⊇(m
c,x) is clear.
Suppose mc⊂n≠A. ∃y(∈n,∉mc
). (mc,x)⊂(n,x) because y∉(m
c,x) and y∈(n,x). (m
c,x)=m
by above. ⇒(n,x)=(1). Let f=a0+a1x+…∈(n,x) be a unit. By (i), a0 is a unit. a0∈n. It
contradicts to n≠(1).
v):Let p be any prime ideal of A. (p,x) is a prime ideal of A[[x]], because A[[x]]/(p,x)
≅A/p. (p,x) c=p is clear.
Ex.1.6
J(A)⊇r(0): trivial.
J(A)⊆r(0):Suppose a∈J(A) and a ∉r(0). By assumption, e=e2 for some e(≠0)∈(a). e=af.
af(1−af)=0. 1−af is a unit⇒af=e= 0. Contradictory.
Ex.1.7
It is enough to show that (p,x)=(1) for any prime ideal p and any x∉p. xn=x⇒
x(xn−1−1)=0. x∉p⇒(x
n−1−1)∈p. Then (p,xn−1
)=(1). (p,xn−1
)⊆(p,x)⇒ (p,x)=(1).
Ex.1.8
Let q be any prime ideal of A. A−q is a multiplicatively closed set. Let Σ be a family of
multiplicatively closed sets which contains A−q. Let the partial order of Σ be inclusion
in the sense of set theory. Let T be any chain of Σ, then the union of all members of T is
also a member of Σ. By Zorn’s lemma, Σ has a maximal element S. Let A−S=p. p is a
prime ideal as following.
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p is an ideal:
Let x∈p. x∪S do not make a multiplicatively closed set by maximality of S. ∃m>0
∃t∈S ,xmt=0. For a ∈ A, (ax)
mt=0∈p⇒ax∉S⇒ax∈p.
x,y∈p⇒0=xmt and 0=y
m’t’. (x+y)∉p⇒x+y∈S⇒(x+y)
m+m’ tt’∈S. (x+y)
m+m’ tt’ =0. It is
contradictory to x+y∈S. ⇒ x+y∈p.
p is a prime ideal:
x,y∉p⇒x,y∈S⇒ xy∈S⇒xy∉p.
If p’(⊂p) is a prime ideal, S’ (=A−p’ )⊃ S. This is contradictory to maximality of S. p is
a minimal prime ideal.
Ex.1.9
⇒:Let a=r(a). r(a)=∩p⊇ap⇒a is an intersection of prime ideals.
⇐:Let a=∩pi. a⊆r(a) always holds. x∉a⇒ x∉∃pj in pi. xn∉pj ⇒x
n∉a.⇒x∉r(a). ⇒
a=r(a).
Ex.1.10
i) ⇒ii) :Let p be the unique prime ideal. Then p is the unique maximal ideal. x∉p⇒x is
a unit. x∈p⇒x is nilpotent becasue r(0)=p.
ii)⇒iii):By assumption, x(∉R=r(0)) is a unit. The image of x in A/R is a unit. All
elements of A/R except 0 are units. A/R is a field.
iii)⇒i): A/R⇒r(0)=R is a maximal ideal of A. Every prime ideal p⊇R⇒ p=R.
Ex.1.11
i): x ∈A ⇒ (1+x)2=1+x. ⇒ 1+2x+x
2=1+3 x=1+ x. 2x=0.
ii):To prove p is a maximal ideal, it is enough to show that for ∀y∉p, p+(y)=(1).
y is an idempotent⇒y(y−1)=0∈p. y∉p⇒ y−1∈p. p+(y)=(1).
A/p consists of 2 elements:
y,z∉p ⇒yz(y−z)=0∈p. ⇒ y−z∈p. A is a union of p and 1+p. Quatient ring is0,1.
iii):It is enough to show the case of 2 terms. (e,f)⊇(e+f−ef) is trivial. e∈e(e+f−ef). f∈ f
(e+f−ef). ⇒ (e,f)=(e+f−ef).
Ex.1.12
e(≠0,1) is an idempotent⇒A=Ae⊕A(1−e), because if x∈Ae∩A(1−e), x=ae=b(1−e) and
xe=ae2=ae=x =b(1−e)e =0. Ae and A(1−e) have maximal ideals ,i.e. m, n repectively.
Ae⊕n and m⊕A(1−e) are maximal ideals. A can not be a local ring.
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Ex.1.14
Let S(0∉S) be any multiplicatively closed set. Let ∆ be a family of ideals which are
disjoint with S. Since ∆ contains 0, ∆≠∅. Let the partial order of ∆ be inclusion in the
sense of set theory. A union of all members of arbitrary totally ordered set of ∆ is a
member of ∆. By Zorn’s lemma, a maximal element ∃p in ∆. p is clearly an ideal. For
∀x,y(∉p), (p,x)∩S≠∅ and (p,y)∩S≠∅ by maximality of p. Then (p,x)(p,y)∩S≠∅, i.e.
xy∉p. p is a prime ideal.
Let S be the set of all of non zero-divisors. S is a multiplicatively closed set. Let ∆ be
a family of ideals which are disjoint with S. Let x be a zero-divisor. (x) is member of ∆.
Let p be a maximal ideal which contains (x) and is disjoint with S. p is a prime ideal by
the above. p consists of only zero-divisors. A−S=∪p⊆A−S p.
Ex.1.15
i): E⊆a⇒V(a)⊆V(E). p∈V(E)⇒p⊇E⇒p⊇a, because a is generated by E. p∈V(a).
V(a)=V(E).
p∈V(a) ⇔p⊇a⇔p⊇r(a) ⇔p∈V(r(a)).
ii):clear by the definition.
iii): p∈V(∪iEi)⇔ p⊇∪iEi ⇔ p⊇Ei for each i ⇔ p∈∩iV(Ei ).
iv):r(a∩b)=r(ab) by(1.13). Then V(a∩b)=V(ab) by (i).
ab⊆p⇒a⊆p or b⊆p⇒V(ab) ⊆V(a) ∪V(b). V(ab)⊇V(a),V(b) ⇒V(ab) ⊇V(a) ∪V(b).
Ex.1.16
Spec(Z) : (p). p is a prime number or 0.
Spec(R) : 0.
Spec(C[x]) : (x−c) or 0. c is a complex number.
Spec(R[x]) : (f) or 0. f is a irreducible polynomial.
Spec(Z[x]) : (f) or (p) or 0. f is a irreducible polynomial, p is a prime number.
(p,f). f mod p is irreducible.
Ex.1.17
O is an open set⇒O=X−V(E) where V(E) = V(a), a is an ideal generated by E.
V(a)=p∈Spec(X) | f∈p for ∀f∈a. O=X−V(a)=p∈Spec(X) | ∃f∈a, f∉p =∪f∈a Xf.
Xf are basic opensets.
i):p∈Xf∩Xg ⇔ f, g∉p ⇔fg∉p⇔p∈Xfg.
ii):Xf=∅⇔f∈p for any p∈Spec(X)⇔f∈∩ p:all prime ideals p = r(0) ⇔f is nilpotent.
iii):Xf=X⇔ f∉p for any p∈Spec(X)⇔ f is a unit.
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iv):Xf=Xg⇔X−V((f))=X−V((g)) ⇔V((f))=V((g)) ⇔V(r((f)))=V(r((g))) ⇔r((f))=r((g)).
(Last⇔:⇐:tivial. ⇒:by r(a)= ∩ a⊆pp=∩p∈V(r(a)) p.)
v):X=∪f Xf =∪f (X−V((f))) = X−∩ f V((f)) ⇔∩ f V((f))=∅⇔ (1)=(f1,f2,…) ⇒ 1=Σi≤n a if i
⇒∩ i≤n V((f i))=∅ ⇒X=∩ i≤n Xfi.
vi):Topology of Xf is induced from X. Let Xf=∪g (X f ∩ X g)(= ∪g X f g). ⇒V(r((f)))=
∩gV((fg))=V(∪g(fg))=V((f)(∪g(g)))=V((f)∩(∪g(g)))⇒r((f))=r((f))∩r(∪g(g)).
⇒ r((f))⊆r(∪g(g)) ⇒ fm∈(g1,g2,…) for some m(>0) ⇒ f
m=Σi≤n h igi ⇒r((f))⊆r(g1,…,gn)
⇒V(r((f)))⊇V(r(g1,…,gn))=V((g1,…,gn))=∩ i≤n V((gi)) ⇒Xf⊆∪ i≤nXgi ⇒Xf=∪ i≤n (Xf ∩Xgi).
Xf is quasi-compact.
vii): ⇒:Xfare basic open sets. Let O be any open set. Suppose Xfα α cover O. If O
can not be a union of finite set of Xf, any finite set of Xfα α can not cover O⇒O is not
quasi-compact.
⇐:Let O=∪i>0Oi. By assumption, O=∪n>j>0Xfj. Then ∪n>j>0Xfj =∪i>0Oi. Each Xfj is quasi-
compact (v), then Xfj is covered by finite set of Oi. Since O is a finite union of Xfi, O is
covered by finite set of Oi. O is quasi-compact.
Ex.1.18
i):⇒: Let x= px. If x is an closed set, then a prime ideal which includes px is only
px, i.e.px is a maximal.
⇐:trivial.
ii): x =∩ x⊆V V =∩ px∈∈∈∈V V=V (px ).
iii):y∈ x =V(px )⇔px⊆py.
iv):Let x, y∈X (x≠y),i.e. px≠py. px⊂py ∨ (¬ px⊂py and px≠py).
case of px⊂py : y∈V (px). Let open set O=X−V (py). Then (y∉O) ∧(x∈O) .
case of ¬ px⊂py and px≠py:Let open set O=X−V (px). Then (y∈O) ∧ (x∉O) .
Ex. 1.19
X(=Spec(A)) is irreducible
⇔ (for any open sets O≠∅,U≠∅ of X , O∩U≠∅).
⇔ (for any open sets Xf≠∅,Xg≠∅ of X, Xf∩Xg= Xfg ≠∅).
⇔ (Both of f,g are not nilpotent⇒fg is not nilpotent) by (ex. 1.17 ii).
⇔nilradical is a prime ideal.
Ex. 1.20
i): Let O1,O2 be open sets of X such that O1∩Y ≠∅ and O2∩Y ≠∅.
Y∩O1≠∅: if not, X−O1⊇Y. O1∩Y ≠∅⇒Y ⊆ Y ∩ (X−O1) ⊂ Y . It contradicts to that
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Y is the closure of Y. Y∩O1≠∅. Same as Y∩O2≠∅. Y is irreducible⇒Y∩O1∩O2≠∅.
⇒Y ∩O1∩O2≠∅.
ii):Let Σ be a family of irreducible subspaces of X. Let the partially order of Σ be
inclusion in the sense of set theory. Choose any ascending sequence Y of Σ . Let M
be the union of all members of Y. M is a member of Σ . Because if O1 ,O2 are
arbitrary open sets such that O1∩M≠∅ and O2∩M≠∅, O1 meats some Yα(∩O1≠∅) and
O2 meats some Yβ(∩O2≠∅). Let α be greater than β. Yα is irreducible ⇒
Yα∩O1∩O2≠∅ ⇒ M∩O1∩O2≠∅. Then M is irreducible. Then any irreducible subspace
is included by some maximal irreducible subspace.
iii):If F is irreducible, F (⊇F) is irreducible by (i). If F is a maximal irreducible
set, F =F. F is a closed set. x is irreducible for any x∈X. If not, open sets ∃O1,O2 ⊆ X,
x ∩O1≠∅, x ∩O2≠∅ and x ∩O1∩O2=∅. x∉O1∩O2. Assume x∉O1. x∈X−O1
⇒ x ⊆ X−O1⇒ x ∩O1=∅. Contradictory.
Any point of X is a member of irreducible sets which are included by the maximal
irreducible set. Then X is covered by maximal irreducible sets.
Let X be Hausdorff.
A maximal irreducible set consists of one point: For any point x of X, x is irreducible
by the above. If an irreducible set F includes two points x, y(x≠y), opensets ∃O1,O2 of X,
x∈O1, y∉O1, y∈O2, x∉O2 and O1∩O2=∅. Then F∩O1∩O2=∅, F is reducible. Then a
subspace which includes 2 or more points cannot be irreducible.
iv): if r(a)=p is a prime ideal, V(a) =V(r(a)) =V(p) = p is irreducible by iii).
If r(a) is not a prime ideal,V(a)=V(r(a)) is reducible. f,g∉r(a)⇒fg∈r(a). Prime ideals ∃p,
q in V(r(a)), f∉p, g∉q. Xf∩V(r(a))≠∅, Xg∩V(r(a))≠∅, V(r(a))∩Xf∩Xg=V(r(a))
∩Xfg=∅.
Then an irreducible subspace is V(p) where p is a prime ideal. q⊂p⇒V(p) ⊂ V(q). Then
the irreducible component is V(q) where q is a minimal prime ideal.
Ex.1.21
i):Let Xf be an basic open set and U=φ*−1(Xf). U=q∈Y | φ*
(q)∈Xf=q∈Y |
f∉φ−1(q)=q∈Y | φ (f)∉q=Yφ(f).
ii):y∈φ*−1(V(a))⇔ φ*
(y)∈V(a) ⇔ φ−1(qy)⊇a ⇔ qy⊇φ(a) ⇔ qy⊇Bφ(a)=a
e⇔y∈V(ae).
iii): Let E be any set of X and a=∩p∈E p.
)(aVE = :V(a) is a closed set including E. Suppose there exists a closed set F(=V(b))
such that V(a)⊃F⊇E. Then r(a)⊂r(b). ∃x∈r(b), x ∉r(a). ⇒ x∈p for ∀p ∈ F. If x∈pα for
∀pα ∈E, x∈a. It contradicts to x∉r(a). Then ∃ pα∈ E, x∉ pα. This means ¬F⊇E. It
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contradicts to assumption.
))(( bV*φ =V(∩b⊆qφ*(q)) =V(∩b⊆q(φ
-1 (q)))=V(φ-1
(∩b⊆qq))=V(φ-1 (r(b)))=V(r(b)
c)
=V(r(bc)) =V(b
c).
iv):If φ is surjective, A/Kerφ≅B.
φ*: Y → V(Ker φ)(⊆X ) is surjective.
φ* is a homeomorphism:
Since φ* is continuous, it is enough to show φ*
(Yg) is an open set of V(Ker φ). φ is
surjective⇒A/Kerφ ≅ B. The inclusion order of prime ideals in B is preserved in A by φ−1
and there is one to one correspondence between prime ideals of B and prime ideals of A
including Kerφ. Then an open set of Spec(B) is mapped to an open set of V(Kerφ) by φ*.
If φ : A →A/r(0), V(Ker φ)(=V(r(0)) is homeomorphic with Spec(A/r(0)). Since V(r(0))
=Spec(A), Spec(A) is homeomorphic with Spec(A/r(0)).
v):If φ*(Y) is not dense in X, ∃ Xf(≠∅⇔f∉rA(0)), φ*
(Y)∩Xf=∅. ⇒φ*(Y)⊆V((f))
⇒φ*−1(φ*
(Y))⊆φ*−1(V((f)))⇒Y⊆V(φ((f)))⇒Y=V(φ((f)))⇒φ(f)∈rB(0). It contradicts that φ
is injective.
More strictly, φ*(Y) is dense in X
⇔φ*(Y)∩Xf≠∅ for ∀Xf(≠∅)
⇔(f∉rA(0)⇒φ*(Y)∩Xf≠∅)
⇔(φ*(Y)∩Xf=∅⇒f∈rA(0))
⇔(φ*(Y)⊆V((f))⇒f∈rA(0))
⇔Kerφ⊆rA(0).
(the last⇔) ⇐: φ*(Y)⊆V((f)) ⇒f∈φ−1
(q) for ∀q∈Y. ⇒φ(f)∈rB(0). φn(f)=φ(f
n)=0. f
n=0
Kerφ⊆rA(0) ⇒ fnm
=0⇒f∈rA(0).
⇒:x∈Kerφ⇒φ(x)=0. ⇒x∈φ−1(0)⊆φ−1
(q) for ∀q∈Y.⇒φ*(Y)⊆V((x)) ⇒ x∈rA(0) by
assumption ⇒ Kerφ⊆rA(0).
vi): ψ°φ : A→B→C is ring homomorphism. (ψ°φ)*:Spec(C) →Spec(A) is defined by
following: q∈Spec(C) →(ψ°φ)−1
(q)= φ−1 (ψ−1
(q))= φ*°ψ
* (q)⇒(ψ °φ)
*=φ*
°ψ*.
vii):Spec(A)=0,p,Spec(B)=A/p×0,0×K.
φ*:Spec(B)→Spec(A). φ*
(A/p×0)=0. φ*(0×K)=p. φ*
is surjective and injective. The
topology of Spec(B) is discrete. 0 of Spec(A) is open, p is close. They are not
homeomorphic.
Ex.1.22
Let πi:A→Ai be projections. p is a prime ideal of A⇔ πi (p) is a prime ideal of Ai and
πj(p)=Aj for j (≠i). Let Xi=(A1,…,pi,…,An)|pi∈Spec(Ai). Spec(A)=∪i=1,..,nXi.
10
Xi∩Xj=∅(i≠j). πi is surjective⇒πi*: Spec(Ai)≅V(Kerπi) by (ex.1.21iv). Xi≅Spec(Ai)≅
V(Kerπi). Xi is closed. Xi=Spec(A)−∪j≠iXj is open.
i)⇒iii): X is disconnected⇒ ∃U1≠∅,∃U2≠∅ , U1∩U2=∅ and U1∪U2=X. Then U1 and U2
are both open and closed sets. Let U1=V(a1) and U2=V(a2). U1∩U2=V(a1)∩V(a2)
=V(a1+a2)=∅ ⇒a1+a2=(1). ⇒ ∃f∈a1,∃ g∈a2 , f+g=1. V(a1∩a2)= V(a1)∪V(a2)=X
⇒(a1∩a2)⊆r(0). ∃n>0 (fg)n=0. ∃x,(f+g)
n=f
n+g
n+xfg=1. xfg is nilpotent⇒f
n+g
n is a unit
(ex1.1). ∃y(fn+g
n)=1. yf
n≠0,1(If yf n
=0, ygn=1. g is a unit. It contradicts to V(a2)≠∅. If
yfn=1, it contradicts to V(a1)≠∅.). yf
n−y2f2n
=yfn(1−yf
n)=yf
nyg
n=0. yf
n(≠0,1) is an
idempotent. ygn=1−yf
n is an idempotent.
iii)⇒ii): e(∈A)is an idempotent (≠0,1) ⇒A=Ae+A(1−e). If x∈Ae∩A(1−e), x=ae=b(1−e).
xe=aee=b(1−e)e=b(e−e)=0. xe=aee=ae= x=0. A=Ae×A(1−e).
ii)⇒i): If A is a direct product of tow rings, Spec(A) is a disjoint union of two open(and
closed) sets. Then Spec(A) is disconnected.
If Spec(A) is disconnected, A contains idempotents(≠0,1) by above. A cannot be an local
ring by (ex.1.12).
Ex.1.23
i): A prime ideal is a maximal ideal in a Boolean ring by (ex. 1.11). Xf =m:a maximal
ideal | f∉m=m: a maximal ideal |m∋1−f =V((1−f)). Because any element is an
idempotent by (ex.1.11), i.e. (1−f)f=0∈m. 1−f∈m or f ∈m (not both satisfied, because
1∉m).
ii):it is enough to show the case of 2 generators.
Xf∪Xg=(X−V((f)))∪ (X−V((g)))=X− (V((f))∩ V((g)))=X−V((f) ∪ (g))=X−V((f,g)). (f,g) is
a principal ideal by (ex.1.11iii). Then (f,g)=(∃h) and Xf∪Xg=Xh.
iii):Suppose Y be an open and closed set. Since Y is open, Y=∪f Xf. Y is quasi-compact,
because of a closed set of quasi-compact space X. Y=∪f Xf is a finite union. There exists
h∈A such that Y=∪fXf=Xh by (ii).
iv):X is quasi-compact by Zariski topology. Let m and n be two different maximal
ideals. ∃m∈m−n. n∈Xm, m∉Xm. 1−m∉m⇒m∈X1−m. 0=m(1−m)∈n. m∉n⇒1−m∈n.
Then n∉X1−m. Xm∩ X1− m = X0 =∅. X is compact Hausdorff.
Ex.1.28
Let X, Y be affine varieties in kn, k
m and I(X), I(Y) be defining ideals and
P(X)=k[s1,…,sn]/I(X) , P(Y) =k[t1,…,tm]/I(Y) be coordinate rings respectively.
(1) regular mapping φ induces a homomorphism ψ:P(Y)→P(X):
Let φ be a regular mapping: X→Y, which is defined as φ:x=(x1,…,xn) ∈X→ (f1(x),
11
…,fm(x)) ∈Y where f1, …,fm∈k[s1,…,sn]. Let η∈k[t1,…,tm]. η°φ denotes η∈k[t1,…,tm]
whose ti is replaced by fi(x), so η°φ∈k[s1,…,sn]. It is clear that η→η°φ is a
homomorphism by η+η’°φ=η°φ+η’°φ and ηη’°φ= (η°φ)(η’°φ). To prove ψ:P(Y) →P(X)
is a homomorphism which is induced from η→η°φ, it is needed that if ξ∈I(Y) , ξ°φ∈
I(X). Since φ is a regular mapping, φ (x) ∈Y for ∀x∈X. From “ξ∈I(Y) ⇒ξ°φ
(x)=0” ,ξ°φ∈ I(X). Then ψ (0)=0.
(2)the correspondence from a regular mapping to a homomorphism: P(Y) →P(X) is
injective:
Let φ:x=(x1,…,xn) ∈ X→ (f1(x),…,fm(x)) ∈Y,f1,…,fm∈k[s1,…,sn ] and φ’:x=(x1,…,xn) ∈
X→ (f ’1(x),…,f ’m(x)) ∈Y, f ’1,…,f ’m∈k[s1,…,sn ]. Let ψ and ψ’ be homomorhisms
induced by φ and φ’ respectively and ψ = ψ’. For ∀ η ∈P(Y), let η∈k[t1,…,tm] be an
inverse image of η . η°φ≡η°φ’ mod I(X). Replacing η by t i (1≤i≤m), we get f i ≡ f ’ i
(1≤i≤m) mod I(X). ⇒fi−f ’i∈ I(X) ⇒ φ=φ’ on X. This means injecivity of this
correspondence.
(3) the correspondence from a regular mapping to a homomorphism: P(Y) →P(X) is
surjecive:
Let ψ:P(Y) →P(X) be a homomorphism. ψ ( t i) = f i∈P(X) (1≤i≤m). let f i ∈k[s1,…, sn]
be any inverse image of f i. Let φ:(s1,…,sn) → (f1(s1,…,sn),…,fm(s1,…,sn)) be an
polynomial mapping. Let η∈k[t1,…,tm] be an inverse image of η ∈P(Y). By φ, we get
η°φ∈k[s1,…,sn]. Since ψ(0)=0, ∀ξ∈I(Y)⇒ξ°φ∈I(X) ⇒ξ°φ(x)=0 for ∀x∈X.⇒φ(x)∈Y. φ
is a regular mapping.
Chapter 2
Ex.2.1
(m,n)=1⇒ ∃ a,b∈Z ,am+bn=1.
(x mod m) ⊗(y mod n)=(am+bn)((x mod m) ⊗(y mod n))=(amx mod m)⊗ (y mod n)+(x
mod m) ⊗ (bny mod n)=0+0=0.
Ex.2.2
0→a →µ A→A/a→0 (µ:the embedding) is exact. Let M be an A-module.
a⊗ M → ⊗1µ A⊗ M →A/a⊗ M →0 is exact.
(A⊗M)/µ⊗1(a⊗M )≅ (A/a)⊗M. A⊗M≅M. Since µ⊗1(a⊗M ) is subgroup of A⊗M, it is
isomorphic to aM. (A/a) ⊗M≅M/aM.
Ex.2.4
12
Let NNf→′→0 be a exact sequence of A-modules.
⇒ : 0→ Mi⊗N’ ⊕(⊕i'∈≠i Mi’) ⊗N’ →Mi⊗N ⊕(⊕i'∈≠i Mi’) ⊗N is exact by(2.14).
⇒0→Mi⊗N’ →Mi⊗N is exact. Mi is Flat. M= iIi M∈⊕ is flat⇒
NMNM f ⊗→′⊗→ ⊗10 is exact. If ( f⊗1 )(x)=0 for any element x= jnj ij nm '⊗∑ <
of 'NM i ⊗ , x=0. Then NMNM i
f
i ⊗→′⊗→ ⊗10 is exact. Mi is flat.
⇐ : Let M= iIi M∈⊕ and each Mi be flat. Let x= )',(,...,1
NnMmnm kkk
nk
k ∈′∈′⊗∑=
. Each
mk is a finite direct sum of elements of Mi. By reindexing, x=⊕i (Σj<n(i)mi,j⊗n’i,j) . If
1⊗f(x)=0, x=⊕i (Σj<n(i)mi,j⊗ f(n’i,j))=0. Then (Σj<n(i)mi,j⊗f(n’i,j))=0 for each i. Since Mi is
flat, (Σj<n(i)mi,j⊗n’i,j)=0. Then x=0. Then M is flat.
Ex.2.5
A[x]=⊕i≥0Axi as A-module. Ax
i≅A⇒Axi is flat. A[x] is a flat A-module by (ex.2.4). A[x] is
a flat A-algebra.
Ex.2.6
Let φ:A[x]×M→M[x] be Σ1≤i≤n (f i(x),mi) →Σ1≤i≤n (mifi(x)). φ is A-multi-linear. ∃φ’:
A[x]⊗M[x]→M[x] and ∃ψ:A[x]×M→A[x] ⊗M, φ’ψ=φ.
φ is surjective⇒φ’is surjective.
φ’is injective: Suppose φ’(Σ1≤j≤m f j (x) ⊗mj)=0. By reindexing, Σ1≤j≤m fj(x)⊗mj=
Σ1≤i≤nxi⊗m’i. φ’(Σ1≤i≤n x
i⊗m’i)=0.direct sum⇒ m’i=0(1≤i≤n). ⇒Σ1≤j≤m fj(x)⊗mj=
Σ1≤i≤n(xi⊗m’i) =0⇒φ’ is injective.
Ex.2.7
Let f(x)= fnxn +…+f0, g(x)= gmx
m +…+g0∈A[x] and f(x), g(x)∉ p[x]. Let i be the highest
degrees of f(x) such that fr∉p and j be of g(x) as i of f(x). c=Σs+t =i+j figj, the coefficient of
degree i+j of f(x)g(x), c ∉ p, because frgs∉p and other figj∈p. Then f(x)g(x)∉ p[x].
An example of m[x] is not maximal:
In Z[x], (p)[x] ⊂(p,x) ≠(1), for any prime number p. (p) is a maximal ideal in Z.
Ex.2.8
i):Let 0→P’→P be any exact sequence of A-modules. If M is A-flat, 0→P’ ⊗ M →P⊗
M is exact. If N is A-flat, 0→P’ ⊗ M⊗ N →P⊗ M⊗ N is exact. M⊗N is A-flat by (2.19).
13
ii): Let 0→P’→P be any exact sequence of A-modules. If B is a flat A-algebra, B is a
flat A-module. 0→P’ ⊗A B →P⊗ A B is exact. These are B-modules. If N is a flat
B-module, 0→P’ ⊗A B⊗B N →P⊗ A B⊗B N is exact. B⊗BN≅N⇒0→P’ ⊗A N →P⊗ A N
is exact. N is also A-module⇒N is a flat A-module.
Ex.2.9
Let 0→M’ →µ M →ε M”→0 be an exact sequence of A-modules. Let nx,...,x1 be
generators of M”. ∃xi∈M, )( ii xx ε= for i=1,…,n. For ∀ z∈M, ii xaz Σ=)(ε (ai∈A).
ε(z−Σaixi)=0⇒∃y’∈M’, z−Σaixi=µ(y’). Let y1,…,ym be generators of M’. z−Σaixi=
Σµ(ajyj)=ajΣµ(yj) (aj∈A). M is generated by x1 ,…, xn,µ(y1) ,…,µ(ym) over A.
Ex.2.10
Let x1,...,xn be generators of N. If :µ M/aM→N/aN is surjective, ∃mi ∈ M ,µ(mi +aM)⊆
xi +aN for each xi. µ(aM)⊆ aN⇒
µ(mi ) = xi + Σaijxj=Σj(δij+aij ) xj (aij∈a,1≤i,j≤n).
By multiplying adjugate matrix, we get a diagonal matrix of which every diagonal
element is a determinant |ijδ +aij|. We get xj(1+a)=Σbijµ(mi) (bij∈A, a∈a) from the
determinant. a⊆ J(A) ⇒1+a is a unit. Then xj∈µ(M). µ is surjective.
Ex.2.11
If φ:Am→A
n is isomorphic, 0→0→A
m→ An→0 is exact. Let m be a maximal ideal of A.
By tensoring with A/m, 0→A/m⊗Am→A/m⊗A
n→0 is exact. 1⊗φ: Am→A
n is an
isomorphism. Since A/m is a field, A/m⊗Am and A/m⊗A
n are vector spaces of degree
m,n respectively (2.14iii). Isomorphic vectorspaces of finite dimension have same
degree. Then m=n.
If φ is surjective, 0→Kerφ→Am→A
n→0 is exact. A/m⊗ Kerφ→A/m⊗Am→A/m⊗A
n→0
is exact. Since A/m⊗Am→A/m⊗A
n is surjective on a vectorspace of degree n from a
vectorspace of degree m, then m≥n.
An example of m>n where φ is injective:
Let Bi=0,1 be rings (i>0) and A=ΠBi. Define φ:A2→A by φ(a,b)=(a0,b0,a1,b1,…,
ai,bi,…), for a=(a0,…,ai,…) and b=(b0,…,bi,…)∈A. φ is an injective homomorphism.
Ex.2.12
14
Let ei=(0,…,0,1,0,…,0) be basis of An and φ: M→A
n and ui (∈M) be such that φ(ui)=ei.
Define ψ:An→M by ψ(ei)=ui. Since φ°ψ:A
n→M→An
is the identity map of An,
0→Kerφ→M→An→0 splits. Let M’=ψ(A
n). Then M≅Kerφ⊕M’. Since M is finitely
generated, M/M’≅Kerφ is finitely generated.
Ex.2.13
NB = B⊗A N. Define g:N→NB by g(y)=1⊗A y. Define p: NB→N by p(b⊗y)=by.
g is injective: 1⊗Ay=0⇒ y=0, because 1 is a unit. p is surjective⇒following is exact.
0→Ker(p) →NB → pN→0. (*)
p°g:N →gNB →p
N is the identity map of N⇒(*) splits. NB≅N⊕Ker(p). N≅g(N)
⇒NB≅g(N) ⊕Ker(p).
Ex.2.15
Let I be a direct set and C=(⊕i∈I Mi). Let D be a submodule of C generated
byxi−µij(xi)xi∈Mi.
For ∀x∈M, ∃xαj∈Mαj(j=1,...,n), x=µ(Σj=1,...,n xαj). ∃i≥αj for j=1,...,n. By adding Σj=1,...,n
µαji(xαj) − xαj ∈ D to Σj=1,...,n xαj, we get x=µ i(xi) where xi =Σj=1,...,nµαji (xαj) ∈Mi.
µ(xi −xj)=0⇔∃ k ≥i,j,µik (xi) =µjk (xj):
⇐:Let xi ∈Mi and xj∈Mj. ∃ k ≥i,j,µik(xi)=µjk(xj)⇒xi −xj=xi −µik(xi) −( xj−µjk(xj)) ∈ D⇒
µ(xi −xj)=0.
⇒:µ(xi −xj)=0⇒xi −xj∈D. µ(xi −xj)=µi(xi) −µj (xj). ∃ k ≥i,j,µi(xi) −µj (xj) =µk(µik(xi)−µjk
(xj))=0. µik(xi)−µjk (xj) ∈ Mk ∩D⇒∃xk∈Mk, µik(xi)−µjk (xj) = xk−µkk(xk)=0.
Then µi(xi)=0⇒For ∀k∈I,0k (∈Mk ),µ(xi −0k)=0. ∃j(≥k,i),µij(xi)= µkj(0k) =0.
Ex.2.16
For each i∈I, let αi: Mi→N be A-homomorphism such that αi=αjµij. Then there exists
an A-homomorphism from the direct sum C of Mi to N such that its restriction to Mi
equals αi. We name this homomorphism φ (C→N). As αi=αjµij holds for i≤j,
φ((Σi(xi−µij(xi)))=Σi(αi(xi)−αj°µij(xi))=0. Then φ is factored as C →C/D(=M)→N. We
name the latter homomorphism α (M→N). Following is commutative.
↑ →↓
→→i
i NMM i
α
αµ
Then αµi=αi. The uniqueness of α is clear by definition.
Ex.2.17
For ∀Mi and Mj, Mi+Mj⊆∃Mk. Then I is a directed set. For i≤j, let µij: Mi⊆Mj be
15
embedding. µij satisfies axioms of (ex.2.14). M=(Mi, µij) is a direct system. For ∀i∈ I,
αi : Mi→ Σk∈I Mk satisfies αi=αj°µij (i≤j). Let M be the direct limit of M.⇒There exists
the unique homomorphism α: M→Σk∈I Mk and α i =α°µi by (ex.2.16).
α is surjective: C=(⊕i∈I Mi) →Σk∈I Mk is surjective⇒α is surjective(ex.2.16).
α is injective: x∈M⇒ x=µ(Σi=i1,...,in x i) (x i∈Mj). ∃j(≥i1,…,in) ∈I, ∃xj∈Mj ,x=µj(xj).
α(x)=α°µj(xj)=αj(xj). α(x)=0⇒αj(xj)=0. αj is embedding⇒xj=0⇒x=0.
Then α is an isomorphism.
Let N be any A-module and xλλ be its generators. N is the direct limit of finitely
generated submodules of M over A. For ∀x∈N,x=Σaλxλ (finite sum)⇒∃ Nβ which is
finitely generated and includes x. Let Nββ be a family of all finitely generated
submodules of N. ⇒It holds that Nβ+ Nβ’ ⊆∃Nβ“ ⇒By directed set I with embedding
µββ’:Mβ⊆Mβ’, N= (Nβ,µββ’) is a direct system and N is its direct limit.
Ex.2.18
Let M=(Mi,µij) be a direct system on I and M be its direct limit. Let N=(Ni,νij) be a
direct system on I and N be its direct limit. Let µi: Mi→M , νi:N i→N be their
homomorhisms respectively. The following is commutative.
ji
ji
ji
NN
MM
ij
ij
→
φ↓φ↓
→
ν
µ
νi°φi:Mi→N is a homomorhism from each member of M to N. νi°φi satisfies the
condition of αi in ex.2.16 ( i≤j⇒νi°φi=νj°νij °φi =νj°φj°µij). Then A-homomorphism
∃φ:M→N,νi°φi=φ°µi (ex.2.16).
Ex.2.19
Let M=(Mi, µij), N =(Ni, νij), P=(Pi, πij) be direct systems on directed set I. Let M, N
and P be direct limits of M, N and P respectively and µi: Mi→M, νi:N i→N, πi: Pi →P
be homomorphisms from each member to the direct limit respectively.
The following is commutative. φj°µij =νij°φi, ψ j °νij=πij°ψ i.
16
ji
ji
ji
ji
ji
PP
NN
MM
ij
ij
ij
→
ψ↓ψ↓
→
φ↓φ↓
→
π
ν
µ
∃φ= limφi: M→N, νi°φi=φ°µi, ∃ψ= limψ i :N→P, πi°ψ i =ψ°νi(ex.2.18).
Mi → iφ Ni → iψPi is exact for ∀i∈I.
Im(φ)⊆Ker(ψ):
For ∀x∈M, ∃i∈I,∃xi∈Mi,µi(xi)=x(ex.2.15).ψ°φ(x) =ψ°φ°µi(xi)= ψ°νi°φi(xi)=πi°ψ i °φi(xi).
ψ i °φi =0⇒ψ°φ(x) =0.
Im(φ)⊇Ker(ψ):
Let y∈ Ker(ψ).⇒ψ (y)=0⇒∃i∈I,∃yi∈Ni, y=νi(yi) (ex.2.15).ψ (y)= ψ °νi(yi)=πi°ψ i (yi) =0.
πi(ψ i (yi)) =0⇒∃j≥i,πij(ψ i (yi))=0(ex.2.15).πij°ψ i (yi)= ψ j°νij(yi). Set νij(yi)=yj. ψj (yj)=0.
Mj→Nj→Pj is exact⇒∃xj∈Mj,φj (xj)= yj. φ°µj(xj) =νj°φj (xj) =νj(yj)= νj°νij(yi)= νi (yi)=y.
µj (xj) ∈M ⇒y∈Im(φ).
Ex.2.20
Let (Mi×N,(µij,1)) be a direct system on a directed set I and M be its direct limit and
µi:Mi→M be a homomorphism from a member of direct system to the direct limit.
(Mi×N,( µij,1)) is a direct system on I. Set νij=µij⊗1. (µjk⊗1)°(µij⊗1) = (µjk°µij)⊗1=
µik⊗1⇒νik=νjk°νij. (Mi⊗N, νij) is a direct system on I. µi⊗1:Mi×N→M⊗N is factored by
∃gi:Mi×N→Mi⊗N and ∃hi:Mi⊗N→M⊗N and µi⊗1=hi°gi. We get the following
diagrams.
←⊗→
↓↓
⊗ →⊗
↓↓
× →×
ν=⊗µ
µ
NM
hh
NMNM
gg
NMNM
ji
j
)(
i
ji
j
),(
i
ijij
ij
1
1
NM
NMNM
ji
j
),(
i
ij
⊗
⊗µ↓⊗µ↓
× →×µ
11
1
The right side diagram is commutative. µi⊗1=µj⊗1°(µij,1). The upper part of left side
diagram is commutative by νij °gi= gj°(µij,1). gi is surjective⇒ The lower part of left side
diagram is commutative⇒hi=hj°νij.
From upper part of the left ∃g= lim gi:M×N→P,g° (µi, 1) =νi°gi(ex.2.18).
g is bi-linear ⇒∃φ:M⊗N→P, ∃g’:M×N→M⊗N, g=φ°g’(2.12).
17
From lower part of the left ∃ψ= lim hi:P→M⊗N ,νi : Mi⊗N→P,hi=ψ°νi (ex.2.16).
In the follwing we use mi ∈ Mi ,n∈ N for simplicity because elements are finite sum of
those.
φ°ψ=1: Let x∈P. x= νi (mi⊗n) =νi °gi(mi,n)= g° (µi,1) (mi,n)= g(µi(mi),n)
= φ°g’(µi(mi),n)= φ (µi(mi) ⊗n). ψ(x)=ψ°νi(mi⊗n)=hi(mi⊗n)=higi(mi, n)=(µi⊗1) (mi, n)=
µi (mi)⊗n ⇒φ° ψ (x) =φ (µi(mi) ⊗ni) =x.
ψ°φ=1: Let y∈M⊗N. y=m⊗n (m∈M). y=g’(m,n)=g’(µi(mi),n)=g’°(µi,1)(mi,n).
ψ°φ(y)=ψ°φ°g’°(µi,1)(mi,n)=ψ°g°(µi,1)(mi,n)=ψ°νi°gi(mi,n) = hi °gi(mi,n)=µi⊗1(mi,n) =
µi(mi)⊗1 =m⊗n=y.
Ex.2.21
Let x,y∈A. ∃i,j∈I,∃xi∈Ai,∃yj∈Aj,x=αi(xi) and y=αj(yj) as modules by (ex. 2.15).
∃k(≥i,j), x=αi(xi) =αkαik(xi),y=αj(yj) =αkαjk(yj). αik(xi), αjk(yj)∈Ak ⇒αik(xi)αjk(yj)∈Ak
because Ak is a ring. Define z=αk(αik(xi)αjk(yj))∈A as modules. Consider formally αk as
a ring homomorphism. We can define multiplication (•) in A as following. A inherites
multiplication of Ai.
z=αk(αik(xi)αjk(yj))=αk(αik(xi))•αk(αjk(yj)) = x•y
This mulplication does not depend on representatives:
Let xi’ and yj’ be another representatives of x and y respectively. For ∃l(≥i’,j’,k), αil(xi)
=αi’l(xi’),αjl(y)=αj’l(yj’)⇒αil(xi)αjl(yj)=αi’l(xi’)αj’l(yj’)∈Al⇒
αk(αik(xi)αjk(yj))= αlαkl(αik(xi)αjk(yj))= αl (αil(xi)αjl(yj)) =αlαi’l(xi’)αj’l(yj’).
There exists the ring structure in A. αi : Ai→A are ring homomorphisms.
If A=0, ∃ i , αi(1)=0. By (ex.2.15) for ∃≥i, αij(1)=0. ⇒ Aj =0.
Ex.2.22
(Ri,αij) is a direct system(xi∈Ri ⇒∃n>0, xin=0. αij (xi
n)=αij
n(xi)=0⇒αij (xi)∈Rj. Other
conditions of direct system are satisfied because (Ai,αij) satisfies.) Let A= lim Ai and
R= lim Ri. 0→Ri→Ai is exact for each i. 0→R→A is exact(ex.2.19).
R⊆rA(0): x∈R⇒∃i∈I,∃xi∈Ri,x=αi(xi). xi is nilpotent⇒x is nilpotent. 0→R→A is
exact⇒x∈rA(0).
R ⊇rA(0): x∈rA(0) ⇒∃i∈I, ∃xi∈Ai,x=αi(xi) (ex.2.15). ∃n>0,xn=αi(xi
n) =0. xi
n∈Ai,∃j≥i ,
αij(x in) =0 (ex.2.15). Let xj=αij(xi). xj
n=0⇒xj∈Rj. x=αi(xi)= αjαij(xi) =αj(xj).⇒ x∈R.
Suppose xy=0 for some x,y(≠0) ∈ A. ∃xi ∈ Ai , ∃yj ∈ Aj, ∃k≥i,j ,xy = αk(αik(xi)αjk(yj))
(ex.2.21). On the otherhand, xy=0⇒∃k’,xy=αk’(0)(ex.2.15) as modules.
∃l≥k,k’,αkl(αik(xi)αjk(yj))=αk’l(0)⇒ αil(xi)αjl(yj)= 0(∈Al). αil(xi) and αjl(yj) are not 0
because they are inverse image of x,y respectively.⇒ Al is not a domain.
18
Ex.2.23
Let J⊆J’ . Define A-homomorphism µJJ’: BJ →BJ’ by µJJ’(b)=b⊗1⊗… ⊗ 1 for ∀ b∈BJ.
(BJ, µJJ’) satisfies axioms of (ex.2.14)⇒ (BJ, µJJ’) is a direct system. Let B be its direct
limit. It is enough to consider finite tensor product as (ex.2.17). Since B is A-module
and BJ has ring structure, B is a ring by (ex.2.21). B is an A-algebra. A-homomorphism
µJ:BJ →B is an A- ring homomorphism by (ex.2.21).
Ex.2.24
i)⇒ii): Let M be a flat A-module. Following is a tensor product of a free resolution of
any A-module N with M.
... →Fn+1 → +δ 1n Fn →δn Fn-1 → −δ 1n ...→F1 →δ1 F0→N→0
is exact at each n.
So TornA(M,N)=0 for any n(>0). im(δn+1)=Ker(δn). Let Rn= Ker(δn) .
0→R n→Fn→Rn-1→0 is exact. M is Flat⇒following are flat.
0→M⊗Rn+1→M⊗Fn+1→M⊗Rn→0
0→M⊗Rn→M⊗Fn→M⊗Rn-1→0
0→M⊗Rn-1→M⊗Fn-1→M⊗Rn-2→0
Ker(M⊗Fn→M⊗Fn-1) =M⊗Rn,Im(M⊗Fn+1→M⊗Fn)=M⊗Rn.
Then TornA(M,N)=0 (∀n>0).
ii) ⇒iii):trivial.
iii)⇒i): By tensoring a free resolution of any exact sequence 0→N’ →N→N“→0 with
M, we get a long exact sequence. Following is its part.
... →Torn A
(M, N’) → Torn A
(M,N) →Torn A
(M, N“) →Torn-1 A
(M, N’) →…
... →Tor1A(M, N“) →M⊗N’ →M⊗N→M⊗N“ →0
Tor1A(M,N“)=0⇒the above sequence is exact. ⇒ M is flat.
Ex.2.25
Let 0→N’→N→N“→0 be exact and N“ be flat. Tensoring a free resolution of
0→N’→N→N“→0 with any flat A-module M, we obtain a long exact sequence as
following.
…→Tor2A(N“,M) →Tor1
A(N’,M) →Tor1
A(N,M) →Tor1
A(N“,M) →…
N“ is flat⇒0→Tor1A(N’,M) →Tor1
A(N,M) →0 is exact. Tor1
A(N’,M)=0⇔Tor1
A(N,M)
=0 . N’ is flat ⇔N is flat by (ex.2.24).
Ex.2.26
19
⇒: Let a be any finitely generated ideal. N is flat⇒Tor1A(A/a,N)=0 by (ex.2.24).
⇐:
Lemma 1: If Tor1A(M,N)=0 for any finitely generated A-module M, N is flat.
Proof:Let 00 →′′→→′→ LLL βα be any exact sequence of A-modules. Let
l1,...,li,... be generators of L and I be a family of finite subsets of l1,...,li,.... Introduce
the order into I by inclusion in the sense of set theory. Then I is a directed set. For λ
(∈I), let Lλ be a submodule of L generated by λ. For λ’(≥λ), there exists embedding map
µλλ’: Lλ → Lλ’ and (Lλ,µλλ’) is a direct system and L is its direct limit(ex.2.17). Denote
α−1(Lλ) by L’λ. L’λ is a submodule of L’. There exists embedding µ’λλ’: L’λ → L’λ’ and
(L’λ,µ’λλ’) is a direct system over I and L’ is its direct limit. Denote β(Lλ) =Lλ/α(L’λ) by
L“λ. L“λ is a finitely generated submodule of L“. There exists a homomorphism µ“λλ’ :
L“λ → L“λ’ (a∈Lλ⇒a+α(L’λ)→a+α(L’λ’)) and (L“λ,µ“λλ’) is a direct sytem over I and
L“ is its direct limit. For ∀λ∈I, 00 →′′→→′→ λβ
λα
λλλ LLL is a exact sequence.
By the hypothesis, Tor1A(L“λ,N)=0⇒ 0→ L’λ⊗ N →Lλ⊗ N →L“λ⊗ N →0 is exact.
0→ lim (L’λ⊗ N) → lim (Lλ⊗N)→ lim (L“λ⊗N)→0 is exact (ex.2.19).
0→( lim L’λ)⊗ N→( lim Lλ)⊗N→( lim L“λ)⊗N→0 is exact (ex.2.20).
0→L’⊗N →L⊗N→L“⊗N→0 is exact. Then N is flat.♦
Lemma 2: Tor1A(L,N)=0 for any A-module L generated by a single element, N is flat.
Proof:Let M be any finitely generated A-module and x1,...,xn be its generators. Let Mi be
an A-module generated by x1,...,xi. From a free resolution of 0→Mi-1 →Mi →Mi /Mi-1→0
which is exact, we get the following long exact sequence.
... →Tor2A(Mi/Mi-1,N)→Tor1
A(Mi-1,N) →Tor1
A(Mi,N)→Tor1
A(Mi /Mi-1,N)→...
By the hypothesis, Tor1A(Mi/Mi-1,N)=0 and Tor1
A(M1,N)=0. Inductively Tor1
A(Mi,N)=0
for any i. Then Tor1A(Mn,N)=Tor1
A(M,N) =0. By lemma 1, N is flat. ♦
Then it is enough to show that Tor1A(L,N) =0 for A-module L(≅A/ann(x)) which is
generated by a single element x. Then if Tor1A(A/a,N)=0 for any ideal a, N is flat by
lemma 2. To complete the proof, it is enough to show following lemma.
Lemma 3:If Tor1A(A/b,N)=0 for any finitely generated ideal b, Tor1
A(A/a,N)=0 for any
ideal a.
Let x1,...,xi,... be generators of a and I be a family of finite subsets of x1, ...,xi, ....
Introduce the order into I by inclusion in the sense of set theory. Then I is a directed set.
Let aλ be an ideal generated by λ (∈I). For λ’(≥λ), there exists embedding map µλλ’ : a λ
→a λ’ and (aλ,µλλ’) is a direct system over I. Define νλλ’ : A/aλ→ A/aλ’ by a+aλ→ a+ aλ’ .
(A/aλ,ν λλ’) is a direct system over I. For any λ(∈I), 0→aλ→A→A/aλ→0 is exact. Since
aλ is finitely generated, Tor1A(A/aλ,N)=0 by the hypothesis. Then 0→aλ ⊗ N→A ⊗ N→
20
A/aλ ⊗ N→0 is exact.
0→ ( lim aλ)⊗N→ ( lim A) ⊗N→ ( lim A/aλ)⊗N→0 is exact(ex.2.20).
0→a⊗N→A⊗N→A/a⊗N→0 is exact.
Then Tor1A(A/a,N) =0 for any ideal a. ♦
Ex.2.27
i) ⇒ii):(x) ⊗M≅ (x)M for any absolutely flat A-module M:
A absolutely-flat⇒M is Flat. For ∀x∈A,0→ (x) →A→A/(x) →0 is exact⇒0→ (x)
⊗M→A⊗M→ (A/(x))⊗M→0 is exact. (A/(x)) ⊗M≅M/(x)M by (ex.2.2).
0))/(()(0
0))/(()(0
→⊗→→→
≅↓↓≅α↓
→⊗→⊗→⊗→
MxMMMx
MxAMAMx
By (2.10) α is an isomprhism ⇒(x)⊗M≅(x)M.
(x) is A-module⇒(x) is Flat.
0→ (x)⊗(x) →A⊗(x) → (A/(x))⊗(x) →0 is exact. (x)⊗M≅(x)M ⇒
0→ (x2) → (x) →0→0 is exact. (x) = (x
2).
ii)⇒iii):For any x∈A, (x)=(x2).⇒∃a∈A,x=ax
2. Set e=ax. e
2=a
2x
2=ax=e. ⇒e is an
idempotent. (e)⊆(x). ex=ax2=x. (e)⊇(x). (e) =(x) . Then a principal ideal is generated by
an idempotent.
Finitely generated ideal is principal: it is enough to show the case of 2 terms. Let (e,f) be
an ideal. As (e) and ( f) are principal, we can regard e2=e and f
2=f by above. Set x=e’+f’
−e’f’ . ⇒ x is an idempotent. (e,f)⊇(x) is clear. e=ex. f=fx. (e,f)=(x).
Let e be an idempotent. A= A(e)+A(1−e). A(e)∩A(1−e)∋y⇒y=ae=b(1−e) ⇒ye = b(1−e)e
=0⇒y=0. ⇒ A= A(e)⊕ A(1−e).
iii) ⇒i): it is enough to show that any A-module N is flat.
N is flat ⇔ Tor1A(A/a,N)=0 for any finitely generated ideal a by (ex.2.26). By
hypotheses of (iii), A=a⊕b. 0→a→A→A/a→0 is exact⇒ Tor1A(A/a,N)
→a⊗N→A⊗N→A/a⊗N→0 is exact, i.e.,Tor1A(A/a,N)→a⊗N→a⊗N⊕b⊗N→b⊗N→0
is exact.⇒Tor1A(A/a,N)=0.
Ex.2.28
A Boolean ring is absolutely flat, because it satisfies (ex.2.27ii). For ∀x∈A(in ex.1.7),
∃n(>1), that xn=x. ⇒A=(x)⊕(x
n-1−1),( (x) and (xn−1−1) are coprime, and z ∈(x)∩(x
n−1−1)
⇒ z =ax=b(xn-1−1) ⇒ zx =0, zx
n-1=ax=0⇒ z=0)
If finitely generated ideals are principal ideals, the condition of (ex.2.27iii) is satisfied:
21
it is enough to show the case of 2 generators. Let (x,y) be an ideal and xn=x ,y
m=y for
∃n,m(>1). (x,y) = (xn-1
+ym-1−x
n-1 y
m-1 ).⇒ (x,y) is a principal ideal.
Let A be absolutely flat, f be a ring homomorphism and (y),y∈f(A) be a principal ideal of
f(A). ∃x∈A,y=f(x) and y2=f(x
2). (y)=f((x))=f((x
2))=(y
2). f(A) is absolutely flat by(ex.2.27).
Let A be an absolutely flat local ring and m be its maximal ideal. If m≠0, ∃x(≠0) ∈ m.
⇒ (x) is a direct summand of A. ⇒ A contains more than one maximal ideal. It is
contradictory. Then m=0 ⇒ A is a field.
Let A be absolutely flat. For any non unit x ∈ A, (x)=(x2) ⇒ x=ax
2. ⇒x(1−ax)=0. Since
x is not unit,1−ax≠0. ⇒ x is a zero divisor.
Chapter 3
Ex.3.1
⇒: Let x1,…,xn be generators of M. If S−1
M=0, ∃ si ∈ S, sixi=0 for i= 1,…,n. Let
s=s1…sn. For ∀m∈M, m=Σaixi. sm=Σs1…sn aixi=0. sM=0.
⇐:trivial.
Ex.3.2
x∈S−1a. It is enough to show that 1+yx is a unit for ∀y∈S
−1A. Let x=a/(1+a’) and
y=r/(1+a”) where r∈A and a,a’,a”∈a. 1+yx=1+ar/(1+a’)(1+a”)=((1+a’)(1+a”)+ar)/
(1+a’)(1+a”)=s/t,(s,t∈S). 1+yx is a unit. ⇒ S−1a⊆J(S
−1A).
Proof of (2.5): Let M be a finitely generated A-module, a be an ideal of A and M=aM.
Let S=1+a. S is a multiplicative closed set. S−1
M =(S−1a) S
−1M. S
−1a⊆J(S
−1A). S
−1M is a
finitely generated S−1
A-module⇒S−1
M=0. ∃s∈S=1+a such that sM=0 by (ex.3.1.) Then
s≡1(mod a).
Ex.3.3
ST=st | s∈S,t∈T is a multiplicative closed set because A is a commutative ring. Let
f: A→S−1
A be a homomorphism defined by f(a)=a/1 for a∈A. Let f (T)=U. U=t/1∈
S−1
A | t∈T. Let g: U−1
(S−1
A)→(ST)−1
A be a homomorphism defined by g((a/s)/(t/1))=
a/st.
g is clearly surjective.
g is injective: g((a/s)/(t/1))=a/st=0⇒∃s’t’∈ST,as’t’=0.
(a/s)/(t/1) = (as’/ss’)(t’/1)/(t/1)(t’/1) = (as’t’/ss’)/(tt’/1)=0.
22
Ex.3.4
T(=f(S)) is a multiplicatively closed set (possively including 0). The multiplication of
elements of A and B is defined by a⋅b=f(a)b. Let a/s∈S−1
A and b/t∈T−1
B. Define S−1
f :
S−1
A→T−1
B by S−1
f(a/s)=f(a)/f(s). a/s⋅b/t=S−1
f(a/s)(b/t)=(f(a)/f(s))(b/t)=f(a)b/f(s)t ∈
T−1
B. Then T−1
B is a S−1
A-module.
S−1
B is a set of classes by the relation:(b,s)≡(b’,s’)⇔∃s”∈S,(s’⋅b−s⋅b’)⋅s”=0. Denote
(b,s)=b/s. The product of a/s∈S−1
A and b/s’∈S−1
B can be defined as a⋅b/ss’. S−1
B is an
S−1
A-module.
Define g : S−1
B→T−1
B by b/s→ b/f(s). g is clearly an S−1
A-homomorphism.
g is clearly surjective .
g is injective : g(b/s)=b/f(s)=0⇒ ∃s’∈S , f(s’)b=s’⋅b=0. ⇒ b/s=0 in S−1
B.
Ex.3.5
Let R=rA(0). The nilradical of Ap is Rp by (3.12). If Rp=0 for prime ideal ∀p, R=0 (3.8).
Counter example: Let e(≠0,1) be idempotent and K be a field. A=Ke⊕K(1−e) is not an
integral domain. Prime ideals of A are p=Ke⊕0 and q=0 ⊕ K(1−e). Elements of A−p are
of the form k’e+k(1−e) (k≠0). Ap=b(1−e)/(k’e+k(1−e)) | b,k,k’ ∈K,k≠0 is an integral
domain. Also Aq is an integral domain.
Ex.3.6
1∈Σ⇒Σ ≠∅. Let the inclusion in the sense of set theory be partilly order of Σ. A
union of all members of any chain in Σ belongs to Σ. By Zorn’s lemma, there exists a
maximal element S in Σ.
⇒: p=A−S is a prime ideal:
p is an ideal: x∈p⇒the multiplicatively closed set genereated by S∪x includes 0
because of maximality of S. ∃s∈S,∃n>0,sxn=0. Ax∩S≠∅⇒∃ax=s’,ss’
n= s(ax)
n=0∈S. It
is contradictory. Then Ax∩S=∅⇒Ax⊆p.
x,y∈p⇒∃s,t∈S,∃n,m>0,sxn=0,ty
m=0. x+y∈S⇒st(x+y)
n+m=0∈S. contradictory. x+y∈p.
p is a prime ideal: Since S is a multiplicatively closed set, a product of elements of S (i.e.
not belong to p) belongs to S (i.e. not belong to p).
The complement p of the maximal multiplicatively closed set S is a prime ideal by
above. If there exists a prime ideal q (⊂p), A−q is a multiplicatively closed set and
A−q⊃S. It contradicts to maximality of S. p is a minimal prime ideal.
⇐:Let p be a minimal prime ideal and S be the complement of p. If S is not maximal,
let Σ’ be a familiy of members of Σ which strictly include S. Σ’ is an inductively ordered
set. By Zorn’s lemma, there exists a maximal element S’ (⊃S). S’ is a maximal in Σ.
23
q(=A−S’) is a prime ideal as above. It contradicts to minimality of p.
Ex.3.7
i):⇒: x∈A−S⇒x∉S⇒ax∉S for ∀ax∈(x). ⇒(x)⊆A−S. By lemma (below), there exists a
prime ideal which includes (x) and is disjoint with S. ⇒ A−S is a union of prime ideals.
⇐: Let pii be prime ideals and S=A−∪ipi. xy∈S ⇔ xy∉∀ pi⇔ x∉∀pi ∧ y∉∀pi ⇔
x∈S ∧y∈S. S is saturated.
ii): Letpii be maximal ideals which are disjoint with S. pi is a prime ideal by lemma
(below). Let 'S =A−∪pi . 'S is a saturated multiplicatively closed set by (i). If there
exists a saturated multiplicatively closed set S’’ )'( S⊂ which contains S, A−S’’ contains
a prime ideal p which is not contained in ∪ipi and disjoint with S. It contradicts to that
pii are all of maximal ideals which is disjoint with S. 'S = S .
Let S=1+a. S is a mutiplicatively closed set by (ex.3.2). Saturation of S is S = A−∪pi
where pi is a maximal ideal which is disjoint with S. pi ⊇a. Because if not, (pi+a)∩S ≠∅.
⇒p+a=1+a’ ⇒pi∩S ≠∅. Contradictory. p⊇a⇒p∩S =∅. Because p∩S ≠∅⇒p=1+a,
⇒¬p⊇a. Contradictory. ⇒∃pi ⊇p. S is the complement of a union of all prime ideals
containing a.
Lemma: Let S be a multiplicatively closed set and I be an ideal which is disjoint with S.
Let Σ be a family of ideals which contains I and is disjoint with S. There exists a
maximal element in Σ and it is a prime ideal.
Proof: I is a member of Σ⇒Σ ≠∅. Let inclusion in the sense of set theory be partially
order of Σ. A union of all members of any chain is a member of Σ. By Zorn’s lemma,
there exists a maximal element p. a,b⊃p⇒∃s(∈a∩S), t(∈b∩S),. st∈ab∩S. Then p does
not contain ab. p is a prime ideal.
Ex.3.8
i)⇒ii): For t/1, 1/t ∈T−1
A. t/1 is a unit of T−1
A . φ is bijective⇒ t/1 is a unit in S−1
A.
ii)⇒iii): t/1 is a unit in S−1
A⇒∃a∈A,∃s∈S,1/t=a/s⇒∃s’∈S,(at−s)s’=0. Set as’=x. xt∈S.
iii)⇒iv): S =A−∪p∩S=∅ p (ex.3.7ii).
T⊆ S : it is enough to show that if prime ideal p∩S=∅, p∩T=∅.
p∩T≠∅⇒∃p(=t∈p∩T). By assumption of (iii), ∃x∈A, xt∈S, xt=pt∈p∩S. p∩S ≠∅.
Contradictory.
iv)⇒v): T⊆ S =A−∪p∩S=∅p. Suppose q be a prime ideal and q∩T≠∅. ⇒ q∩ S ≠∅.
¬q⊆∪p∩S=∅p⇒q∩S ≠∅.
v)⇒i): S−1
A and T−1
A are S−1
A-module. If φ q: (S−1
A)q →(T−1
A)q is bijective for any
q∈Spec(S−1
A), φ is bijective by (3.9). q=S−1p for some prime ideal p such that p∩S=∅
24
by (3.11). p∩T=∅ by hypothesis. The element of (S−1
A)q is of the form (a/s)/(r/s’)
where r∉p.
φq is injective: φq((a/s)/(r/s’))=0⇒∃r’/s“(∉q),(a/s)(r’/s“)=(ar’/ss“)=0 in T−1
A⇒
∃t∈T,ar’t =0. t∉p⇒ (a/s)/(r/s’)= (ar’t/s)/(rr’t/s’)=0.
φq is surjective: The element of (T−1
A)q is of the form (a/t)/(r/s’) where r∉p.
(a/t)/(r/s’)=(a/1)/(rt/s’). rt∉p⇒ (a/t)/(r/s’)=φ q((a/1)/(rt/s’)).
Ex.3.9
xy∈S0 ⇔x,y∈S0. S0 is a saturated multiplicatively closed set. D=A−S0 is a union of prime
ideals (ex.3.7). This coincides with ex.1.14.
A minimal prime ideal consists of only zero-divisors: A minimal prime ideal p is the
complement of a maximal multiplicatively closed set S which does not contain 0(ex.3.6).
If non-zero-divisor ∃x∉S, S∪x forms a multiplicatively closed set (⊃S). It contradicts
to the maximality of S. ⇒S contains all zon-zero-divisors⇒p consists of only
zerodivisors.
i): Let S be a multiplicatively closed set and A→S−1
A be injective⇔S does not contain
zero-divisors. Since S0 is the largest multiplicatively closed set which does not contain
zero-divisors, S0 is the largest multiplicatively closed set such that makes A→S−1
A be
injective .
ii):a/s∈S0−1
A is a zero-divisor⇔∃b/t∈S0−1
A,(a/s)(b/t)=0⇔∃u∈S0,abu=0⇔a is a
zerodivisor. Then, a/s is not a zerodivisor⇔a∈S0⇔s/a∈S0−1
A⇔a/s is a unit in S0−1
A.
iii): A→S0−1
A is injective: by (i).
Surjective: it is enough to show that for ∀ a/s∈S0−1
A, ∃ x∈A,x/1=a/s. To gain x, we
solve x/1=a/s. ∃s’ ∈S0,(a−sx)s’=0. s’ is not a zero-divisor⇒a−sx=0. s∈S0 is a unit by
hypotheses⇒there exists inverse s−1
of s in A. Then x=as−1∈A.
Ex.3.10
i):it is enough to show (a/s)=(a2/s
2) for ∀ a/s∈S
−1A by (ex.2.27). (a/s)⊇(a
2/s
2). Suppose
(x/t)(a/s)∈(a/s) (x, a ∈A,t∈S). xa∈(a). A is absolutely flat⇒(a)=(a2) ⇒∃y∈A,xa=ya
2.
(x/t)(a/s)=(ya2/ts)=(ys/t)(a
2/s
2) ∈ (a
2/s
2).
ii):⇒: Am is absolutely flat by (i). Am is a field by (ex.2.28).
⇐: It is enough to show (a)=(a2) for any principal ideal of A by (ex.2.27) .
Since these are A-modules and (a) ⊇(a2), it is enough to show (a)m=(a
2)m for any
maximal ideal m of A. Since Am is a field, (a)m=0 or (a)m=(1). If (a)m=0,(a)m⊇(a2)m=0.
Then (a)m=(a2)m . If (a)m=(1), (a
2) m=(1). Then (a)m=(a
2)m.
25
Ex.3.11
i)⇒ii):Let R=r(0) and m be any maximal ideal of A and n be the maximal ideal of A/R
corresponding m. (A/R)n is a field by(ex.3.10ii). n(A/R)n =0. There exists no prime ideal
strictly contained by n. In A, there exists no prime ideal which is strictly contained by m
and is containing R. Prime ideals of A are maximal ideals.
ii)⇒iii): x(∈Spec(A) ) corresponds to a maximal ideal m. m =m. One point is a
closed set.
iii)⇒iv): let m be a prime ideal which corresponds to x∈Spec(A). x is a closed set⇒
m =m. m is a maximal ideal. Spec(A) consists of only maximal ideals. Let m,n(≠m)
∈Spec(A) and S1=m−n, S2=n−m. ∃m∈S1, ∃n∈S2 and mn=0. Because if mn≠0 for
∀m∈S1 and ∀n∈S2, (A−m)∪(A−n) generates a multiplicatively closed set, say S. 0 is an
ideal which disjoints with S. Maximal element p among ideals which are disjoint with S
is a prime ideal and p⊆m∩n. ⇒ p⊂m or p⊂n. It contradicts. Then n∈Xm, m∉Xm,
m∈Xn, n∉Xn and Xm∩Xn=Xmn=X0=∅. Then Spec(A) is Hausdorff.
iv)⇒i):Suppose A/R is not absolutely flat. ⇒ a maximal ideal ∃n of A/R , (A/R)n is not a
field (i.e. n(A/R)n≠0) (ex.3.10). If there is not a prime ideal in (A/R)n (=B) except a
maximal ideal, rB(0)=n(A/R)n. rB(0)=rA/R(0)n(3.11v). Since rA/R(0)=0, rA/R(0)n=0. Then
n(A/R)n=0. It is contradictory. Then there is a prime ideal (not maximal) in (A/R)n. Then
there exist a prime ideal p⊂m which is the maximal ideal corresponding to n. In
Spec(A), the open set which does not contain p does not contain m. Spec(A) is not
Hausdorff. Contraditory.
Since Spec(A) is Hausdorff and quasi-comapct, Spec(A) is compact. Spec(A)=X is a
totally disconnected space if any two point of Spec(A) are not connected. Let m,n(m≠n)
be any points of Spec(A). X is Hausdorff⇒ ∃Xf , ∃Xg , n∈Xf, m∉Xf, m∈Xg, n∉Xg, Xf∩Xg
=∅. X−Xf∪Xg =V((f,g)). A/R=A’ ⊕(f mod R,g mod R) (ex.2.27iii). A’ and (f mod R,g
mod R) have ring structures. Since Spec(A)≅ Spec(A/R), Spec(A) is a disjoint union of
Spec(A’) and Spec((f mod R,g mod R)). Spec(A’) ≅V((f mod R,g mod R)) (≅ V((f,g))) is
an open set of Spec(A/R). Then V((f,g)) is an open set of Spec(A). Spec(A) is a disjoint
union of Xf,Xg and V((f,g)).
Ex.3.12
T(M)=x∈M|∃a(≠0)∈A,ax=0. x,y∈T(M)⇒ ∃a,b(≠0) ∈A,ax=0,by=0⇒ab(x+y)=0. A is
an integral domain⇒ab≠0⇒x+y∈T(M). For ∀r∈A, arx=0. ⇒rx∈T(M).
i): Let x∈M. Suppose x∉T(M). If ∃ a(≠0)∈A and a(x+T(M))⊆T(M), ax∈T(M).
∃b(≠0)∈A,bax=0, i.e. x∈T(M). It is contradictory. Then M/T(M) is torsion free.
ii): x∈T(M) ⇒ ∃ a(≠0) of A ,ax=0. af(x)=f(ax)=0. f(x) ∈T(N).
26
iii):Suppose MMM gf ′′→→′→0 is exact.
f ’:T(M’)→ T(M) is injective: f ’ can be defined by (ii) and is an A-homomorphism by (i).
f ’ is injective, because of restriction of f.
g’ :T(M) →T(M”) is restriction of g to T(M). g’ can be defined by (ii).
Im(f ’)⊆Ker(g’): g°f=0⇒g’ °f ’=0.
Im(f ’) ⊇Ker(g’): If g’(y)=0 for y∈T(M), g(y)=0,i.e. ∃x’ ∈M’,y=f(x’). ∃a(≠0) ∈ A,
ay=0⇒ ay=af(x’)=f(ax’)=0. ax’=0 because f is injective. x’ ∈T(M’). y=f’(x’).
iv): Let κ (A)=K and µ:x→1⊗Ax.
T(M)⊆Ker(µ): Suppose x∈T(M), i.e. ax=0 for some a(≠0) ∈A. 1⊗Ax = (a(1/a)) ⊗ Ax
=1/a⊗A ax=0. x ∈ Ker(µ).
T(M) ⊇ Ker(µ): K is the direct limit of direct system K=(Ki,αij) on directed set I ordered
by inclusion where Ki is an A-module generated by 1/ai and αij is the embedding.
K⊗AM= )(lim MKi ⊗ by (ex.2.20). If 1⊗Ax=0 in K⊗AM, there exists i such that 1⊗x=0
in Ki⊗AM by (ex.2.15). Let S=ainn. Ki=S
-1A. Ki⊗AM≅S
-1M and b/ai
n ⊗x is isomorphic
to bx/ain(3.5). 1⊗x=ai/ai⊗x→aix/ai=0. Then ai
nx=0. x∈T(M).
Ex.3.13
T(S−1
M)⊆S−1
(T(M)): m/s∈T(S−1
M)(s∈S,m∈M)⇒∃a/s’(≠0)∈S−1
A,am/s’s=0⇒∃t∈S, atm
=0 ⇒since at≠0,m∈T(M) ⇒m/s∈S−1
T(M).
T(S−1
M)⊇S−1
(T(M)): m/s∈S−1
T(M) (s∈S,m∈T(M)) ⇒∃a(≠0) ∈A, am=0⇒asm/s=0⇒
(as/1)m/s=0⇒since as/1≠0,m/s∈T(S−1
M).
i)⇒ii): T(M)=0 if only if (T(M)) p=0 for any prime ideal p of A. (T(M)) p=T(Mp)=0
means that Mp is torsion free for any prime ideal p of A.
ii)⇒iii):trivial.
iii) ⇒i):T(Mm)=0⇒T(Mm) =(T(M)) m =0. T(M)=0 by(3.8).
Ex.3.14
M/aM is an A/a-module. M/aM=0 if (M/aM) n =0 for any maximal ideal n of A/a by
(3.8). A maximal ideal m containing a of A corresponds to n.
Localize a sequence 0→aM→M→M/aM→0(exact) with m. (M/aM)m≅Mm/(aM)m .
(M/aM)m≅(M/aM)⊗A/a(A/a)m by (3.5). (A/a)m≅(A/a)n by (ex.3.4). ⇒ (M/aM)m≅
(M/aM)n. (M/aM)n≅Mm/(aM)m. Since Mm=0 by hypotheses, (M/aM)n=0.⇒ M=aM
Ex.3.15
Let e1,…,en be a set of canonical basises of F=An. Let x1,…,xn be any generators of F.
27
Define φ(ei)=xi.
φ is surjective: x1,…,xn are generators of F over A.
φ is injective: By(3.9) we prove Kerφ=0. We may assume A be a local ring with the
maximal ideal m. Let k=A/m. The exact sequence 0→Kerφ→F →φ F→0 gives an
exact sequence 0→ k⊗Kerφ→k⊗F → ⊗1φk⊗F→0(ex.2.24). k⊗F=k
n. A surjective
mapping from vector space of n degree to vector space of n degree is isomorphism.
k⊗Kerφ=0. k⊗Kerφ≅ Kerφ/mKerφ (ex.2.2). Kerφ is a finitely generated A-module by
(ex.2.12). By Nakayama’s lemma Kerφ=0.
Let x1,…,xm(m<n) be generators of F and x’1,…,x’m be their images in Fm⊗k=kn.
x’1,…,x’m generate a vector space of n degree. It is contradictory.
Ex.3.16
i)⇒ii): by (3.16).
ii)⇒iii): Let m be a maximal ideal of A. ∃n ∈Spec(B), nc=m. n
cec=m
ec. n
cec=n
c by(1.17).
m=mec
. If me=(1), m=(1). Contradictory.
iii)⇒iv): by hint.
iv)⇒v): by hint.
v) ⇒i): Tensoring 0→a→A→A/a→0 with B(A-flat), 0→ae→B→(A/a)⊗B→0 is exact.
(A/a) ⊗B≅B/ae. Let M=A/a. By hypotheses, A/a→(A/a) ⊗B is injective. A/a→B/a
e is
injective. aec⊆a. ⇒ a
ec=a.
Ex.3.17
Let M be an A-module. If g is faithfully flat, then MB→ MB⊗BC is injective for
MB(B-module) by(ex.3.16v), i.e., M⊗AB→M⊗AC is injective.
For any exact sequence 0→M’ →M of A-modules, following diagram is commutative.
CMCM
ji
BMBM
A
v
A
A
u
A
⊗→⊗′→
↓↓
⊗→⊗′
↓↓
0
00
i,j are injective since g is faithfully flat. v is injective since g°f is flat. u is injective. f is
flat.
Ex.3.18
Let p∈Spec(A),q∈Spec(B) and p=qc. B is A-flat⇒Bp is Ap-flat (3.10). Bq,local ring of
28
Bp, is Bp-Flat (3.6) ⇒Bq is Ap-Flat(ex.2.8). For the maximal ideal pAp of Ap,
(pAp)e⊆qBq ≠(1). Since Bq, Ap-algebra, statisfies prerequisites of (ex.3.16) and
(ex.3.16iii), Spec(Bq)→Spec(Ap) is surjective by (ex.3.16ii).
Ex.3.19
i): M≠0⇔∃p∈Spec(A), Mp≠0 ⇔Supp(M) ≠∅ by(3.8).
ii): p∈Supp(A/a)⇔(A/a) p=Ap/ap≠0⇔a∩(A−p)= ∅⇔a⊆p⇔p∈V(a).
iii): 0→M’→M→M”→0 is exact⇒0→M’ p →M p →M” p →0 is exact.
Since Mp ≠0⇔(M’p ≠0∨ M”p )≠0, Supp(M)=Supp(M’)∪Supp(M”).
iv): ⊇: trivial.
⊆:p∉∪Supp(Mi)p ⇒( Mi)p =0(∀i). For ∀x/s ∈ Mp, x=Σi≤naixi. x/s=Σi≤naixi/s=0.
v): Let x1,…,xn be generators of M. Supp(M)=∪ iSupp (Axi)=∪iSupp(A/Ann(xi)) =
∪V(Ann(xi))=V(∩iAnn(xi))=V(Ann(M)) by (ii)(iv).
vi):(M⊗AN) p=M p⊗ApN p (3.7)=0⇔Mp=0∨Np=0 (ex.2.3) .
Supp(M⊗A N)= Supp(M)∩ Supp(N).
vii):Supp(M/aM)=Supp(M⊗AA/a))=Supp(M)∩Supp(A/a)=V(Ann(M))∩V(a)=V(Ann(M)
+a).
viii): Let x1,…,xn be generators of M. B⊗AM≅ΣB⊗AAxi. Let Ann(xi)=ai. B⊗AAxi=
B⊗AA/ai =B/aie. V(ai
e)=q∈Spec(B) | q⊇ai
e=q∈Spec(B) | f
*(q)⊇ai
ec= q∈Spec(B) |
f*(q)⊇ai. Because. f
*(q) ⊇ai
ec⇒f*(q) ⊇ai: trivial. f
*(q) ⊇ai
ec⇐ f*(q) ⊇ai: q
c⊇ai⇒
qcec⊇ai
ec⇒Since q cec
=qc, f
*(q) ⊇ai
ec.
Then V(aie)=f
*−1(V(ai)). Supp(B⊗AM)= ∪Supp(B/ai
e)=∪V(ai
e)=∪f
*−1 (V(ai))=
∪f*−1
(V(Ann(xi))) = f*−1
(∪V(Ann(xi))) =f*−1
(V(∩Ann(xi)))=f*−1
(Supp (M)) by (v).
Ex.3.20
i):⇒: Let p∈Spec(A) and b be an ideal of B and p=bc. b
c=b
cec⇒p=pec
. Then p is the
contraction of a prime ideal of B by (3.16).
⇐: trivial.
ii):Let q∈Spec(B) and q=ae. q
c=a
ec∈Spec(A). ⇒ aece
=ae=q. Suppose q≠q’ and q’ =a’
e
and qc=q’
c. a
ec= a’
ec⇒aece
=a’ece⇒a
e=a’
e⇒ q=q’. It contradicts. f*
is injective.
Counterexample:
Let f:Z→Z[X]/(X 2
) be embedding. Prime ideals of Z[X]/(X 2
) are not extended ideals.
Let (p,bX+a) be a prime ideal of Z[X]/(X 2
). ⇒p is a prime number or 0.
case of p ≠0: (bX−a) (bX+a)=a2∈(p, bX+a) ⇒ (a
2, p)≠1. p divides a. Then (p, bX+a) is
of the form (p, bX). Then b=1. Since (Z[X]/(X 2
))/(p,X)≅ Z/(p), (p,X) is a prime ideal.
case of p =0: a2∈(bX+a). One of prime factors of a
2,say q, must belong to (bX+a).
29
(bX+a) is of the form (q, bX). Then b=1. If a=0, (X) is a prime ideal, because
(Z[X]/(X2))/(X)≅ Z.
Then prime ideals of Z[X]/(X 2
) are of the form (p,X) where p is a prime number
including 0. f* is injective.
Ex.3.21
i): φ*: Spec(S
−1A)→Im(φ*
)=S−1
X is surjective. There is one to one corresponding
between members of Spec(S−1
A) and members of Spec(A) which are disjoint with S
(3.11). φ*: Spec(S
−1A)→S
−1X is bijective. φ*
is continuous. Let Yf/s be an open
neighborhood of Y=Spec(S−1
A). To claim Spec(S−1
A) is homeomorphic to S−1
X, it is
enough to show that φ*(Yf/s) is an open set of S
−1X.
Yf/s=q∈Spec(S−1
A) | f/s∉q . φ*(Yf/s)=p∈Spec(A) | S
−1p≠(1), f/s∉S
−1p=p∈Spec(A) |
S∩p=∅, f∉p=S−1
X∩Xf. φ*( Yf/s) is open in S
−1X . Then Spec(S
−1A)≅ S
−1X.
Let S=fnn where f∈A. S
−1A=Af. φ
*(Spec(Af))=p∈Spec(A) | f∉p =Xf is a basic open
set of X.
ii): S−1
X is a subspace of X whose members are disjoint with S by (i). S−1
Y is a subspace
of Y whose members are disjoint with f(S). Let q∈S−1
Y and f−1
(q)=p. p is a prime ideal
of A and p∩S=∅. Then f*(S
−1Y) ⊆S
−1X. ⇒S
−1Y ⊆ f
*−1(S
−1X). Let S
−1f* be restriction of f
*
to S−1
Y. S−1
f*: S
−1Y→S
−1X.
Let p∈S−1
X. p is a prime ideal of A and p∩S =∅. If f*(q)≠p for ∀q∈Spec(B), nothing to
do. If ∃q∈Spec(B) and f*(q)=p, f
−1(q)=p. p∩S=∅⇒q∩f(S)=∅. Then q∈S
−1Y. f
*−1 (S
−1X)
⊆S−1
Y. Then f*−1
(S−1
X)=S−1
Y.
iii):Let b=ae. *f : Spec(B/b)=V(b)→Spec(A/a)=V(a): Let q∈Spec(B/b). Let q’ be a
prime ideal of B(q’⊇b) corresponding to q. Since f−1
(q’)⊇aec⊇a, *f (q) corresponds to
f−1
(q’)/a =f*(q’)(⊇a)∈Spec(A/a)=V(a). *f is the restriction of f
* to V(b).
iv): same notation as (iii).
Let f : A→B and f*: Spec(B)→Spec(A) and S
−1f*:Spec (Bp)→Spec(Ap) and *1
fS− : Spec
(Bp /(pAp) e
) →Spec (Ap/pAp). Let Ap/pAp=κ(p). There is one to one correspondence
between f*−1
(p) and (S−1
f*)−1 (pAp). Inverse images of p by *1 fS −
correspond to inverse
images of Ap/pAp, i.e. a prime ideal 0 of κ(p). Then inverse images of p is
Spec(Bp/(pAp)e). Bp=B⊗AAp and (pAp)
e=B⊗ApAp⇒Bp/(pAp)
e=(B⊗AAp)/(B⊗A pAp)= B⊗A
(Ap/pAp)= B⊗Aκ (p) by (ex.2.2).
Ex.3.22
The canonical image of Spec(Ap) in Spec(A) is Y=q∈Spec(A) | q⊆p.
Let Xff∈A be basic open sets of X=Spec(A).
30
Y⊆∩f∉p Xf : q∈Y⇒q⊆p. f∉p⇒ f∉q . q∈∩f∉p Xf.
Y⊇∩f∉p Xf : q∉Y⇒¬p ⊇q. ∃ f∈q , f∉p. For this f, q∉Xf. q∉∩f∉pXf.
Ex.3.23
i):Since U is a basic open set, it is enough to show that if U=Xf=Xg, Af=Ag. Xf=Xg⇒
r((f))=r((g)) (ex.1.17iv). ∃m,n(>0) and ∃u,v∈A, gn=uf and f
m=vg. ∀x/f
s∈Af
⇒(gns−u
sfs)=0⇒ x/f
s=xu
s/g
ns⇒Af ⊆ Ag. ∀y/gs ∈Ag⇒ (f
ms−vsg
s)=0⇒y/g
s=yv
s/f
ms ⇒ Ag ⊇
Af. Af=Ag.
ii): Let U’=Xg,U=Xf and U’⊆U. Xg⊆Xf⇒V((g))⊇V((f)).⇒V(r((g)))⊇V(r((f)))⇒r((g))
⊆r((f)). ∃n>0 and ∃u∈A, gn=uf. Define a homomorhism ρ: Af →Ag by x/f
r → xu
r/g
rn.
ρ is a ring homomorphism: ρ(x/fr+y/f
s)=ρ((xf
s+yf
r)/f
r+s)=(xf
s+yf
r)u
(r+s)/g
(r+s)n=(xf
su
(r+s)+
yfru
(r+s) )/g
(r+s)n= (xg
snu
r+ yg
rnu
s)/g
(r+s)n= xu
r/g
rn + yu
s/g
sn= ρ(x/f
r)+ρ (y/f
s). ρ(x/f
r•y/fs)=
ρ(xy/fr+s
)=xyur+s
/g(r+s)n
=xur/g
rn•yus/g
sn=ρ(x/f
r) •ρ(y/f
s).
Let Xf=Xf ’ and Xg=Xg’. ρ’:Af ’ →Ag’ is same of the form as ρ: Xf=Xf ’ ⇒∃l>0 and ∃w ∈A,
fl=wf’. Xg⊆Xf ⇒∃n>0 and ∃u∈A,g
n=uf. From Xg=Xg’ ⇒∃m>0 and ∃v∈A,g’
m=vg. ⇒
g’mnl
= vnu
lwf’. ⇒ρ’:Af ’ →Ag’ is given by x/f’
r→xvn r
ul r
w r/g’
mnlr. ρ’ is equal to ρ by
following iii). ρ depends on only U and U’.
iii):Let U=Xf, U’=Xg and U=U’ . g n
=uf and fm=vg. Define ρ:Af →Ag by a/f
m→aum/f
mn.
ρ is surjective: It is enough to show the inverse image of 1/g ∈Ag. ρ(v/fm)= vu
m/g
nm=1/g.
ρ is injective: ρ(x/fs)=xu
s/g
ns=0⇒∃i>0,g
ixu
s=0⇒g
ixu
sfs=0⇒g
i+nsx=0⇒g
i+nsxv
i+ns=0⇒
fm(i+ns)
x=0⇒ x/fs=0.
ρ(x/fs) =xu
s/g
sn. x(g
sn−usfs) =0⇒x/f
s=xu
s/g
sn. ρ is the identity map of Af.
iv):Let U=Spec(Af), U’=Spec (Ag) and U“= Spec (Ah) and U⊇ U’ ⊇ U”. fu=gn,gv=h
l and
fw=hm. Define homomorphisms ρ:Af→Ag, ρ’:Ag→Ah and ρ”:Af→Ah. as ii). For ∀x/f
s ∈Af,
ρ’ρ(x/fs)=ρ’(xu
s/g
ns)=xu
sv
ns/h
nsl. and ρ”(x/f
s)=xw
s/h
ms.
x(hms
usv
ns−wsh
nsl) =x(f
sw
su
sv
ns−wsg
nsv
ns) =x(f
sw
su
sv
ns−wsfsu
sv
sn)=0. Then ρ’ρ=ρ”.
v): Let U p∈U be the family of all basic open sets which include p. U p∈U forms a
directed set by inclusion(⊇) of the set theory. Let X=A(U), ρUU’. Since
ρUU’ :A(U)→A(U’) is satisfying (iii) and (iv), X is a direct system. There exists the
direct limit, i.e. )(lim UAU
→∈p
.
For each U=Xf, p∈U. Since fn
⊆A−p, ∃αU: Af→Ap. ∃αU =αU’ρUU’ ⇒ ∃ α: )(lim UAU
→∈p
→Ap by (ex.2.16).
αU=αU’ρUU’: let U’(=Xg)⊆U(=Xf). fu=gm⇒ρUU’(x/f
s)=xu
s/g
sm for ∀x/f
s∈Af. αU’ρUU’((x/fs))
=αU’(xus/g
sm)=xu
s/g
sm. αU(x/f
s)=x/f
s. Because g
n,f
n ⊆A−p, xu
s/g
sm =x/f
s holds in Ap .
31
Then αU =αU’ρUU’ .
α is surjective: for ∀a/p’∈ Ap (p’∉ p ), ∃Xp’ (=U) in the directed set Up∈U and αU
(a/p’)=a/p’. The equivalent class of a/p’ ∈Ap’ in the direct system is the inverse image of
a/p’ ∈Ap by α.
α is injective: let Xs∈Up∈U and a/s∈As. Suppose a/s=0/1 in Ap. ∃t∉p,at=0. Xt∈Up∈U.
Xs∩Xt =Xst∈Up∈U. The image of a/s∈As in Ast is at/st =0. This means that a/s belongs
to the same class of 0 in the direct system. Then it is 0 in )(lim UAU∈p
. α is injective.
Ex.3.24
Let Uii∈I be basic open sets of Spec(A)=X and Ui =Xfi. X is quasi-compact⇒∃ f1,…,fn
∈A, X=∪fi≤n Xfi. si is of the form ai/fini
(ai∈A) in A(Xfi). By hypothesis, ρi,ij(si)= ρj,ij(sj) in
A(Xfi ∩ Xfj) =A(Xfifj). Then
ρi,ij(si) =bi /(fifj)mij
=ρj,ij(sj) =bj /(fjfi)mji
(bi, bj ∈A).
Let N>>maxni,mij. si is represented as si=ci/fiN (ci∈A) and cifj
N/fi
Nfj
N = cjfi
N/fj
Nfi
N.
X=∪i≤n Xfi⇒(f1,…,fn)=1⇒(f1N,…,fn
N)=1. (Because if (f1
N,…,fn
N) ≠1, maximal ideal ∃m
⊇(f1N,…,fn
N). fi ∈m ⇒(f1,…,fn) ≠1 is contradictory. ) ∃hi ∈A, Σ hi fi
N=1.
s=Σi=1,…,ncihi is an inverse image of si in A:
for 1≤j ≤n, in A(Xfi)
s/1=Σi=1,…,ncihifjN/fj
N
=Σi(≠j) cihifiNfj
N/fi
Nfj
N + cjhjfj
N/fj
N
=Σi(≠j) cjhifiNfi
N/fi
Nfj
N + cjhjfj
N/fj
N
= Σi=1,…,ncjhifiN/fj
N =cj/fj
N= sj/1.
s is unique : Let t be an another inverse image of si in A. For any prime ideal p,
p∈∃Xfi. s/1− t /1 =0 in A(Xfi) by hypothesis. Then (s−t)/1=0 in A(Xfi). Since fi ∉p, (s−t)p
=0. ⇒(s−t)=0⇒s−t=0.
Ex.3.26
Let (Aα,iαβ) be a direct system on the same directed set of direct system(Bα,g αβ) and
Aα=A, iαβ=1A. Since fα : A→Bα satisfies gαβ°fα=fβ (α≤ β), fα induce a homomorphism f :
A→B where A is the direct limit of direct system(Aα,iαβ) and B is the direct limit of
direct system(B α,g αβ) by (ex.2.18). Following is a commutative diagram.
BB
g
BA
f
f
→→↓
↓↓↓
→→
ββ
αα
µβ
αβ
µα
32
Let iα:Aα→A. Since f °iα=µα°fα where iα=1A by (ex.2.18). For ∀α, f=µα°fα.
f*(Spec(B))⊆ ∩ αfα
*( Spec (Bα)): p∈f
*(Spec(B)) ⇒∃q∈Spec(B), p = f
-1(q) .
For ∀α, µα-1
(q)∈Spec(Bα) and f*
αµα-1
(q) =p. ⇒ p ∈∩ αfα*( Spec (Bα)).
f*(Spec(B))⊇∩αfα
*( Spec (Bα)): p∉ f
*(Spec(B)) ⇒f
*-1(p)=∅ ⇒ B⊗Aκ(p)=0.
B⊗Aκ(p)=( lim Bα)⊗Aκ(p) = lim Bα⊗Aκ(p). ∃α, Bα⊗Aκ(p)=0 (ex.2.21).
⇒ f*
α-1
(p)=∅. p∉fα*( Spec (Bα)).⇒ p∉∩αfα
*( Spec (Bα)).
⇒ p ∈∩ αfα*( Spec (Bα)).
Ex.3.27
i) : Let Bα⊗…⊗Bβ be a tensor product of finite set of Bαα. Define
µβγ:Bα⊗…⊗Bβ→Bα⊗…⊗Bβ⊗…⊗Bγ by µβγ:bα⊗…⊗bβ→bα⊗…⊗bβ⊗1⊗…⊗1
(1∈Bi(β<i≤γ)). X=(Bα⊗…⊗Bβ,µβγ) is a direct system. Let B be the direct limit of X. B
is a tensor product of byBαα by (ex.2.23). There exists an A-homomorphism f : A→B
as (ex.3.26). following is a commutative diagram.
BB...B...B
g
B...BA
)f....,f(
)f...,f(
→⊗⊗⊗⊗ →↓
↓↓↓
→⊗⊗ →
µγβα
⊗⊗
µβα
⊗⊗
γα
βα
f*(Spec(B)) =∩α,…,β ∈∆ (fα⊗…⊗ fβ)
*( Spec (Bα⊗…⊗Bβ)).(∆ run through finite sets) by
(ex.3.26). (fα⊗…⊗fβ)*(Spec(Bα⊗…⊗Bβ))=Spec(Bα)∩…∩Spec(Bβ) by (ex.3.25). Then
f*(Spec(B)) =∩α fα
*( Spec (Bα)).
ii):Let B=ΠBα(finite product). Spec(B)=∪(Bα,Bβ,…,pλ,i,…,Bµ) where Spec(Bλ)= pλ,ii.
Define f: A→B by f(x)=(fα(x),…, fµ(x)). f * (Spec(B))= ∪fα
*( Spec (Bα)) by (ex.1.22).
iii): f*(Spec(B)) satisfies axioms for closed sets: Let X=Spec(A).
X is closed: Take B as A and f as idA : A→A. X=idA* (Spec(A)).
∅ is closed: Take B as 0 and f : A→0(1=0 is not denied in chapter 1). f *
(Spec(0))= ∅.
Finite union is closed: Let X’=∪ifi* (Spec(Bi)). Let B=ΠBi and Define f: A→B by
f(x)=(f1(x), f2(x),…, fn(x)). X’=f *(Spec(B)) by(ii).
Intersection is closed: LetY αbe closed sets (including infinite case). Let Y=∩αY α. For
each α, ∃ fα: A→Bα,Yα=fα* (Spec(Bα)). Build a direct system of tensor products of Bαα
as (i). B be its direct limit B. ∃f:A→B, Y=f *(Spec(B))(=∩αfα
* (Spec(Bα)=∩αYα) by(i).
Constructible topology is finer than Zariski topology:
A closed set in Zariski topology is defined by V(a) where a is a ideal. Let f : A→A/a.
Since V(a)=f*(Spec(A/a)), V(a) is a closed set in the constructible topology. Following
example show that the constructible topology is strictly finer than Zariski topology. Let
p be a prime ideal of A and f : A→Ap. f*(Spec(Ap)) is closed in the constructible
33
topology by definition. But f*(Spec(Ap))=∩ a∉pXa (ex.3.22) is not always closed in
Zariski topology. For example, f*(Spec(Zp))=0,p is not closed in Zariski topology,
because the smallest closed set including 0 is X.
iv): Suppose X be covered by countable open sets Oi in the constructible topology.
Xi=X−Oi are close and ∩ iXi=∅. There exists a ring Bi and a homomorphism fi: A→Bi
such that Xi=fi*(Spec(Bi)) for each i. We can define a direct system (B1⊗AB2⊗A…⊗ABi,
µij) where B1⊗AB2⊗A…⊗ABi is a finite tensor product and µij : B1⊗B2⊗…⊗Bi
→B1⊗B2⊗…⊗Bj is defined by µij (bB1⊗…⊗ bBi)= bB1⊗…⊗bBi⊗1⊗…⊗1
(1∈Bk(i+1≤k≤j)). Let B be the direct limit of this direct system. B is a tensor product
of Bii. ∃f : A→B. f *(Spec(B))=∩iXi=∅. B=0. ∃n , B1⊗B2⊗…⊗Bn=0 by (ex.2.21). Let
C=B1⊗B2⊗…⊗Bn and gn=f1⊗f2⊗…⊗fn: A→C. gn*(Spec(C))=∩fi
*(Spec(Bi))= ∅. Then
∩i=1,..,nXi=∅. X=∪i=1,…,nOi. XC is quasi-compact.
Ex.3.28
i): Let Xg=p∈Spec(A) | g∉p. Since Xg is open in Zariski topology, Xg is open in the
constructible topology. Let f: A→Ag. Xg=f*(Spec(Ag)). Xg is closed in the constructible
topology.
ii): Let p,q∈Spec(A) and p≠q. ∃g∈p, g∉q(no loss of generality). By hypotheses, Xg is
open and closed by C’ topology. p∉Xg, q∈Xg. X−Xg is open and closed. q∉
(X−Xg),p∈(X−Xg). Xg∩(X−Xg)=∅ ⇒ XC’ is Hausdorff.
iii): identity map 1X : XC→XC’ as sets is a homeomorphism:
Since XC’ is the smallest topology where Xg is open and close, we can define
X−Xgg∈A ∪Xg g∈A as open basis of XC’. Since Xg is a open and close by (i) in XC,
X−Xg is open and close in XC . Then 1X is continuous.
Remains to prove is to show any closed set in XC is closed in XC’.
Let E be a closed set of XC. ∃ B and ∃f:A→B,E=f*(Spec(B)). Set Spec(A)=X. Decompose
f = i°j where j : A→f(A) and i: f(A) →B. f*=j
*°i
*. Since j is surjective, j
*(Spec(f(A))) is
homeomorphic to V(Ker(j))∩X by (ex.1.21iv). In the following, X denotes both
Spec(f(A)) and Spec(A), do not confuse. V((g)) is a closed set in Spec(f(A)).
p∈i*(Spec(B))⇔ p
ec=p by (3.16).
i*(Spec(B))=p∈Spec(f(A))|Bp∩ f(A)=p=p∈Spec(f(A)) |∀g∈p, B(g)∩ f(A)⊆p
=X− p ∈Spec(f(A)) | ∃g∈p,¬B(g)∩ f(A)⊆p =X−∪g∈f(A) (V((g)) ∩∪d∈B(g) ∩ f(A) Xd)
=∩ g∈f(A) (X− (V((g)) ∩∪d∈B(g) ∩ f(A) Xd))= ∩ g∈f(A) ((X− V((g)) ∪(X−∪d∈B(g) ∩ f(A) Xd)
=∩ g∈f(A) (Xg∪∩ d∈B(g) ∩ f(A) (X−∪ Xd)).
f*(Spec(B))=V(Ker(j))∩ ∩ g∈f(A) (Xg∪∩ d∈B(g) ∩ f(A) (X−∪ Xd)).
V(a)=∩ a∈a V(a)= ∩ a∈a (X− X a) is closed in C’ topology. Then V(Ker(j)) is closed in XC’.
34
f*(Spec(B)) is closed in XC’.
iv): XC is quasi-compact by (ex.27iv). XC’ is Hausdorff by (ii). XC is homeomorphic to
XC’ by (iii). XC is Hausdorff. By (ii), Xg is open and close in XC’. XC is homeomorphic to
XC’. Xg is open and closed in XC. Then for p≠q, open set ∃Xg of XC, p∈Xg, q∉Xg. Since
Xg is closed, X−Xg is an open set, p∉X−Xg, q∈X−X. Xg∩(X−Xg)= ∅ and Xg∪(X−Xg)=X.
The subspace consisting of more than one point is a union of 2 open sets which are
disjoint. Then it is not connected. Then connected component consists of only one
point,i.e.totally disconnected.
Ex.3.29
f * is closed mapping in constructible topology:
Let E be a closed set of Spec(B). ∃C and ∃h:B→C, E=h* (Spec(C)). f
*(E)=f
*°h
*
(Spec(C))=(h°f) * (Spec(C)). f
*(E) is closed in Spec(A).
f * is continuous:
Open basis of X=Spec(A) in constructible topology are Xa=p∈Spec(A)| a∉p and
X−Xa=p∈Spec(A) | a∈p by(ex.3.28). f*−1
(Xa)=Yf(a)⊆Y(=Spec(B)) by (ex.1.21i). Yf(a) is
open in constructible topology. f*−1
(X−Xa) =Y−Yf(a)⊆Y. Y−Yf(a) is open in constructible
topology.
Ex.3.30
⇒: ∀p is closed in XC, because p= f*(Spec(Ap/pAp)) where f: A→Ap→ Ap/pAp. If
p is closed in Zariski topology, there is not a prime ideal strictly containing p. Any
prime ideal is a maximal ideal. A/R is absolutely flat (ex.3.11).
⇐: If A/R is absolutely flat, every principal ideal is idempotent(ex.2.27ii). For ∀f∈A,
∃a∈A,f−af2=r∈R. For open neighbourhood Xf of Spec(A), Xf∩X1-af=Xr=∅ and
Xf∪X1-af=X. Because f(1−af) =r∈∀p.1∉p⇒f ∉p or 1−af ∉p. Then Xf is an open and
closed set. Zariski topology coinsides with constructible topology in this case(ex.3.28).
Chapter4
Ex.4.1
Let a=∩q i (1≤i≤n) be a minimal primary decomposition of a where r (q i)= p i.
Let pj be minimal prime ideals containing a. pj⊆pi by (4.6). ⇒Number of pj is
finite. pj/a are minimal ideals of Spec(A/a). Spec(A/a)=∪jV(p j/a). Since each V(pj/a)
is a closed set corresponding to a minimal ideal, it is an irreducible component of
Spec(A/a) by (ex.1.20). Spec(A/a) has finite irreducible components.
35
Ex.4.2
Let a=r(a) and a =∩qi (1≤i≤n) be a minimal primary decomposition of a where r(qi)=pi.
If there exists an embedded prime ideal pj in pii, a=r(a)=∩i≠jpi. a=∩i≠j pi is an another
primary decomposition whose prime ideals are pii≠j. This contradicts to the
uniqueness theorem (4.5).
Ex.4.3
Let q be a p-primary ideal and m (⊇p) be a maximal ideal. qm is a pAm-primary ideal
by (4.8). A is absolutely flat⇒Am is a field (ex.3.10). Then mAm=0. pAm=0. p=m by
(3.11). qm is a mAm-primary ideal. There is one to one correspondence between
primary ideals which are disjoint with A−m and primary ideals of Am by (4.8). 0 is the
unique primary ideal of Am⇒qm=0. q=(qm) c=0
c=(mAm)
c = m.
Ex.4.4
Z[t]/m =Z/2Z is a field. m is a maximal ideal. Z[t]/q=Z/4Z. zero-divisor (≠0) (only 2
mod 4) is nilpotent⇒q is a primary ideal. Z/4Z has the only one prime ideal, i.e.(2). 0 is
(2)-primary ideal in Z[t]/q. Inverse images of 0 and (2) by Z[t]→Z[t]/q, are q and m
respectively. q is m-primary. m⊃q⊃m2=(4,2t,t
2) ⇒q is not power of m.
Ex.4.5
p1 and p2 are prime ideals⇒they are primary. m is maximal⇒m2 is primary.
a=(x2,xy,xz,yz). p1∩p2=(x,yz), m
2=(x
2,y
2,z
2,xy,xz,yz). a=p1∩p 2∩m
2 is a minimal primary
decomposition. p1 and p2 are isolated, m is embedded.
Ex.4.6
Let I,J be ideals of C(X).
Lemma1: If I⊇J, V(I)⊆V(J).
Proof: By definition, V(I)=x∈X| f(x)=0, ∀ f ∈I . For ∀x∈V(I) and ∀f∈I, f(x)=0. Then
for ∀g∈J(⊆I), g(x)=0. x∈V(J). ♦
Lemma2:V(I∩J)=V(I) ∪V(J).
Proof:⊇: by Lemma1.
⊆: If x∉V(I)∪V(J), ∃f∈I and ∃g∈J , f(x)≠0 and g(x)≠0. fg∈I∩J and f(x)g(x)≠0⇒
x∉V(I∩J)).♦
Suppose 0=∩qi (1≤i≤n) be a minimal primary decomposition of 0 where r(qi)=pi. X is
compact⇒V(m)≠∅ for any maximal ideal of C(X) by (ex.1.26i). Then for any I≠(1),
36
V(I) ≠∅ by (lemma1). V(0)=X. X= V(0)=∪1≤i≤n V(q i) by (lemma2).
X is an infinite set. ∃V(qi) is an infinite set from compactness. For x0,y0 (x0≠y0)∈V(q i),
∃open sets Ux0 (∋x0) and Uy0 (∋y0) of X ,Ux0∩Uy0=∅, because X is Hausdorff. Uf=x∈X |
f(x)≠ 0 f ∈C(X) is an open basis of X by (ex.1.26iii). ∃f,g∈C(X), Ux0⊇Uf(∋x0), Uy0⊇Ug(∋y0)
and Uf∩Ug=∅. f(x)g(x)=0 for ∀x∈X⇒fg=0. ⇒fg∈∩qi (1≤i≤n), especially fg∈qi.
Because ¬(X−Uf)⊇V(qi) and f is a real-valued function, ¬( nfUX − )⊇V(qi) for ∀n>0.
Then fn∉qi. As the same, g
n∉qi. Then qi is not a primary ideal. it contradicts. 0 is not
decomposable.
Ex.4.7
i): f(x)∈aA[x] ⇒every coefficient of f(x) belongs to a⇒ f(x)∈a[x].
f(x)∈a[x] ⇒ ai∈a (ai is the coefficient of xi of f(x)). aix
i∈aA[x]. f(x)=Σaixi∈aA[x].
ii):Suppose f(x),g(x)∉p[x]. Let a i be the coefficient of highest degree of f(x) which does
not belong to p and b j of g(x) be same as ai. Let ck be a coefficient of f(x)g(x).
ci+j=Σs+r=i+j asbr ∉p, because a ib j∉p and other asbr ∈p. Then f(x)g(x)∉p[x].
iii): A[x]/q[x] ≅(A/q)[x]. Let A/q be A, q be 0 (no loss of generality). Let f(x),g(x)∈A[x].
f(x)g(x)=0 and g(x)≠0⇔∃c(≠0)∈A, cf (x)=0 by (ex.1.2iii). For every coefficient ai of f(x),
cai=0. Since 0 is a primary ideal and c≠0, ai is nilpotent. f(x) is nilpotent by (ex.1.2).
Then 0 ideal of A[x] is a primary ideal. Let rA(0)=p. For ∀f(x)∈rA[x](0), f(x) is
nilpotent⇔every coefficient of f(x) is nilpotent (ex.1.2) ⇔ every coefficient of f(x)
belongs to rA(0)=p⇔f(x)∈p[x]. Then rA[x] (0)=p[x]. p[x] is a prime ideal by (ii). 0 is
p[x]-primary.
iv): a=∩qi ⇒a[x]=∩qi[x]. qi[x] is a primary ideal by (iii). If a[x]=∩qi[x] is not a
minimal primary decomposition, there exists redundant qk[x]. If a[x]=∩i≠k qi[x], we get
a=∩ i≠k q i in constants. It contradicts to minimality of a=∩q i .
v):Suppose a prime ideal a[x]⊆∃p’(⊆p[x]). p’∩A is a prime ideal. ⇒a⊆ (p’ ∩A) ⊆p. p
is a minimal prime ideal of a⇒p’ ∩A=p.
f(x)∈p[x]⇒ for every coefficient ai of f(x),ai∈p=(p’∩A)⊆p’ ⇒ f(x)∈A[x]p’=p’ ⇒p[x]
⊆ p’. p’=p[x]. p[x] is a minimal prime ideal of a[x].
Ex.4.8
Let A=k[x1,...,xn] and p=(x1,…,xi). A/p(≅k[xi+1,...,xn]) is an integral domain⇒p is a prime
ideal. Let Ai=k[x1,…,xi]. Ai/p≅k ⇒p is a maximal ideal of Ai. q=ps is a primary ideal of
Ai. The extention of q to A is q[xi+1,...,xn] by (ex.4.7i ) and induction. q[xi+1,...,xn] is a
primary ideal of A by (ex.4.7iii) and induction.
37
Ex.4.9
⇒ ∃a(≠0),xa=0⇒x∈(0: a). (0 : a)≠(1) ⇒ a minimal prime ideal ∃p (⊇ (0 : a))⇒x∈p. ⇒
p∈D(A).
⇐: p∈D(A) ⇒ for ∃a∈A,p is a minimal prime ideal containing (0 : a). Suppose x(∈p)
is not a zero-divisor. Then x∉(0 : a). (A−p)∪xii>0 generates a multiplicatively closed
set ,say T. A maximal ideal p’ which contains (0 : a) and is disjoint with T is a prime
ideal. p’⊂p . It contradicts minimality of p. x(∈p) is a zero-divisor.
D(S−1
A)⊆ D(A) ∩Spec(S−1
A):A prime ideal of S−1
A is S−1p where p is a prime ideal of A
which is disjoint with S. S−1p ∈D(S
−1A) ⇒S
−1p is a minimal prime ideal of (0 : a/s) for
∃a/s≠0. (0 : a) ⊆p. a prime ideal ∃p’,(0 : a) ⊆p’⊂p, ⇒ (0 : a/s) ⊆S−1p’ ⊂S
−1p. It
contradicts to minimality of S−1p. p∈D(A) ∩Spec(S
−1A).
D(S−1
A)⊇D(A)∩Spec(S−1
A): p∈D(A) ∩Spec(S−1
A) ⇒S−1p≠(1) and p is a minimal prime
ideal of (0 : a) for ∃a/s≠0. S−1p is a minimal prime ideal containing (0: a/s). If not,
∃S−1p’ such that (0: a/s) ⊆S
−1p’⊂S
−1p. By contraction, (0 : a) ⊆p’ ⊂p. It contradicts to
minimalty of p. Then p∈D(S−1
A).
Let 0 =∩q i (1≤i≤n) be a minimal primary decomposition of 0 where r (q i)= p i. Since pi
is of the form r(0: a) by (4.5), p i is a minimal prime ideal containing (0: a). pi∈D(A).
p∈D(A) ⇒ p is a minimal prime ideal containig (0 : a) for some a(≠0)∈A.
r(0 : a)=r(∩q i: a)=r(∩(q i : a))=∩r(q i : a)=∩p j(j run part of (1≤i≤n) by (4.4)). p⊇r(0 :
a)=∩p j⇒∃pj ⊆p. p is a minimal prime ideal containig (0 : a) ⇒p=pj.
Ex.4.10
i): a∈Sp(0)⇒a/1=0/1⇒ for ∃ s∉p, as(=0)∈p ⇒a∈p.
ii):⇒: In the canonical homomorphism A→Ap, r(0)c=r(0
c) (1.18). r(Sp(0))=p⇒
r(Sp(0))=r(0c)=r(0)
c=p. p is not a minimal prime ideal⇒ a prime ideal ∃p’(⊂p).
⇒r(0) ⊆p’Ap ⊂pAp. ⇒p=r(0)c⊆p’⊂p. It contradicts.
⇐: p is a minimal prime ideal of A⇒rAp(0)=pAp⇒p= rAp(0)c=r (0
c) = r(S p(0)).
iii): a∈Sp(0) ⇒ a/1=0/1. for ∃s∉p,as=0⇒ s∉p’ ⇒a/1=0/1 in Ap’⇒a∈Sp’(0).
iv):⊇: trivial.
⊆: Suppose x≠0. (0 : x) ≠ (1). If (0 : x) ≠0, x is a zero-divisor. a minimal prime ideal ∃p
⊇ (0 : x), p∈ D(A) by (ex.4.9). For this p, x∉Sp(0).
If (0 : x)=0, x is a non zero divisor. For prime ideal ∀q, x/1≠0/1 in Aq. x∉∩Sq(0).
Ex.4.11
Let p be a minimal prime ideal of A. pAp is the unique maximal ideal in Ap.
38
r(0)=pAp⇒0 is the smallest pAp-primary. There is one to one correspondence between
primary ideals of Ap and primary ideals of A which are disjoint with A−p by (4.8). Sp(0)
is contraction of 0. it is the smallest primary ideal in A.
Let a=∩Spα (0) where pα run through the minimal primary ideals. x∉r(0) ⇒a minimal
primary ideal x∉∃p. x/1≠0 in Ap,i.e. x∉Sp (0). ⇒ a⊆r(0).
⇒: Suppose 0=∩q i (1≤i≤n) be a minimal primary decomposition of 0 where r(q i)=p i.
Let a be as above. Spα (0) is the smallest pα-primary ideal. a=0⇒0=(∩q i)=(∩Spα (0)).
⇒r(0)= ∩pi=∩pα. For each α, ∩pi⊆pα. ∃p i⊆pα. pα is minimal prime⇒pi=pα. p i is a
finite set⇒pα is a finite set. ⇒ a=0=∩Spα (0) is a primary decomposition of 0. Then
pα are isolated. pα is a finite set⇒pα =p ii by(4.5).
⇐:0=∩q i (1≤i≤n) be a minimal primary decomposition of 0 where r (q i)= p i. Let p i
be isolated. r(0)= ∩p i=∩pα (pα run through the minimal prime ideals) ⇒∩p i ⊆pα⇒
∃p i =pα, because pα is a minimal prime ideal. Then pα is a finite set. Finite prime
ideals hold ∩pi=∩pα. For ∀pi, there exists ∃pα=pi, because pi is isolated. Then
pα=pi. qi⊇Spα(0) where qi corresponds to pα and Spα (0) is the smallest pα -primary .
Then 0=∩q i ⊇∩S pα (0)=a.
Ex.4.12
i):⊆: x∈S(a)∩S(b) ⇒ ∃a∈a, ∃b∈b, ∃s,t∈S, x/1=a/s=b/t. ∃v∈S,(at−sb)v=0. Let y=
atv=sbv. y∈a∩b.∃u∈S,(xs−a)u=0.⇒tvxsu−atvu=0.(tvsx−y)u=0. x/1=y/tvs⇒x∈S(a∩b).
⊇:trivial.
ii):⊇:x∈S(r(a))⇒x/1=r/s(s∈S,r∈A,∃n>0, rn=a∈a). x
n/1=r
n/s
n=a/s
n∈S−1a. ⇒x
n∈S (a). ⇒
x∈r(S (a)).
⊆: x∈r(S(a)) ⇒∃n>0, xn/1=a/s(s∈S,a∈a)⇒∃t∈S,(x
ns−a)t=0.(tsx)
n∈a.tsx∈r(a). tsx/ts=
x/1∈S−1
(r (a)) . x∈S(r(a)).
iii):⇒: S(a)=(1). ⇒1/1=a/s(a∈a,s∈S). ⇒ ∃t∈S, (s−a)t=0. st∈a⇒a∩S≠∅.
⇐: s∈S∩a. ⇒ ∀x∈A, x/1=sx/s∈S−1a. x∈S(a).
iv):⊆: x∈S1(S2(a))⇒x/1=y/s(y∈S2(a) and s∈S1)⇒∃s’∈S1,(xs−y)s’=0. y/1=a/t(t∈S2,a∈a)
⇒ ∃t’∈S2, (yt−a)t’=0⇒ xss’tt’ −as’t’=0⇒x/1=a/st∈ (S1S2)−1a⇒x∈(S1S2) (a).
⊇: x∈(S1S2) (a) ⇒ x/1=a/st(a∈a,s∈S1,t∈S2). ∃s’∈S1 and ∃t’∈S2 ,(xst−a)s’t’=0. xss’tt’∈a.
⇒xss’/1= S2-1
(a) ⇒xss’∈ S2 (a) ⇒xss’/ss’=x/1∈ S1−1
S2(a). x∈ S1 (S2(a)).
Let a=∩qi be a minimal primary decomposition of a where r(q i)=pi. S−1a=∩S
−1qi.
S(a)=∩S(qi) by (i). S∩p i ≠∅⇒ S−1qi =S
−1A⇒S(q i)=A. S∩pi=∅⇒S(q i)=q i by (4.8). Set
of ideals S(a) is finite because they are intersection of q i.
Ex.4.13
39
i): r(pnAp)=pAp is a maximal ideal⇒p
nAp is a pAp-primary ideal. p
(n) (=(p
nAp)
c) is a
primary ideal by (4.8). r(p(n)
)=r((pnAp)
c)=r(p
nAp)
c=(pAp)
c=p ⇒p
(n) is a p-primary ideal.
ii):Let pn=∩q i be a minimal primary decomposition of p
n where r(q i)=pi. r(p
n)=∩pi.
r(pn)=p. pi is a finite set. ∃pi=p. pi are unique, pj(j≠i)⊃p. Set Sp=A−p.
Sp−1p
n=∩(Sp
−1qi). pj∩Sp≠∅(j≠i). Sp
−1qj=(1). Sp
−1p
n=Sp
−1qi. Sp
−1qi is a Sp
−1p-primary
ideal and Sp(pn)=qi by (4.8). p
(n)=Sp(p
n) =q i is a p-primary component of p
n.
iii): Let p(m)p
(n)=∩qi be a minimal primary decomposition of p
(m)p
(n) where r(qi)=pi.
r(p(m)p
(n))=r(p
(m))∩r(p
(n))=p by(1.13) and (i). ∃pi=p. pj(j≠i)⊃p. Sp
−1(p
(m)p
(n)) =Sp
−1qi.
The p-primary component of p(m)p
(n) is qi. On the other hand, Sp
−1(p
(m)p
(n))=Sp
−1p
(m)
Sp−1p
(n)=p
mp
nAp=p
m+nAp. qi=p
(m+n).
iv):⇒:p(n) =p
n⇒pn is a primary ideal by (i).
⇐:Let pn
be a p-primary ideal. pn∩Sp(=A−p)=∅. Sp
−1p
n is a pAp-primary ideal. By(4.8),
Sp(pn)=p
n⇒p(n)
=pn.
Ex.4.14
Let a=∩q i be a minimal primary decomposition of a where r(qi)=pi and p be a maximal
element of the ideals (a : x).
p is a prime ideal: Let p=(a : x). If not, ∃a,b ∉p,ab∈p. ax∉a ,bx∉a,abx∈a⇒a∈(a : bx)
and a∉ (a : x). ⇒ (a : bx)⊃(a : x) is contradictory.
p=(a : x)⇒p=r(a : x)=∩r(qi : x) by (1.12) and (1.13). x∉a⇒x∉∃qj. Let suffix j indicate
primary ideals not containing x and i indicate primary ideals containing x. r(qi : x)=A.
x∉q j⇒r(q j : x)=pj by (4.4). p=∩pj. p=∃pj by(1.11).
Ex.4.15
Let a=∩qi be a minimal primary decomposition of a where r(qi)=pi. Since f∈p⇔p∉Σ,
Sf = f nn
is a multiplicatively closed set which disjoints with ∪p∈Σp. Sf (a)=∩Sf (q i)
(ex.4.12). For Sf∩pi(∉Σ)≠∅, Sf (qi)=A. For Sf∩pj (∈Σ)=∅, Sf (qj)=q j. Sf (a)= ∩p∈ΣSf (q)
=qΣ. (a : f n)=∩ (qi: f
n). For pi(∉Σ), f
n∈q i for large n(>0),because f∈pi. Then (qi:fn)=(1).
pi∈Σ⇔for any n(>0), f n∉p i. (q i: f
n)=q i by (4.4). qΣ=( a : f
n) for large n(>0).
Ex.4.16
An ideal of S−1
A is of the form S−1a where a is an ideal of A by (3.11). Let a=∩qi be a
minimal primary decomposition of a where r(qi)=pi. S−1a=∩S
−1qi. S∩pi≠∅⇒
S−1qi=S
−1A. S∩pi=∅⇒ S
−1q i is a S
−1p i-primary ideal(4.8). S
−1a is decomposable.
Ex.4.17
40
Let a be an ideal of A and p1 be a minimal prime ideal of a. q1= Sp1 (a) is a p1-primary
ideal (ex.4.11). By (L1) property, ∃x∉p1, q1= Sp1(a)=(a : x).
a= q1∩(a+(x)):
⊆:trivial.
⊇:y∈q1∩(a+(x)) ⇒y=a+rx∈q1(a∈a, r∈A) ⇒yx∈a⇒rxx∈a⇒rx∈(a: x). Sx(=xnn>0) is
a multiplicatively closed set. Sx⊆Sp1. r∈Sx(a)⊆ Sp1(a)=(a : x) ⇒ y∈a.
Remaining part is proved in the text.
Ex.4.18
i)⇒ii): Let a=∩q i be a minimal primary decomposition of a where r(q i)=pi. Let S=A−p
where p is a prime ideal. Σ =p i | S∩pi=∅ is an isolated set. ∃f∈p, f∉p i for pi∈Σ and
Sf(a)=qΣ= (a: fn) for large n(>0) by(ex.4.15). On the other hand, Sp(a)= Sp(∩qi)= ∩Sp(qi)
=∩pi∈Σqi=qΣ=Sf(a)=(a: f n). (L1) is satisfied. (L2) is satisfied because set of ideals S(a) is
finite by (ex.4.12).
ii)⇒i): If (L1) is satisfied, any ideal is intersection of primary ideals by the method in
(ex.4.17). Let a=∩q i where q i is a primary ideal q i and r (q i)=p i (possibly infinite).
Let Sn= A−(p1∪...∪p n). Sn is a descending sequence of multiplicatively closed sets.
By (L2), ∃n(>0), Sn(a)=Sn+1(a).⇒There is not a new prime ideal in processing of the
method in (ex.4.17). ⇒an=(1). ⇒a=∩q i is a finite intersection of primary ideals. By
reducing redundant primary ideals, we get a primary decomposition of a.
Ex.4.19
There is one to one corresponding between p-primary ideals of A and pAp-primary
ideals of Ap. Every pAp-primary ideal contains 0. Every p-primary ideal, i.e. contracted
pAp-primary ideal, includes Sp(0).
Remains to proof are in the text.
Ex.4.20
x∈rM(N)⇔ ∃q>0, xq∈(N : M) ⇔x∈r(N : M). (N : M)=Ann(M/N) by (2.2).
⇒r(N : M)=r(Ann(M/N)). It is clearly an ideal.
Ex.4.21
Let Q be primary in M. xy∈(Q: M ) and x∉(Q: M ) ⇒ ∃m∈M such that xm∉Q. y(xm)∈Q.
xm(∈M, ∉Q) ⇒y is a zero-divisor of M/Q. Since Q is primary in M, y is a nilpotent in
M/Q ⇒ynM/Q=0⇔y
n∈(Q : M ). (Q: M ) is a primary ideal.
Lemma(anologue to(4.3)) : if Qi(1≤i≤n) are p-primary in M, ∩Qi is p-primary in M.
41
Proof: x is a zero-divisor of M/∩Qi⇒x is a zero-divisor M/Qi (1≤i≤n). Then xni
is
nilpotent of M/Qi. Let n=maxni. xnM/∩Qi=0⇒∩Qi is primary in M. r(∩Qi : M)=
∩r(Qi : M)=p. ∩Qi is p-primary. ♦
Lemma(anologue to(4.4)) : Let Q be primary in M, (Q : M)=q and r (Q : M)=p.
i) if x∈q, (Q : x)=M.
ii) if x∉q, (Q : x)=Q’ is p-primary in M.
iii) if x∉p, (Q : x)=Q.
Proof. i): x∈q= (Q : M)⇒(Q : x)=M is clear.
ii): if y is a zero-divisor of m∈M/Q’ , ymx∈Q. If mx∈Q, then m∈Q’. ⇒y is not a
zero-divisor. Then mx∉Q. ∃n>0,ynM⊆Q⊆Q’. Q’ is primary in M. y∈(Q’ : M) ⇒yx∈q.
x∉q⇒∃n>0,yn∈q. Then y∈p. Q’ is p-primary.
iii): (Q : x) ⊇Q: clear.
(Q: x) ⊆Q : m∉Q⇒if mx∈Q, ∃n>0,xn∈(Q: M)⇒it contradicts to x∉p. ♦
Ex.4.22
Lemma: Let Q be p-primary in M. x∉Q, (Q : x)≠ 0⇒ r(Q : M)=r(Q : x) =p.
Proof: x∉Q, (Q : x)≠0⇒a∈(Q : x), a n∈(Q : M)⊆p ⇒ r(Q : x)⊆p.
(Q : x) ⊇ (Q : M) ⇒r(Q : x) ⊇r(Q : M) =p⇒r(Q : M)=r(Q : x)=p. ♦
Lemma(analogue to (4.5)) : Let N= Q1∩…∩Qn be a primary decomposition of N in M
where (Qi : M)=qi and rM(Qi)=r(qi)=pi. pi occurs in ideals of the form r(N : x) for some
x∈M. If r(N : x) is a prime ideal, it occurs in pi.
Proof : there is no redundancy in Qi ⇒¬Qi⊇∩j≠i Qj. ∃x∉Qi and x∈∩j≠i Qj and (Qi :
x)≠0. r(N : x)=r(Qi : x)=pi (above lemma). If r(N : x)=p is a prime ideal, x∉N. x∉∃Qi.
p=r(N : x)=∩x∉Qipi. p=∃pi. ♦
Uniqueness of primary decomposition of N in M: by Lemma(analogue to (4.5)). ♦
0= Q1/N∩…∩Qn/N is a primary decomposition of 0 in M /N:
Qi/N is a primary in M/N: x∈(M/N)/(Qi/N) is a zero divisor⇔∃m∈M such that x(m+N)
⊆Qi+N⇔xm⊆Qi⇔x is a zero divisor of M/Qi⇒xnM⊆Qi ⇒x
n (M/N)/(Qi /N)=0.
x∈(Qi /N : M /N)⇔x (M/N)⊆Qi/N⇔xM⊆Qi⇒(Qi/N : M/N)=(Qi : M) ⇒pi=r(Qi/N : M/N).
Ex.4.23
(4.6) of modules
’Let N be decomposable in M and N=Q1∩…∩Qn be a minimal primary decomposition
where Qi is primary in M and rM(Qi)=pi. If P(⊇N) is a p-primary, p includes at least one
minimal prime ideal belonging to N in M. Then a minimal prime ideal in the set of
rM(Q) where Q (⊇N) is primary in M occurs in pii’.
42
Proof : (N : M)⊆(P : M). By taking radicals,p1∩…∩pn⊆p. ∃pi⊆p by (1.11). ♦
(4.7) of modules
’ Let N be decomposable and N=Q1∩…∩Qn be a minimal primary decomposition of N
in M where rM(Qi)=pi.
∪i=1,...,npi=x∈A | (N : x) M≠N.
Especially, in the case of N=0, the set of zero-divisors of M is a union of prime ideals
belonging to 0 in M.’
proof:⊇: (N : x)M≠N. ⇒∃m∈(N : x) M,m∉N,xm∈N⇒m∉∃Qi. x is a zero-divisor of M/Qi.
Qi is primary⇒ ∃s(>0), xs∈ (Qi : M). ⇒ x∈rM(Qi)=pi.
⊆: x∈pi=r(Qi : M) ⇒. ∃s>0, xs∈(Qi:M). x
sM⊆Qi . There is not redundant Qi⇒
∃m∈∩Qj (j≠i),m∉Qi(m∉N). Since Qi is primary, xsm∈∩Qi =N. Let k be the smallest
number which mxk∈N. x is a zero-divisor of M/N for mx
k-1∉N.
N=0⇒x∈A|(0: x) M≠0 is a set of zero divisor of M ⇒∪i=1,...,npi is a union of prime
ideals belonging to 0 in M. ♦
(4.8) of modules
’Let S be a multiplicatively closed set of A and Q be a p-primary in M.’
i) S∩p≠∅⇒S−1
Q= S−1
M.
ii)S∩p=∅⇒S−1
Q is a S−1p-primary in S
−1M . Contraction of S
−1Q is Q.
There is the one to one correspondence between primary submodules in S−1
M and
primary submodules in M of which prime ideal disjoints with S.’
Proof:
i):S−1
Q⊆S−1
M is trivial.
⊇: Suppose m/s∈S−1
M. For ∀p∈S∩p, ∃n>0,mpn∈Q. m/s =mp
n/sp
n∈S−1
Q.
ii): Let a/s∈S−1
A be a zero-divisor of S−1
M/S−1
Q. ⇒∃m/t(≠0)∈S−1
M,(a/s)(m/t)
=q/u∈S−1
Q. ∃v∈S,(amu−stq)v=0. amuv∈Q. Q is primary⇒∃n>0,(auv)nM⊆Q.
⇒(a/s)nS
−1M⊆S
−1Q. ⇒S
−1Q is primary in S
−1M. r(S
−1Q : S
−1M)=S
−1q where q is a prime
ideal of A.
p⊆q : Since S−1
r(Q : M)⊆r(S−1
Q : S−1
M).
p⊇q : If x∉p, xm∉Q for ∀m∉Q (∈M). Suppose x/1 is a zero-divisor of S−1
M/S−1
Q.
∃m/s(≠0), (x/1)(m/s)=q/s’∈S−1
Q. (xms’−sq)t=0. For ms’t∉Q, xms’t∈Q. Contradictory.
x/1 is not a zero-divisor of S−1
M/S−1
Q. S−1
Q is primary in S−1
M⇒ x/1∉ S−1q⇒ x∉q
Then p=q. S−1
Q is S−1p-primary. (a∈A,m∈M,s,t,u,v∈S).
Correspondence from S−1p-primary to p-primary:
Let N’ be S−1p-primary in S
−1M and N=m∈M | ∃s∈S, m/s∈N’. S
−1N=N’ .
N is p-primary in M: If x(≠0)∈A is a zero divisor of M/N, ∃m∈M(∉N),xm∈N. ⇒∃s∈S,
43
xm/s(=(x/1)(m/s))∈N’. (m/s)∉N’ ⇒ ∃n>0, (x/1)nS
−1M⊆N’. This means that for ∀m’∈M,
xnm’/1∈N’ ⇒x
nm’∈N. ⇒N is primary in M. x/1∈S
−1p. Since S∩p=∅, x∈p.
Let r(N : M)=p’. Since extention of p’ is S−1p by above, p’=p. Correspondence from
S−1p-primary to p-primary is injective. By (i) and (ii), this correspondence is one to one.
For p-primary Q where p disjoints S, Q=(S−1
Q) c is clear by above. ♦
(4.9) of modules
’Let S be a multiplicatively closed set of A and N=Q1∩…∩Qn be a primary
decomposition of N in M where rM(Qi)=pi. Suppose S ∩ pi≠∅ for 1≤i≤m and S ∩ pi=∅
for m+1≤i≤n . Let S(N) denote contraction of S−1
N to M.
S−1
N =∩S−1
Qj, S(N )= ∩Qj (m +1≤j≤n).
This form is a minimal primary decomposition of S(N ) in M.’
Proof:
By localizing, S−1
N=∩1≤i≤nS−1
Qi (3.4). For pi1≤i≤m, S−1
Qi=S−1
M. For pi m+1≤j≤n,
S(S−1
Qj)= Qj by ((4.8) of modules). S(N)=∩ m+1≤j≤n Qj. ∩Qj is not redundant because it is
a part of Q1∩…∩Qn.♦
(4.10) of modules
’Let N be decomposable and N=Q1∩…∩Qn be a minimal primary decomposition of N
in M where Qi is primary in M and rM(Qi)=pi. Let Σ(=pii=i1,…,im)be an isolated set.
N Σ=Qi1∩…∩Qim does not depend on decomposition.’
Proof:
Let N=Q’1∩…∩Q’n’ be an another minimal primary decomposition of N in M where Q’j
is primary in M and rM(Q’j)=p’j. pi=p’j (n=n’) by(ex.4.22). pii=i1,…,im=p’j
j=j1,…,jm. p’j is isolated.
Let S=A−∪i=i1,…,impi. S−1
(Q1∩…∩Qn )=S−1
(Q’1∩…∩Q’n). ⇒ (S−1
Qi1∩…∩S−1
Qim )=
(S−1
Q’i1∩…∩ S−1
Q’im) by ((4.9) of modules). S(N)=NΣ=Qi1∩…∩Qim=Q’j1∩…∩Q’jm.♦
(4.11) of modules
’Especially ,isolated primary component Qi only depends on N.’
Proof : Take S=A−pi . Trivial.♦
Chapter 5
Ex.5.1
Let f =φ°ψ where ψ:A→f(A) be surjective and φ: f(A)→B be injective. f *= ψ*
°φ*.
Let V(b) be a closed set of Spec(B). B is integral over f(A)⇒B/b is integral over
f(A)/bc(5.6). For ∀p∈Spec(f(A)/b
c), ∃q∈Spec(B/b), φ*
(q)=p(5.10) ⇒Spec(f(A)/bc)⊆
φ*(Spec(B/b)). Spec(f(A)/b
c) ⊇φ*
(Spec(B/b)) is trivial. φ*(V(b)) =φ*
(Spec(B/b)) =
44
Spec(f(A)/bc)= V(b
c) is a closed set of Spec(f(A)).
ψ is surjective. ψ*:Spec(A/Ker(f))=Spec(f(A))→V(Ker(f)) is a surjective
homeomorphism (ex.1.21iv). ψ*°φ
* (V(b))=f
*( V(b)) is a closed set in V(Ker(f)). A closed
set of V(Ker(f)) is closed in Spec(A). Then φ*(V(b)) is a closed set in Spec(f(A)).
Ex.5.2
Let f:A→Ω be a homomorphism. p=f−1
(0) is a prime ideal of A. B is integral over A⇒
∃q∈Spec(B), q∩A=p (5.10). Let K=κ(A/p) and L=κ(B/q). ∀b(∈B) is integral over A,
⇒ b ∈B/q is integral over A/p (5.6i). ⇒b is integral over K. b/1 is integral over K.
L is an algebraic extension over K.
Define f : K→Ω by f ( yx / )=f(x)/f(y) for x,y∈A(y∉p) and x =x mod p and y =y mod
p. This definition does not depend on representatives. )()1/( xfxf = . f is the
composition of homomorphisms A→A/p→K. Regard f as the embedding. K⊆L⊆Ω.
Let g : L→Ω be the embedding. g |K= f . Define g be the composition of
homomorphisms B→B/q→L →g Ω. g(b)= g ( 1/b ) for b∈B. For any a ∈A,
g(a)= )1/(ag = )1/(af = 1/)(af =f(a). Then g| A=f.
Ex.5.3
x∈B’⊗AC ⇒ x=Σb’i⊗ci (finite sum). By(5.2), it is enough to show each b’⊗c is integral
over f⊗1(B⊗C). b’ satisfies a monic polynomial.
b’ n+u1b’
n−1+...+un=0 (ui∈f(B)).
Multiplying by 1⊗ c n
,
(b’ ⊗ c) n
+(u1⊗ c ) (b’
⊗ c) n-1
+...+un⊗ c n
=0 (ui⊗ c i ∈ f⊗1(B⊗C)).
Then f ⊗1: B⊗C →B’⊗AC is integral .
Ex.5.4
A counterexample.
Let B=k[x]⊃A=k[x2−1]. n=(x−1) is a maximal ideal of B. A∩n=m=(x
2−1) is a maximal
ideal of A. A−m=g(x2−1) ∈A | g(0)≠0. If 1/(x+1)∈Bn is integral over Am,
(1/(x+1)) n
+(h1(x2−1)/g1(x
2−1))(1/(x+1))n−1
+…+hn(x2−1)/gn(x
2−1)=0.
( gi and hi are polynomials of (x2−1). gi(x
2−1)∈A−m.)
Multiplying by (x+1)nΠgi(x
2−1),
Πgi(x2−1)+h1(x
2−1)Πi≠1gi(x2−1)(x+1) +…+ hn(x
2−1)Πi≠ngi(x2−1) (x+1)
n=0.
Substitute x by −1, Πgi(0)=0. It is contradictory. Bn is not always integral over Am.
Ex.5.5
45
i) : x∈A, x−1∈B⇒x
−1 is integral over A.
x−n
+a1x1−n
+…+an=0 (ai∈A).
Multiply by xn−1
.
x−1
= −(a1+ a2x+…+an x n−1
) ∈A.
ii): J(A)=∩ m:maximal idealm. J(B)=∩ n:maximal ideal n. For ∀mα, ∃nβ ∩A=mαby (5.10) and
(5.8). ⇒ J(A) ⊇J(B) ∩A.
J(B) ∩A=∩ n:maximal ideal (n∩A). n∩A is a maximal ideal of A by (5.8). J(A) ⊆J(B)∩A.
Ex.5.6
By induction, it is enough to prove the case B×C.
x∈B×C is a finite sum of (bi,ci). It is enough to show that each (b,c) is integral over
A(5.2). b is integral over A⇒A[b](⊆B) is finitely generated module over A. A[c] (⊆C) is
finitely generated modul over A. A[b] ×A[c] (⊆B×C) is finitely generated module over A.
(b,c) ∈A[b] ×A[c] is integral over A.
Ex.5.7
Suppose x(∈B) is integral over A. Let xn+a1x
n−1+…+an=0 (ai∈A) be the least degree
of monic polynomial which x satisfies. Then x(xn−1
+a1x n−2
+…+an−1)=−an∈A.
B−A is a multiplicatively closed set. If x∈(B−A), (xn−1
+a1x n−2
+…+an−1) ∉(B−A). Then
xn−1
+a1x n−2
+…+an−1=a ∈A. It contradicts to minimality of n. Then x∈A.
Ex.5.8
i) : Let f(x),g(x)∈B[x] be monic polynomials. Suppose f(x)g(x)∈C[x]. Let ζi, ηj ∈K be
roots of f(x) and g(x) respectively(K be a field (⊇B) which split f(x) and g(x) into linear
forms). ζi, ηj are integral over C because they are roots of monic polynomial
f(x)g(x)∈C[x]. Coefficients of f(x) (resp. g(x)) are represented by finite sums and
products of ζ i (resp. ηj) which are integral over C. Then Coefficients of f(x) and
g(x) which are elements of B are integral over C. Then f(x), g(x)∈C[x](5.5).
ii) : Let f(x)=x n
+s1x n-1
+…+ sn∈B[x], g(x)=x m
+t1x m-1
+…+tm∈B[x] and f(x)g(x)∈C[x].
Consider )()( xgxf (=f(x)g(x) mod (q∩C))∈C/(q∩C)[x] for any prime ideal q of B.
C/(q∩C) is integral over A/(q∩A). C/(q∩C) is included by integral closure of A/(q∩A)
in B/q. By(i), ji ts , are integral over A/(q∩A).
Suppose s∈B. s mod q is integral over A/(q∩A) for prime ideal ∀q of B⇒s∈C :
If not, s∉C. T=h(s) |h : any monic polynomial over A is a multiplicatively closed set,
because 0∉T. a maximal ideal ∃p which disjoint with T is a prime ideal of B.
46
T∩p=∅⇒0∉T mod p⇒ s mod p is not integral over A/(A∩p). Contraditory. Then s∈C.
Coefficients of f(x) and g(x) belong to C.
Ex.5.10
i):
(a)⇒(b): Let f : A→B be a ring homomorphism. Let p=f*(q) where p∈Spec(A) and
q∈Spec(B). V=q’∈Spec(B) | q’⊇q is a closed set. f*(V) is a closed set by (a). p’⊇p⇒
p’∈f *(V) ⇒ ∃q’∈V , p’=f
*(q’) ⇒ q’⊇q.
(b)⇒(c): For ∀q∈Spec(B), set f*(q)=p. For ∀ 'p (∈Spec(A/p))=p’ mod p where p’ ∈
Spec(A), ∃q’∈Spec(B),q’⊇q, p’= f*(q’) by (b). Let q′ = q’ mod q. Then )(
** | qVff = :
Spec(B/q)→Spec(A/p) is surjective.
(c)⇒(b) : Let p,p’∈Spec(A), p⊆p’ and ∃q∈Spec(B), f*(q)=p. *f :Spec(B/q)→
Spec(A/p) is surjective by (c). ⇒ for p′ ∈Spec(A/p), ∃q′ ∈Spec(B/q), p′ = )(* q′f . p’ =
f*(q’). f has goingup property.
ii):
(a’) ⇒ (b’): as in the text.
(b’) ⇒ (c’): Set f*(q) = p∈Spec(A) for ∀q∈Spec(B). A member of Spec(Ap) is of the
form p’p where p’∈Spec(A) and p⊇p’. q⊇∃q’∈Spec(B),p’=f*(q’) by (b’). q’q∈Spec(Bq).
Then f*
: Spec(Bq)→Spec(Ap) is surjective.
(c’) ⇒ (b’): Let p⊇p’ for p, p’∈Spec(A) and p=f*(q) where q∈Spec(B). p’p∈Spec(Ap).
Since f*q
: Spec(Bq)→Spec(Ap) is surjective by (c’), ∃q’q∈Spec(Bq), p’p = f
* q (q’q).
q’q∩B=q’ ∈Spec(B). Then p’= f*(q’).
Ex.5.11
Suppose f : A→B is a flat homomorphism. Set p=qc for ∀q∈Spec(B). f
*: Spec(Bq)
→Spec(Ap) is surjective by(ex.3.18). f has the goingdown property(ex.5.10ii).
Ex.5.12
G is a finite group. For ∀x∈A, f(t)=Πσ∈G (t−σ(x)) is a monic polynomial which x
satisfies. Coefficients of f(t) are G-invariants because they are symmetric polynomials
of σ(x)σ∈G. ⇒f(t) is a monic polynomial over AG. x is integral over A
G.
σ(S)⊆S for ∀σ∈G. Define the action of G on S−1
A by σ(a/s)=σ(a)/σ(s)∈S−1
A for
a/s∈S−1
A. σ(1) σ(1)= σ(1). Multiplying σ(1) (σ(1)−1)=0 by σ−1, we get σ−1
(1)=1⇒
σ(1)=1. This definition is a canonical extention of the action of G on A because
σ(a/1)=σ(a)/1.
47
Define a homomorphism φ:(S−1
A) G
→(SG) −1
A (⊇(SG) −1
AG ) as following.
For a/s∈(S−1
A) G
,set t=Πσ∈G σ(s) (∈SG).
φ(a/s)=a(Πσ≠1σ(s))/t (∈ (SG) −1
A).
For ∀τ∈ G , a/s=τ (a/s)⇒
φ(a/s)=a(Πσ≠1 σ (s))/t=φ(τ (a)/ τ (s) )= τ (a) (Πσ≠τσ (s))/t .
∃ sτ∈SG ,(a(Πσ≠1 σ (s))− τ (a) (Πσ≠τσ (s))) sτ=0.
Set Πτ∈G sτ=t’ (∈SG) and b= t’aΠσ≠1σ (s). For ∀τ ∈G,
τ (b)= t’ τ (a) Πσ≠τσ (s)=(Πσ≠τsσ) τ (a) Πσ≠τ (σ (s)) sτ=t’aΠσ≠1 σ (s) =b.
⇒b∈AG.
φ(a/s)=τ (a) Πσ≠τσ (s)/t= t’ τ (a) Πσ≠τσ (s)/tt’= b/tt’ ∈(SG)−1
AG.
φ(a/s)= b/tt’=0⇒ ∃u∈SG(⊆S), bu=t’aΠσ≠1σ(s)u=0⇒t’tau=0⇒a/1=0 in S
−1A. Then a/s
=0 in (S−1
A)G. φ:(S
−1A)
G →(S
G) −1
AG is injective. It is clear that any element of (S
G)−1
AG
has an inverse image in (S−1
A)G. φ is an isomorphism.
Ex.5.13
p∈Spec(AG). Let P=q∈Spec(A) | q∩A
G=p. For ∀q∈P, σ(q) is a prime ideal (If
xy∈σ(q), σ−1(x)σ−1
(y)∈q⇒ (σ−1(x)∈q)∨(σ−1
(y)∈q)⇔(x∈σ (q))∨(y∈σ(q))). q∩AG=p⇒
σ(q)∩AG=p. σ(q)∈P.
Let q’(≠q)∈P. For ∀x∈q, Πσ∈Gσ(x)∈(AG∩q) =p⊆q’. ∃σ(x)∈q’ ⇒x∈σ−1
(q’)
⇒q⊆∪σ∈Gσ (q’). q⊆∃σ (q’) (1.11). q=σ (q’) by (5.9). G acts transitively on P. Since G
is finite, P is finite.
Ex.5.14
Let A be an integrally closed domain, K =κ(A), L be finitely normal separable extention
of K and B be the integral closure of A in L. x(∈B) is integral over A. Let f(t) be the
monic polynomial over A of the least degree which x satisfies. x∈L. Let g(t) be the
monic polynomial over K of the least degree which x satisfies. f(t)∈A[t](⊆K[t]). There
exists h(t)∈K[t] such that f(t)=g(t)h(t). Since A is integrally closed, g(t) is a monic
polynomial over A by (ex.5.8). f(t) is the monic ploynimial of least degree over A.
f(t)=g(t). Let G=G(L/K). For ∀σ∈G, σ(x)∈L. σ(x) satisfies g(t). Since σ(x)∈B, σ(B)⊆B.
As σ−1(B)⊆B, σ (B)=B.
Let y∈BG(⊆L). y is invariant by G=G(L/K). y∈K. y∈B
G(⊆B) is integral over A. A is
integllay closed. y∈A. BG⊆A. A⊆B
G is clear. A=B
G.
Ex.5.15
(a)L is a finite separable extension field of K.
48
Let A be an integrally closed domain, K =κ(A), L be a finitely separable extention of K
and B be the integral closure of A in L. Let M be the smallest finitely normal separable
extension field including L and G=G(M/K). G is a finite group. Denote by M(B) the
integral closure of A in M. (M(B))G =A by (ex.5.14).
For ∀p∈Spec(A), ∃r∈Spec(B) , r∩A=p by (5.10). Since M(B) is integral over B(⊇A),
∃q∈Spec(M(B)),r=qc
by (5.10). q∩A=q∩B∩A=r∩A=p. For p∈Spec(A),
q∈Spec(M(B)) | p=qc is finite (apply (ex.5.13) for G and (M(B))
G =A ). r∈Spec(B)
such that r∩A=p is of the form r=qc. Then q∈Spec(B) | p= r
c is finite.
(b) L is a finitely purely inseparable extension field over K.
By (5.10), ∃q∈B,p=qc. Since L is a finitely purely inseparable extension field of K, for
∀z∈L , ∃e ≥0 , Kzep ∈ .
We show that Q=y∈B| ∃e≥0,epy ∈p is q. Q∩A=p is clear.
Q is an ideal: x,y∈Q. ⇒ ∈epx p, ∈
fpy p. Let g=max(e,f). ∈+gpyx )( p⇒x+y∈Q. b∈B⇒
∃f ≥0, ∈f
pb K. x ∈Q⇒ ∈e
px p. Let g=max(e,f). Since g
pb ∈K is integral over A and A is
integrally closed,g
pb ∈A. ∈gg
pp xb p. Then bx∈Q.
Q is a prime ideal: Let x,y∈B and xy∈Q. ∈epxy)( p, ∈
fpx A and ∈gpy A for some e, f,
g≥0. Let h=max(e,f,g). Since ∈hh pp yx p, ∈
hpx p or ∈hpy p. Then x∈Q or y∈Q.
Suppose z∈q. ∃e≥0,e
pz ∈K. Since e
pz ∈K is integral over A and A is integrally closed, e
pz ∈A. Since e
pz ∈q, e
pz ∈p. Then z ∈Q. q⊆Q. q=Q(5.9). Then Spec(B)→ Spec(A)
is bijective.
A finite extension filed is composition of a finitely separable extension and a finitely
purely inseparable extension. p has finite fibers.
Ex.5.16
Let A be a finitely generated algebra over infinite field k. Then there exist y1,…,yr in A,
which are algebraically independent over A, such that A is integral over k[y1,…,yr]:
Proving by induction on n, the number of generators. If n=r, nothing to prove. Then we
assume n>r.
The case of n=1: r=0. x1, which is a generator, is algebraic over k . Then k[x1] is integral
over k.
Suppose the case n −1 is true.
The case of n: We can renumber xi if needed so that x1,…,xr are algebraically
49
independent over k and xr+1,…,xn are algebraic over k[x1,…,xr]. Since xn is algebraic
over k[x1,…,xn−1], there exists a nonzero polynomial f such that f(x1,…,xn−1,xn)=0. Let s
be the highest degree of f and F be the homogeneous part of degree s of f and F’ be
others (f=F+F’). Since k is infinite, ∃α1,…,αn−1 ∈k, F(α1,…,αn−1,1)≠0. Let x'i=xi−αixn
for each i(<n). f=F(α1,…,αn−1,1)xns + F”(homogeneous polynomial of the degree s, but
degree of xn<s) +F’ (polynomial of the degree<s)=0. By dividing F(α1,…,αn−1,1)(≠0),
xn is integral over k[x’1,…,x’n-1]. By induction hypothesis, ∃y’1,…,y’r ∈k[x’1,…,x’n-1],
which are algebraically independent over k , such that k[x’1,…,x’n-1] is integral over
k[y’1,…,y’r]. xn is integral over k[y’1,…,y’r](⊆A)(5.4). k[x’1,…,x’n-1, xn]=k[x1,…, xn]=A is
integral over k[y’1,…,y’r]. Substitute x'i by xi−αixn . y’1,…,y’r are linear combinations of
x1,…, xn and we rewrite them y1,…,yr. Then k[x1,…,xn] is integral over k[y1,…,yr].
Let k be an algebraically closed field. Let X be an affine algebraic variety in kn and
A=k[x1,…,xn] be its coordinate ring. From the above, we get y1,…,yr which are
algebraically independent over k and x’r+1,…,x’n which are integral over k[y1,…,yr,],
by a linear transformation of (x1,…,xn). Followings are described in terms of
coordinate system (y1,…,yr, x’r+1,…,x’n). Regard y1,…,yr as indeteminates because of
their algebraic independence. Let L be the r-degree linear subspace in kn which y1,…,yr
generate. The coordinate ring of L is B=k[y1,…,yr]. The coordinate ring of X is
A=k[y1,…,yr,x’r+1,…,x’n]. A is integral over B. Let P=(k1,…,kr) be an arbitrary point of L.
Let f:B→k be a homomorphism defined by (y1,…,yr)→(k1,…,kr). Since k is an
algebraically closed field, a homomorphism ∃g:A→k which is an extension of f by
(ex.5.2). Let g(x’i)=ki∈k (r<i≤n). (k1,…,kr,kr+1,…,kn) satisfies polynomial equations by
which x’r+1,…,x’n are integral over k[y1,…,yr]. Then (y1−k1,…, yr−kr, x’r+1−kr+1,…,
x’n−kn) is a maximal ideal of A and Q=(k1,…,kr,kr+1,…,kn) is a point of X. Let γ:X→L be
(y1,…,yr,yr+1,…,yn)→(y1,…,yr). γ is a linear transformation. γ is surjective.
Ex.5.17
Let k be an algebraically closed field and A=k[x1,…,xn] be a polynomial ring over k. Let
I be an ideal of A and X be V(I). If I≠(1), P=A/I≠0. P is a finitely generated k-algebra.
Let t1,…,tn be images of x1,…,xn in P. By some linear transformation, we can get
z1,…,zm, zm+1,…,zn such that z1,…,zm are algebraically independent over k and zm+1,…,zn
are integral over k[z1,…,zm] (ex5.16). The monic polynomials over k[z1,…,zm] which
zm+1,…,zn satisfy generate I. We can solve zm+1,…,zn after substituting any values of k
for z1,…,zm, because k is an algebraically closed field. Thus We get (z1,…,zn), which is a
zero point of I, i.e. ,the coordinate of a point of X after the linear transformation. Then
50
there exists a map from X onto km. Then X≠∅.
Let m be any maximal ideal. P=A/m is a finitely generated algebra over k and a field. If
m≠0, P is integral over k[z1,…,zm]. Since P is a field, k[z1,…,zm] is a field by (5.7). It is
contradictory. Then m=0. P is an algebraic extention over k, then P=k, because k is
algebraically closed. Let αi be an image of ti in k. The image of αi is also αi. Then
(t1−α1,…, tn−αn)⊆m. (t 1−α1,…, t n−αn) is a maximal ideal. Then (t 1−α1,…, t n−αn) =m.
Ex.5.19
I(X)≠(1) ⇒ a maximal ideal ∃m∈Spec(k[t1,…,tn]),I(X)⊆m. Since X⊇ zero points of m,
it is enough to show that the set of zero points of m is not empty. A=k[t1,…,tn]/m is a
field. A is a finitely generated k-algebra. A is a finite algebraic extension of k by (ex.
5.18). Since k is an algebraically closed field, A=k. t1,…,tn are algebraic over k. Let φ:
k[t1,…,tn]→k=k[t1,…,tn]/m be a homomorphism. Let ai (∈k) be the image of ti by φ.
Kerφ=m⊇(t1−a1,…,tn−an). I=(t1−a1,…,tn−an) is a maximal ideal. Then m=I. Since
(a1,…,an) is a zero point of m, X is not empty.
Ex.5.20
Let B be an integral domain, A be a subring of B and B be a finitely generated algebra
over A, i.e. B=A[z1,…,zm]. Let K=κ(A) and S=A−0. K=S−1
A. S−1
B is a finitely
generated algebra over K. ∃x1,…,xn∈S−1
B such that they are algebraically independent
over K and S−1
B is integral over K[x1,…,xn] by (ex.5.16). Let xi=b i /s i (bi∈B,s i∈S). Let
gj be a monic polynimial having coeffients in K[x1,…,xn] which zj/1(∈S−1
B) satisfies.
)1(][ 1
1
1 mj,x,...,xKff...YfYg nj,ij,n
n
j,
n
j j
jj ≤≤∈+++= −
Let s be the product of sii=1,…,n and denominators of coefficients of all of
fα,j 1≤α≤nj,1≤j≤m. s∈S. Let xi=b i /s i=y i /s ∈S−1
B. y i are algebraically independent over
K. (Because if not, xi are not algebraically independent over K.) Let mj be the highest
degree of fi,j 1≤i≤nj. Let
][ 1
1
1 nj,ij,n
n
j,
nm
j
m
j y,...,yAuu...YuYsgshj
jjjj ∈+++== −.
Multiply hj by )( jj
nmnmY jj >−
or )( jj
mnnms jj <−
and denote it by h’j. h’j is a monic
polynomial over A[y1,…,yn] which szj satisfies. A[sz1, …,szm] is integral over B’=A[y1,
…,yn]. A[sz1,…,szm]s = Bs is integral over B’s by (5.6ii).
51
Ex.5.21
Let A be a subring of B which is an integral domain and finitely generated over A.
∃s(≠0)∈A and ∃y1,…,yn(∈B), Bs is integral over B’s where B’=A[y1,…,yn] and y1,…,yn
are algebraically independent over A (ex.5.20). Let f : A→Ω(an algebraically closed
field) be a homomorphism such that f(s)≠0. Define a homomorphism h: B’→Ω by
h(yi)=0 for i≤n and h(a)=f(a) for ∀a∈A. h is an extention of f. Define a homomorphism
hs: B’s→Ω by hs(b’/si)=h(b’)/f(s
i). ∃ gs: Bs→Ω which is an extension of hs (ex.5.2).
Define a homomorphism g : B→Ω by g(b)= gs (b/1), i.e., g(b)= gs |B. g is an extension
of f.
Ex.5.22
Let B be an integral domain, A be a subring of B and B=A[x1,…, x n]. For ∀v (≠0)∈B, Bv
is a finitely generated algebra over A (x1,…,xn and 1/v generate Bv ). Then there exists
∃s(≠0)∈A, (Bv)s is integral over (Bv’)s where Bv’ =A[y1,…, yr] in which y1, …, yr (∈Bv)
are algebraically independent over A by (ex.5.20). Since J(A)=0, a maximal ideal ∃m,
s∉m. A/m=k is a field. Let f : A→Ω (Ω is an algebraically closed field including k(y1,…,
yr)) be a homomorphism. f(s)≠0. f can be extended to an homomorphism g : Bv →Ω by
(ex.5.21). Let g(yi)= yi. g(v)≠0, because v is a unit of Bv. To claim J(B)=0, it is enough to
show that Ker(g)∩B=n is a maximal ideal of B and v∉ n.
Since g(Bv) includes k(y1,…, yr) and is algebraic over k(y1,…, yr), g(Bv) is a field (5.7).
Then Ker(g) is a maximal ideal of Bv. There is one to one correspondence between
prime ideals of Bv and prime ideals of B which are disjoint with vi(3.11iv). Since
v∉Ker(g), Ker(g)∩B=n is a maximal ideal of B such that v∉n.
Ex.5.23
i) ⇒iii):Suppose a prime ideal p is not a maximal ideal and p⊂∩ p⊂q∈Spec(A) q .
p⊂(∩ p⊂q∈Spec(A)q)⊆(∩ p⊂m:maximal ideal m). It contradicts (i).
ii) ⇒i) : If a prime ideal p is not an intersection of maximal ideals, J(A/p)≠0. Since A/p
is a integral domain, rA/p(0)=0. J(A/p)≠rA/p(0). There exists a homomorphic image in
which the Jacobson radical does not equal to the nilradical. Contraditory.
iii)⇒ii): Suppose a homomorphism ∃f, J(f(A))≠rf(A)(0). ∃x∈J(f(A)), x∉rf(A)(0). Let
S=xn(0∉S). S is a multiplicatively closed set in f(A). A maximal element q in the
family of ideals which are disjoint with S is a prime ideal. Every maximal ideal of f(A)
contains x. Since x∉q, q is not a maximal ideal of f(A). Every prime ideal strictly
containing q joints with S and contains x. There is one to one correspondence between
prime ideals of f(A) and prime ideals of A/Ker(f). f−1
(q)=p is not a maximal ideal of A.
52
p∩f−1
(x)=∅. p=∩p⊂pα∈Spec(A) pα by hypothesis. Each f(p α) is a prime ideal of f(A)
strictly containing q⇒ it contains x. p∩f−1
(x) ≠∅. Contradictory.
Ex.5.24
i): the case that B is integral over A: Assume B is not a Jacobson ring. A prime ideal ∃q
of B which is not an intersection of maximal ideals containing q by (ex.5.23i). J(B/q)≠0.
B/q is an integral domain, i.e.rB/q(0)=0. ∃ f (≠0)∈J(B/q). Let qc=A∩q=p. Let f∈B be an
inverse image of f . f∉q. f is integral over A.
f n+a1f
n-1+…+an=0 (ai∈A).
In B/q,
)(01
1 p/Aaa...faf in
nn ∈=+++ − .
fa...faf n
nn
1
1
1 −− +++ ∈J(B/q) and na ∈A/p.
B/q is integral over A/p⇒J(B/q)∩ A/p = J(A/p) by (ex.5.5ii).
fa...faf n
nn
1
1
1 −− +++ ∈J(A/p).
A is a Jacobson ring. J(A/p)=r(0)=0 by (ex.5.23ii). na =0. fa...faf n
nn
1
1
1 −− +++ =0.
B/q is an integral domain and f ≠0. ⇒ 01
2
1
1 =+++ −−−
n
nn a...faf .
Consequently,∀ ia =0. nf = 0. Contradictory. B is Jacobson ring.
ii):the case that B is a finitely generated A-algebra: Let q be any prime ideal of B. B/q is
a finitely generated A/p-algebra where A∩q=p. Since A is a Jacobson ring, J(A/p)=0.
J(B/q)=0 by (ex.5.22). q is an intersection of maximal ideals. B is a Jacobson ring.
Especially, the ring which is finitely generated over a Jacobson ring is a Jacobson ring
by (ii). Since a field is a Jacobson ring, an algebra finitely generated over a field is a
Jacobson ring,
Ex.5.25
i)⇒ii): Let B =A[x1,...,x n] be a field . ∃y∈B is an algebraically independent over A⇒
for an irreducible polynomial h(y),1/h(y) ∈ B. Since 1/h(y) is algebraically independent
over A, it cannot be represented by a product or sum of other elements of B. Number of
irreducible polynomials is infinite. It contradicts to that B is a finitely generated
A-algebra. Then B is algebraic over A. A is a Jacobson ring and an integral domain, i.e.
53
J(A)=0. Let s(≠0)∈A be as in ex.5.21. A maximal ideal ∃m, s∉m. Let f : A→A/m→Ω (Ω
is the algebraically closed field including A/m=k) be a composition of homomorphisms.
f(s) ≠0. f can be extended to a homomorphism g : B→Ω (ex.5.21). g(B) is a field
because B is a field. Since B is finitely generated algebra and algebraic over A, g(B) is a
finitely algebraic extention of k. g(B) is finite over k. g is injective. Suppose that B is not
finite over A. Since B=A[x1,…,x n], ∃x∈x1,...,x n,A[x] is not finite over A. Then x is not
integral over A (5.1). g(A[x]) is finite over k, i.e., g(A[x])=k+kg(x)+…+kgr-1
(x). gr(x)=kr-1
gr-1
(x)+…+k1g(x)+ k0. xr is an inverse image of g
r (x) and ar-1 x
r-1 +…+a1g(x)+ a0 is an
inverse image of kr-1 gr-1
(x)+…+k1g(x)+ k0. But xr≠ ar-1 x
r-1 +…+a1g(x)+ a0 because x is
not integral over A. It is contradictory to that g is injective. Then B is finite over A.
ii)⇒i):Let p be any non-maximal prime ideal of A. B=A/p is a domain, not a field. ∀f
(≠0)∈ A/p is not a zero-divisor. Bf≠0. Bf (=B[1/f]) is a finitely generated A-algebra. If Bf
is a field, Bf is finite over A by (ii). Bf is finite over B. 1/ f is integral over B. B is a field
by(5.7). Contadictory. Then Bf is not a field. A prime ideal ∃q(≠0) ⊆ Bf, f/1∉q. f∉qc
(⊆B=A/p). For any f (≠0), There exists a prime ideal which does not contain f. ⇒0, a
prime ideal of B, is the intersection of all non 0 prime ideals in B. In A, p is an
intersection of all prime ideals which strictly contain p. A is a Jacobson ring by
(ex.5.23iii).
Ex.5.26
(1)⇒(2) : Let E be a closed set of X.
0XE ∩ =X−∪O∩E∩X0=∅O.
O∩E is locally closed. If O∩E≠∅, O∩E∩X0≠∅ by (1).
O ∩E∩X0=∅ ⇔O ∩E=∅. ⇒ 0XE ∩ = X−∪O∩E=∅O = EE = .
(2)⇒(3):
surjective: Cleary.
injective: Let V,U be open sets of X and (X0∩V)=(X0∩U).
X0−(X0∩V)=X0∩ (X−V)=X0∩ (X−U).
X−V and X−U are closed sets of X. By (2) ,
UXUXXVXVXX −=−∩−=−∩ )(,)( 00 .⇒V=U.
(3) ⇒(1): If (1) is not satisfied, a locally closed set ∃L=O∩F, L≠∅ and L∩X0=∅ where
F=X−V and O and V are open sets of X. O∩(X−V)∩X0=∅⇒O∩X0⊆V⇒
O∩X0⊆O∩V∩X0. ⇒O∩X0=O∩V∩X0. O=O∩V by (3). Then O⊆V. O∩(X−V)= ∅. It
54
contradicts to L≠∅.
i)⇒ii): Let X0 be the set of all maximal ideals of A which is a Jacobson ring.
Let L=O∩F (≠∅) be any locally closed set of Spec(A) where O=p∈Spec(A) |¬a ⊆p
and F=p∈Spec(A) | b⊆p. It is enough to show L∩X0≠∅ by (1).
∃p∈L⇔p∈O and p∈F. If p is a maximal ideal, L∩X0 ≠∅. If p is not a maximal ideal,
p=∩ p⊂m:maximal ideal m, because A is a Jacobson ring. Since ¬a⊆p =∩ p⊂m m, ¬a⊆∃m’.
Since b⊆p=∩ p⊂m m, b⊆m’. m’ ∈O∩F. Then L∩X0≠∅.
ii)⇒iii): Let L=p be a locally closed set of Spec(A) and X0 be a set of all maximal
ideals of A. By hypotheses, X0 is very dense⇒ L∩X0≠∅. p∈X0. p is a maximal ideal
⇒L is a closed set of Spec(A).
iii) ⇒i): If A is not a Jacobson ring, a non-maximal prime ideal ∃p ⊂∩ p⊂q∈Spec(A) q.
∃f∈∩pα, f∉p. F=q∈Spec(A) | q⊇p is a closed set. O= q∈Spec(A) | f ∉q is an open
set. Let L=F∩O. L= p . Since p is not a maximal ideal, L is not a closed set..
Ex.5.27
0∈Σ⇒ Σ≠∅. Any relation of domination in Σ forms a chain. The union of all members
of a chain belongs to Σ. By Zorn’lemma, there exists a maximal element A in Σ.
⇒: Let x(≠0)∈K and A[x] be the subring of K generated by x over A. Let m be the
maximal ideal of A and m[x] be the extension of m in A[x]. By (5.20), A[x] ≠ m[x] or
A[1/x] ≠ m[1/x] holds, since A is a local ring. Now suppose A[x] ≠ m[x]. A maximal
ideal ∃n⊇m[x]. n∩A=m, since n is a proper ideal and includes m. A[x]n is a local ring
in K and includes A, because they are in K. Since nA[x]n⊇ n⊇ m, A[x]n dominates A. By
maximality of A, A[x]n =A, hence x ∈ A. If A[1/x] ≠ m[1/x], 1/x∈ A. A is a valution ring.
⇐ : Let A is a valuation ring of K. If a local ring (B,n) dominates A, ∃x∈B, x∉A. Then
1/x∈A and 1/x∈m. 1/x∈n because B dominates A. x is a unit in B. It contradicts.
Ex.5.28
(1)⇒(2): If a and b does not include each other, ∃x∈a,∃y∈b, x∉b and y∉a. A is a
valuation ring⇒ x/y∈A or y/x∈A. x∈Ay⊆b or y∈Ax⊆a. It contradicts.
(2)⇒(1): For ∀k(=f/g)∈K(f,g∈A), (f)⊆(g) or (g)⊆(f). ⇒ f=∃ag or g=∃bf ⇒k∈A or 1/
k∈A.
Ap is a integral domain and the field of fraction of Ap is K (=κ(A)). Let I,J are ideals of
Ap, I= ap,J = bp where a and b are ideals of A. a⊇b or a⊆b ⇒ I⊇J or I⊆J⇒ Ap is a
valuation ring by (2).
Let κ(A/p) be the field of fraction of A/p. If I,J are ideals of A/p, I c and J
c have relation
Ic⊆J
c or I
c⊇Jc in A which is a valuation ring. By the canonical homomorphism, I⊆J or
55
I⊇J.
Ex.5.29
Let B be a ring such that A⊆B⊆K. Let n be the set of all non units of B. If n is an ideal,
n is the unique maximal ideal. Let x,y∈n. For ∀b∈B, bx is not a unit⇒bx∈n. x/y∈K.
x/y∈A or y/x∈A. If x/y∈A, x+y=y(1+ x/y) is not a unit.Then x+y∈n. If y/x∈A, same as
the case of x/y∈A . n is an ideal.
Ex.5.30
Let A be a valuation ring of K. Let U be the group of all units of A (⊆K*). Let Γ=K
*/U.
x’∈xU, y’∈yU⇒x’/y’=x/y⋅u(u∈U). If x/y∈A,x’/y’∈A. Then the total order designated in
the text does not depend on representatives.
Let ξ,η,ω∈Γ and x,y,z∈K* be representatives of ξ,η,ω respectively. xz (resp.yz) is a
representative of ξω (resp. ηω). ξ≥η⇒x/y∈A⇒ xz/yz=x/y∈A, ξω≥ηω.
Let v : K*→Γ be a canonical homomorphism. v(x+y) ≥min(v(x), v(y)): For x,y∈A, x/y∈A
or y/x∈A⇒(x+y)/y = x/y +1∈A or (x+y)/x∈A⇒v(x+y) ≥ v(x) or v(x+y)≥v(y) ⇒v(x+y)≥
min(v(x),v(y)).
Ex.5.31
Let A=0∪x∈K*| v (x) ≥0.
A is a ring: x,y∈A⇒ v(xy)=v(x)+v(y)≥0, xy∈A. v(x+y)≥ min(v(x),v(y))≥0⇒x+y∈A.
v(1)=v(1)+v(1)⇒v(1)=0⇒1∈A.v(1)=v(−1)+v(−1)=0⇒−1∈A.v(x) ≥0⇒v(−x)= v(−1)+v(x)
≥0. −x∈A.
A is a valuation ring: x(≠0)∈K⇒0=v(x⋅1/x)=v(x)+v(1/x). v(x)≥0 or v(1/x) ≥0⇒x∈A or
1/x∈A.
Ex.5.32
Proof is according to the definition of the valuation ring in ex.5.31.
Let A=0∪x∈K*| v(x)≥0 and p be a prime ideal of A.
v(A−p) is a set of elements ≥0 of an isolated subgroup of Γ:
Let X be a group which x| x∈A−p ∪1/x | x∈A−p generate by multiplication in K*.
∆ is a subgroup of Γ:
1∈X⇒0∈∆. a∈∆⇒∃x∈X,v(x)=a. 1/x∈X⇒v(1/x)=−a∈∆. a,b∈∆⇒∃x,y∈X v(x)=a,v(y)=b
⇒xy∈X⇒v(xy)= v(x)+v(y), a+b∈∆.
v(A−p) =α∈∆| α≥0 :
⊆ :For ∀x∈A−p, v(x)≥0. ⇒v(A−p) ⊆α∈∆| α≥0 .
56
⊇:For ∀α(≥0)∈∆, inverse images of α are in A. v(A−p)∩v(p) =∅ (∃x∈A−p, ∃y∈p,
v(x)=v(y)⇒v(x/y)=0. x/y∈A⇒x∈(y)⊆p. Contradictory.). Suppose the inverse image of α
is in p. ∃p∈p,v(p) =v(x) −v(x’) ∈∆ where x,x’∈A−p. Then p=x/x’ ∈A. It contradicts. The
inverse image of α∈∆| α≥0 is A−p.
v(A−p) =α∈∆|α≥0:
∆ is isolated: Let x∈A−p,y∈K,0≤v(y)≤v(x)( v(x)∈∆). y∈A. v(x/y)≥0⇒x/y=a∈A⇒x=ay.
x∉p⇒y∉p⇒y∈A−p. v(y)∈∆.
Spec(A)→ isolated subgroup of Γ is surjective:
Let ∆ be any isolated subgroup of Γ and p=x∈A|v(x)≥0,v(x) ∉∆ .
p is a ideal of A:
x,y∈p⇒v(x+y)≥min(v(x),v(y))≥0. Suppose v(x+y)∈∆. v(x+y)≥v(x)≥0 or v(x+y)≥v(y)≥0,
v(x) or v(y) ∈∆(∆ is isolated). It contradicts to v(x) and v(y) ∉∆. ⇒v(x+y)∉∆ ⇒x+y∈p.
x∈p,a∈A⇒v(ax)=v(a)+v(x)≥0. Suppose v(ax)=v(a)+v(x)∈∆. v(a)+v(x)≥v(x)≥0. ⇒
v(x)∈∆(∆ is isolated). It contradicts to v(x)∉∆. ⇒ v(ax)∉∆. ax ∈p.
p is a prime ideal:
x,y∈A, xy∈p and x∉p⇒v(x)≥0,v(y)≥0,v(xy)=v(x)+v(y)∉∆,v(x)∈∆.v(y)∈∆. ∆ is an
additive group⇒(v(x)+v(y))∈∆. It is contradictory. Then v(y) ∉∆⇒y∈p.
Spec(A)→ isolated subgroup of Γ is injective:
Suppose p and q(≠p) and ¬q⊇p. ∃x(∈p,∉q). v(x)∉v(A−p), v(x)∈v(A−q) ⇒v(A−p)≠
v(A−q).
The valuation group of A/p:
A/p is a valuation ring by ex.5.28. We define the valuation group of Ap by ex.5.30 and
ex.5.31. Let k*=κ(A/p)−0=x+p/y+p | x,y∈A−p. Let K’
* be a subgroup of K
* which
x∈K* | x∈A−p∪1/x | x∈A−p generate. Define a group homomorphism γ : K’
*→k*
by γ (x/y)=x+p/y+p. Let m be the maximal ideal of A. Since m/p is the maximal ideal of
A/p, the subset of all units of A/p is U’=x+p| x∈A−m. γ is clearly surjective.
k*/U’=K’
*/γ−1
(U’). γ−1(U’)=x| x∈A−m coincides with the subset of all units of A.
γ−1(U’)=U. k
*/U’=K’
*/U is the image in Γ of the group whichx|x∈A−p ∪1/x| x∈A−p
generates. Then k*/U’= K’
*/U=∆.
The valuation group of Ap:
Ap is a valuation ring (ex.5.28). κ(Ap)= κ(A)=K. We define the valuation group of Ap by
ex.5.30 and ex.5.31. The subset consisting of all units of Ap, i.e.U’=(x/y)|x,y∈A, x∉p,
y∉p forms a subgroup of K* and includes U. Γ’=K
*/U’=(K
*/U)/(U’/U)=Γ/(U’/U). U’/U
is a subgroup of Γ which x|x∈A−p ∪1/x| x∈A−p generates by product operation of
K*. U’/U is ∆ in Γ. Γ’= K
*/∆.
57
Ex.5.33
Γ is a totally ordered abelian group⇔If α,β,γ∈Γ and α≤β, then α+γ≤β+γ.
Let u1, u2∈A and u1= Σ1≤i≤n λiX αi (λi ∈k, λ1λn≠0,αi<αj for 1≤i<j≤n) and u2=Σ1≤i≤m δiXβi
(δi ∈k, δ1δm≠0,δi<δj for 1≤i< j≤m ).
A is an integral domain:The coefficient of a term Xαn+ βm of u1u2 is λnδm(≠0).
v0 satisfies condition(1): u1u2=λ1Xα1δ1Xβ1+…=λ1δ1Xα1+β1+… .
Xα1+β1 is the smallest term of u1u2. v0 (u1u2)=α1+β1=v0 (u1)+v0 (u2).
v0 satisfies condition (2): v0 (u1+ u2)= v0 (Σ1≤i≤n λiX αi +Σ1≤i≤m δiX βi).
Reordering by the order of Γ, v0 (u1+u2)=minv0(u1),v0(u2) for α1≠β1, v0(u1+u2)≥
minv0 (u1), v0 (u2) ,possibly λ1+δ1=0 for α1=β1.
Let u1/ u2∈K (u1, u2∈ A). Define v (u1/ u2)=v0 (u1)−v0 (u2).
well-define:u1/u2=w1/w2⇒u1w2=w1u2⇒v0(u1)+v0(w2)=v0(w1)+v0(u2)⇒v(u1/u2)=v(w1/w2).
v satisfies condition(1):v(u1/u2⋅ w1/w2)=v(u1w1/u2w2)=v0(u1w1)−v0(u2w2)=(v0(u1)+ v0(w1))
− (v0(u2) +v0(w2))= v(u1/u2) +v(w1/w2).
v satisfies condition(2):v(u1/u2+w1/w2)=v((u1w2+u2w1)/u2w2))= v0(u1w2+u2w1)− v0(u2w2)
≥minv0(u1w2),v0(u2w1)− v0 (u2w2)=min(v0(u1w2) −v0(u2w2)),(v0(u2w1) −v0(u2w2)) =
minv(u1/u2), v(w1/w2).
The valuation group of v=Γ: The valuation group of v is included by Γ. For α,β of∈ Γ,
XαXβ corresponds to α+β and Xα/Xβ corresponds to α−β.
The valuation group of v =Γ.
Ex.5.34
Let A be a valuation ring of K and f*: Spec(B)→Spec(A) be a closed mapping. If g:B→K
is a A-homomorphism, g°f : A→K is the embedding. Because g(f(a))=g(a⋅f(1A))
=ag(1B)=a1K=a. Set g(B)=C. f(A)⊆B⇒gf(A) ⊆g(B) ⇒A⊆C. Let n be a maximal ideal of
C. g is surjective⇒ g−1
(n)=n’ is a maximal ideal of B. n’ is a closed set of Spec(B). f*
is a closed mapping⇒f*(n’)=m is a closed set of Spec(A). m is a maximal ideal of
A. A is a valuation ring and m is the maximal ideal. m=f−1
g−1
(n) ⇒ gf(m)=m⊆n. Cn
dominates A (nCn∩C=n⊇m). By (ex.5.27), A=Cn⊇C. Then A=C⇒g(B)=A.
Chapter 6
Ex.6.1
i): Let u: M→M be a homomorphism.
Ker(u)⊆...⊆Ker(ui-1
)⊆Ker(ui ) ⊆...
M is Noetherian. Ker(un)=Ker(u
n+1)=.... for any n ≥∃n0.
58
u is injective: Let x∈M,u(x)=0. u is surjective⇒un is surjective. Im(u
n)=M. ∃y ∈M,
x=un(y). u(x)=u
n+1(y)=0. y∈Ker(u
n+1) = Ker(u
n). u
n(y)=0. x=0.
ii): Let u: M→M be a homomorphism.
Im(u) ⊇...⊇Im(ui-1
) ⊇Im(ui) ⊇...
M is Artinian. Im(un)= Im (u
n+1)=… for any n≥∃n0.
u is surjective: For ∀x∈M, un(x)∈Im(u
n)=Im (u
n+1). ∃y∈M,u
n(x)=u
n+1(y) . u
n(x−u(y))=0.
u is injective⇒un is injective. x=u(y).
Ex.6.2
If M is not Noetherian, a submodule ∃N of M which is not finitely generated (6.2). Let
xi be generators of N and N i =Ax1+…+Axi (i>0). N i ≠ N, because if ∃i ,N i = N, N is
finitely generated. Then N i ⊂ N. ∃xi+1∈N (∉Ni). Let N i+1= N i +Axi+1. Then N i⊂N i+1.
Nii is a chain of finitely generated submodules that has not a maximal element.
Ex.6.3
Let M/N1, M/N2 be Noetherian A-modules (resp. Artinian). 0→N2/(N1∩N2) →
M/(N1∩N2) →M/N2→0 is exact. N2/(N1∩N2) ≅(N1+N2) /N1. (N1+N2) / N1 is a submodule
of M/N1⇒it is Noetherian(resp. Artinian). By(6.3) M/(N1∩N2) is Noetherian(resp.
Artinian).
Ex.6.4
Let M=Au1+...+Aun be a Noetherian A-module and a=Ann(M). Define an
A-homomorphism φ:A→ M ⊕…⊕ M by φ(x)=xu1⊕…⊕xun. Kerφ=a. A/a→→→→ M ⊕…⊕ M
is injective. A/a is an A/a-module. M is an A/a-module according to explanation
following (2.1). A/a-submodules of M are A-modules⇒A/a-submodules satisfiy a.c.c..
Let Iii be any ascending sequence of ideals of A/a. Isomorphic images of Iii in M
⊕…⊕ M (Noetherian A/a-module(6.4)) satisfies a.c.c.. Then A/a is Noetherian.
A counter example: G in chapter 6 example 3(page74) is an Artinian. G is Z-module
and a=Ann(G) is 0. Z/a=Z is not Artinian.
Ex.6.5
Let Y be a subspace of X. If (Y∩O1) ⊆…⊆(Y∩On) ⊆… is an ascending sequence of
open sets of Y where Oii are open sets of X, it satisfies a.c.c., because ∪n≥i Oin
satisfies a.c.c. in X. Y is a Noetherian topological subspace.
Let Oii be an open covering of X. ∪n≥i Oin is an ascending sequence of open sets. It
satisfies a.c.c., since X is Noetherian. ∃n>0, ∪n≥i Oi =∪n+1≥i Oi=…=X. Then a finite set
59
of Oi covers X. X is quasi-compact.
Ex.6.6
i) ⇒iii): Let Y be a subspace of X, Oii be open sets of X and Y∩Oii be an open
covering of Y. ∪i≤n Y∩Oin, an ascending sequence, satisfies a.c.c., because ∪i≤nOin
satisfies a.c.c.. Finite set of ∪i≤n Y∩Oin covers Y. Y is quasi-compact.
iii) ⇒ii):trivial.
ii) ⇒i): Let On n>0 be any ascending sequence of open sets of X. Let Y=∪n>0On. Y is an
open set. By hypotheses, Y is quasi-compact. A finite set of On n>0 covers Y. ∃n >0,
On= Y. Then ascending sequence On n>0 satisfies a.c.c..
Ex.6.7
Suppose that Noetherian space X is not a finite union of irreducible closed components.
Let Σ=F: a closed set of X | F is not a finite union of irreducible closed subspaces..
X∈Σ⇒Σ≠∅. Σ satisfies d.c.c.. A minimal element ∃F’∈ Σ. F’ is not irreducible⇒ open
sets ∃U,V of X ,F’∩U≠∅, F’∩V≠∅, and F’∩U∩V=∅. F’∩(X−U) and F’∩(X−V) are
proper closed sets (not empty) of F’, they are not members of Σ. They are finite unions
of irreducible closed components. F’ is a union of them⇒F’ is a finite union of
irreducible components. It contradicts.
Ex.6.8
Let Vii be a descending sequence of closed sets of Spec(A). Vi=V(r(ai)) for each i,
where ai is an ideal by (ex.1.15). r(ai)i is an ascending sequence of ideals, because if
V(r(a))⊇V(r(b)), r(a)⊆r(b), since r(a)=∩p∈V(r(a)) p. A is Noetherian⇒∃n>0,r(an)=
r(an+1)=…. . ⇒ Vn= Vn+1=… . ⇒ Vii satisfy d.c.c..
A counterexample:
Let k be a field. Let xii be infinite indeterminates. Let A=k[x1,…,xi,…]/(x12,…,xi
2,…).
m=(x1,…,xi,…) is a maximal ideal. (x1) ⊂(x1,x2) ⊂…⊂ (x1,…,xi) ⊂… is a strictly
ascending chain of infinite ideals of A⇒A is not Noetherian.
every element of A is of the form k1+Σiaixi (ki∈k,ai∈A) (finite sum) ⇒it is nilpotent if
k1=0 and is a unit if k1≠0. Since any element of m is nilpotent, m is the unique prime
ideal. Spec(A)=m. Spec(A) is a Noetherian space.
Ex.6.9
Let A be a Noetherian ring. Spec(A) is a Noetherian space by (ex.6.8). The set of
irreducible components of Spec(A) is finite (ex.6.7). An irreducible component of
60
Spec(A) is V(p) where p is a minimal prime ideal (ex.1.20iv). Prime ideals p≠p’⇔ V(p)
≠ V(p’). ⇒The number of minimal prime ideals is finite.
Ex.6.10
M is finitely generated over A by (6.2). Supp(M)=p∈Spec(A) | Mp≠0=V(Ann(M)) by
(ex.3.19v). Supp(M) is a closed set of Spec(A). V(Ann(M)) is homeomorphic to
Spec(A/Ann(M)). A/Ann(M) is a Noetherian ring by (ex.6.4). Spec(A/Ann(M)) is a
Noetherian space. Supp(M) is a Noetherian space.
Ex.6.11
⇒: If f * is a closed mapping, f has the going-up property by (ex.5.10i).
⇐:Let E=V(b) be a closed set of Spec(B). Set F= f*(E). Minimal prime ideals of V(b)
correspond to minimal prime ideals of B/b. Spec(B) is a Noetherian space⇒Spec(B/b)
is a Noetherian space by (ex.6.5). A Noetherian space is a finite union of irreducible
components by (ex.6.7). An irreducible component is a closed set of a minimal prime
ideal by (ex.1.20iv). The set of minimal prime ideals qi of B/b is finite. E=∪i≤nV(qi).
Let pi=ψ*(qi)(∈Spec(A)). For ∀p(⊇pi)∈Spec(A), ∃q∈Spec(B), qi⊆q and p=q∩A, by the
going-up property of f. E is closed⇒q∈E. Then p∈F. F=∪i≤nV(pi). F is a closed set.
⇒f* is a closed mapping.
Ex.6.12
Suppose pi is an infinite strictly ascending sequence of prime ideals of A. The
sequence of Vi where Vi=p∈Spec(A) | p⊇pi is an infinite strictly descending
sequence of closed sets of Spec(A). It contradicts to that Spec(A) is a Noetherian space.
Then any set of prime ideals of A satisfies a.c.c. .
Counterexample:Let Ai=Z/(2)[xi]/(xi2−xi) and A=A1⊕A2⊕…⊕Ai… . A is Boolean
(ex.1.11). Prime ideals are maximal ideals(ex.1.11). Then prime ideals satisfy a.c.c. .
Ideals of the form (A1, A2,…, Ai-1, xi, Ai+1,…) are maximal⇒Spec(A) consists of infinite
maximal ideals. f≠0,1⇒ f and f−1 are idempotent and not nilpotent. ⇒ Xf ≠∅ , X1-f≠∅,
Xf∩X1-f=∅⇒Spec(A)= Xf∪X1-f is a disjoint union. Then they are open and closed set.
One of them is an infinite set. Let it be Xf . Let m,n be different maximal ideals of Xf .
∃g∈m,∉n. Spec(A)=Xg∪X1-g is a disjoint union. Xg∩Xf and X1-g∩Xf are both proper
closed subset of Xf and one of them is an infinite set. Repeating this, we get a properly
descending sequence of infinite closed sets. Spec(A) is not Noetherian.
Chapter 7
61
Ex.7.1
Suppose A is not a Noetherian ring. Let Σ be a family of ideals of A which is not finitely
generated. Σ is not vacant. Let the order of Σ be the inclusion in the sense of set theory.
Let aii be any chain of members of Σ. Let a=∪iai. If a∉Σ, a is finitely generated and
let x1,…,xn be generators of a. xi belongs to some of ai. Let aj(i) be the first ideal of
ai to which xi belongs. Let m=max j(i). am=a. am is finitely generated. It contradicts.
Then a∈Σ. By Zorn’s lemma, a maximal element ∃a∈Σ.
a is a prime ideal:
If not, ∃x∉a, y∉a and xy∈a. Let I=a+(x) (⊃a). I is a finitely generated. We can choose x
and x1,…,xn∈a as generators of I. Let a0=(x1,…,xn)(⊆ a). I=a+(x)=a0+(x). Let J=a0+x⋅(a:
x). J⊆a: trivial. J⊇a: ∀z∈a⇒⇒⇒⇒z+rx=a0+sx (a0∈a0, r,s∈A)⇒⇒⇒⇒(s−r)x=z−a0⇒⇒⇒⇒ (s−r)∈(a: x).
z∈J. Then J=a. a⊂(a: x) (y∉ a) ⇒⇒⇒⇒ (a: x) is finitely generated. Let y1,…,ym be generators
of (a: x). J=(x1,…,xn, xy1,…, xym). a(=J) is finitely generated. It contradicts.
If A is not a Noetherian ring, there exists a prime ideal which cannot be finitely
generated.
Ex.7.2
⇒: If f is nilpotent, then an, coefficient of xn of f, is nilpotent for any n≥0 by (ex.1.5ii).
⇐: Let Ji =(a1,…,ai) where an is the coefficient of xn of f. Ji is an ascending sequence
of ideals. ∃m>0,Jm=Jm+1=Jm+2 =…. an(n≥m) is generated by a1,…,am. a1,…,am are
nilpotent (airi=0). Let r=1+Σi=1,…,m (ri−1). f
r=0.
Ex.7.3
Let a be an irreducible ideal.
i)⇒ii): Let r(a)=p(a prime ideal). If S∩p≠∅, (S−1a)=S
−1A⇒ (S
−1a)
c=A. For ∀x∈S∩p,
∃n>0,xn∈a∩S. ⇒(a : x
n)=A=(S
−1a)
c. If S∩p=∅, (S
−1a)
c=a by (4.8ii). For ∀s∈S, (a :
s)=a by (4.4iii).
ii)⇒iii): Set S=xnn>0 for ∀x∈A. If 0∈S, ∃m≥0,x
m=0. (a: x
m)=A. The ascending
sequence (a: xn)n is stationary for n(≥m). If 0∉S, ∃m≥0,(S
−1a)
c=(a: x
m) by hypotheses.
(S−1a)
c=∪s∈S(a : s) by (3.11ii). The sequence (a : x)⊆…⊆ (a : x
n) ⊆ (a : x
n+1)
⊆…satisfies a.c.c. because (S−1a)
c=(a: x
m).
iii)⇒i): There is no loss of generality in taking a=0. Suppose xy=0 and x≠0(x,y∈A). By
hypotheses, ∃n>0,(0:yn)=(0:y
n+1)=… . z∈(x)∩(y
n)⇒z=ax=by
n (a,b∈A). zy=axy=by
n+1=0.
b∈(0 : yn+1
)=(0 : yn). z=by
n=0. (x)∩(y
n)=0. 0 is an irreducible ideal and (x)≠0, (y
n)=0. y
is nilpotent. 0 is primary.
62
Ex.7.5
Let A be a Noetherian ring and B be a finitely generated A-algebra. A⊆BG⊆B. B is
integral over BG by (ex.5.12). B
G is finitely generated A-algebra by (7.8).
Ex.7.7
Let X=x∈kn
| fα(x)=0 for fα(t1,…,tn) (∀α∈I ) . X=x∈kn
| f (x)=0 for ∀f∈I(X) where
I(X) is an ideal of k[t1,…,tn] which fα(t1,…,tn)∀α∈I generate. k[t1,…,tn] is Noetherian.
I(X) is generated by a finite set of fα (t1, …,tn).
Ex.7.8
Yes. A is isomorphic to A[x]/(x). A is Noetherian by (7.1).
Ex.7.9
Let a≠0 and x0(≠0)∈a. The set of maximal ideals including (x0) is finite. The set of
maximal ideals including a(⊇ (x0)) is finite. Let m1,…,mr be maximal ideals including a
and mr+1,…,mr+s be maximal ideals including (x0) except m1,…,mr. For each j(1≤j≤s),
∃xj∈a,xj∉mr+j. For each i (1≤i≤r), Amia is finitely generated because Ami is Noetherian.
There exist finite elements of a which generate Amia for each i (i=1,…,r). Then the set of
a union of those generators is finite. Let them be xs+1,…,xt. Set a0=(x0,…,xt)(⊆a). To
claim a=a0, it is enough to show that Ama0=Ama for every maximal ideal(3.9). For mi
(i=1,…,r), Amia=Amia0 (mi includes generators of Amia). For mr+j (j=1,…,s), Amr+ja=(1)
and Amr+ja0=(1) (xj∉mr+j). For other m,¬a⊆m. ¬a0⊆m(x0∉m). Then Ama=Ama0=(1).
Ex.7.10
Let M be a Noetherian A-module and L be any A[x]-submodule of M[x]. Let N=a∈M |
∃f(x)∈L, a is the coefficient of the highest degree of f(x). N is an A-submodule of M. M
is Noetherian⇒N is finitely generated. Let a1,…, ak be generators of N and fi(x) be a
polynomial of degree ri having ai as the coefficient of the highest degree. Let
r=max ri .
Let L’ be an A[x]-module generated by fi(x).
Let f(x)∈L and deg(f(x))=n ≥r. a, the coefficient of xn of f(x), is Σi=1…,kbiai (bi∈A). Then
f(x)− Σi=1…,kbifixn−ri∈L. Consequently it is reduced to the case n<r.
In the case of n<r, M’ =M+Mx+....+Mxr-1
is a finitely generated A-module⇒M’ is
Noetherian(6.5). L’∩M’ is finitely generated. Let g1(x),…,gj(x) be its generators. L is
generated by f1(x),…,fk(x), g1(x),…,gj(x).
63
Ex.7.11
Counterexample: Let A= Z/(2)[x1,…,xi,…]/(x12−x1, x2
2−x2,…,x1x2, x1x3,…,x2x3, x2x4,…).
A is not Noetherian. A is absolutely flat. All prime ideals of A are maximal(ex.3.11
r(0)=0). Let m be any maximal ideal of A. Then Am is a field by ex.3.10, which is
Noetherian.
Ex.7.12
Let a1⊆a2⊆…⊆an⊆…be a sequence of ideals of A. a1e⊆a2
e⊆…⊆ane⊆…is its extension
in B. B is Noetherian ⇒∃n>0,ane=an+1
e=… . B is faithfully-flat⇒a i
ec=ai (ex.3.16). Then
an=an+1=… . A is Noetherian.
Ex.7.13
Let f : A→B be finite-type and f(A)=C. B is a finitely generated C-algebra.
∃b1,…,bn(∈B), B=C[b1,…,bn]. Let x1,…,xn be indeterminates and define a
homomorphism φ : C[x1,…,xn]→B by φ(xi)=bi. B≅C[x1,…,xn]/Kerφ. For ∀p∈Spec(C),
set κ(C/p)=k. The fiber of p is Spec(Bp/pBp)(=Spec(k⊗CB)) which is a subspace of
Spec(B). Tensor 0→Kerφ→C[x1,…,xn]→B→0(exact) with k over C. Tensoring is right
exact⇒k⊗CB is isomorphic to an quotient ring of C[x1,…,xn]⊗Ck(=k[x1,…,xn]).
k[x1,…,xn] is Noethrian⇒k⊗CB is Noethrian. Spec(Bp/pBp) is a Noetherian Space.
Spec(C) is isomorphic to V(Ker(f)) in Spec(A). Let q∈Spec(A). If q⊇Ker(f), the fiber of
q is the fiber of p=f(q)(∈Spec(C)). It is a Noethrian subspace of Spec(B) by above. If
¬q⊇Ker(f), the fiber of q is vacant.
Ex.7.15
i)⇒ii): trivial.
ii)⇒iii): If M is A-flat, 0→m⊗M→A⊗M→k⊗M→0 is exact.
iii)⇒iv): Tor1A(A/m,M)→m⊗M→A⊗M→k⊗M→0 is exact. Since m⊗M→A⊗M is
injective by hypotheses, Tor1A(k,M)=0.
iv) ⇒i):Proof is in the text.
Ex.7.16
Let A be a Noetherian ring and M be a finitely generated module over A.
i)⇒ii): Let 0→N’ →N be any exact sequence of Ap-modules. N’ = N’p, N = Np.
Mp(=Ap⊗AM) is an Ap-module. M is A-flat by hypotheses. N’ and N are also A-modules.
0→N’⊗AM →N⊗AM is exact. For any prime ideal p, 0→(N’⊗AM) p→(N⊗AM) p is exact,
64
i.e., 0→ N’ p ⊗Ap M p → N p ⊗ Ap M p is exact by (3.7).
0→ N’ ⊗Ap M p → N ⊗ Ap M p is exact by (3.5). Mp is Ap-flat.
Ap is a Noetherian local ring. Mp is finitely generated. Mp is free by (ex.7.15).
ii) ⇒iii):trivial.
iii) ⇒i): Mm is Am-flat by (ex.7.15). M is A-flat by (3.10).
Ex.7.17
(7.12 for modules)
’If M is a Noetherian A-module, an irreducible A-submodule N of M is primary in M’
proof: there is no loss of generality in taking N =0. Suppose y (∈A) be a zero divisor of
M. ∃m0(≠0)∈M, ym0=0. Regard y : M→M as a homomorphism such that m→ym.
Ker(yn) n>0 is an ascending sequence of subgroup (≠0) of M. M is Noetherian⇒
∃n>0,Ker(yn)=Ker(y
n+1) =… . For ∀z∈(y
n)M∩Am0, ∃m1∈M and ∃a∈A,z=y
nm1=am0 .
yn+1
m1=yam0=0. m1∈Ker(yn+1
)=Ker(yn). z=0. (y
n)M∩Am0=0. 0 is irreducible and Am0≠0.
(yn)M=0. y is nilpotent. 0 is primary.♦
(7.11 for modules)
’If M is Noetherian, any subgroup of M is a finite intersection of irreducible modules.’
Proof: If not, let Σ=L:a submodule of M | L cannot be an intersection of finitely many
irreducible submodules.. Σ≠∅. A maximal element ∃N in Σ, because Noetherian. N is
reducible⇒ ∃N1, N2 ,N1⊃N, N2⊃N and N=N1∩N2. N1 and N2 do not belong to Σ. N1 and
N2 are intersections of finite many irreducible modules. Then So is N. It contradicts.♦
Ex.7.18
i)⇒ii): If M is Noetherian, any submodule of M has primary decompositions by
(ex.7.17). Let 0=∩Ni be a primary (irredundant) decomposition of 0 in M. (0 :
M)=∩(Ni : M)=∩qi. (Ni : M)=qi is a primary ideal. Let r(qi)=pi. A is Noetherian ⇒
∃n>0, pin⊆qi. ∃y∈∩j≠iNj,∉Ni. pi
ny⊆Ni⇒pi
ny=0. ∃m>0,pi
m−1y≠0,pi
my=0. For
x(≠0)∈pim−1
y, Ann(x)⊇pi. x∈Ay⇒x∈∩j≠i Nj. x≠0⇒x∉Ni. For ∀a(≠0)∈Ann(x), ax∈Ni. Ni
is primary. ∃l>0,alM⊆Ni. a
l∈qi. a∈pi. Ann(x)=pi.
ii)⇒iii): Ax is a subgroup of M. Let Ann(x)=p. Ax ≅A/Ann(x)=A/p.
iii)⇒i): If a subgroup ∃N of M is isomorphic to A/p, ∃x(∈N) ≅ 1 mod p. r(0 : x)=p⇒p is
a prime ideal beloging 0 in M by (ex.4.22).
Let 0=∩Ni be a primary decomposition of 0 in M where (Ni : M)=qi and r(qi)=pi.
Let p1 be a prime ideal belonging 0 in M. A subgroup ∃M1≅A/p1 by (ii). M/M1 is a
Noetherian A-module. 0 is decomposable in M/M1. ∃M2/M1≅A/p2 where p2 is a prime
65
ideal of A. As long as Mi≠M, 0 is decomposable in M/Mi. Consequently, M1⊂…⊂Mi⊂….
and Mi/Mi-1≅A/pi(pi is a prime ideal of A). M is Noetherian⇒∃Mr and M=Mr
Ex7.19
There is no loss of generality in taking a=0. Let 0=b1∩…∩br=c1∩…∩cs be two
minimal decompositions by irreducible ideals and be r≥s. It is enough to prove the case
where s is the minimum number of irreducible ideals among decompositions of 0.
If s=1, 0 is irreducible. ⇒∃bi, after reindexing b1, is a zero ideal. so r=1. And
r(b1)=r(c1)=r(0).
Let s≥2. c1∩…∩cs=0⇒φ:A→Πj(A/cj) is injective (1.10iii). For i=1,…,r, let Ii= c1∩…∩
cj-1∩ bi∩ cj+1…∩cs. Suppose any Ii ≠ 0. ⇒ ∃bi∈Ii, ∉ cj. φ(bi)=(0,…,0, bi mod cj,0,…,0).
(bi)≅φ((bi))=(0,…,0,(bi) mod cj,0,…,0)≅(0,…,0, cj+(bi) mod cj,0,…,0). ∩i Ii =0 ⇒
∩i(bi)=0⇒(0,…,0,∩icj+(bi) mod cj,0,…,0)=0. ⇒∩i(cj+(bi) mod cj) =0. ⇒ ∩i(cj+(bi))=cj.
This contradicts to that cj is irreducible. Then c1∩…∩cj-1∩∃bi∩ cj+1…∩cs=0. Let
X=c1∩…∩ cj-1 ∩ cj+1…∩cs. X≠0⇒ ∃x∈X , x ∉cj and x ∉bi. (0: x)= (cj∩X: x) = (c j: x) ∩
(X: y)= (c j: x). As the same (0: x)=(bi: x). By (4.4), (cj: x) is a primary ideal whose prime
ideal is r(cj). (bi: x) is as the same. Then r(cj) = r(bi). Replace cj by bi. We need
only s of bi. If r>s, 0=b1∩…∩br is redundant. From minimality of s, r =s.
(ex.7.19) of modules
” Let M be a Noetherian A-module and H be a submodule of M. If H has minimal
decompositions of irreducible submodules as following
H=N1∩…∩Nr=L1∩…∩Ls ,
then r=s and after reindexing, rM(Ni) = rM (Li)”
proof: there is no loss of generality in taking H=0.
An irreducible submodule is a primary submodule(ex.7.17).
r=s and after possible reindexing rM(Ni) = rM (Li):
Let 0=N1∩…∩Nr=L1∩…∩Ls be minimal decompositions by irreducible submodules
and r≥s. It is enough to prove the case where s is the minimum number of irreducible
submodules among decompositions of 0.
If s=1, 0 is irreducible. So N1=0 after reindexing Ni. Then r=1. rM(L1) = rM(N1)=rM(0).
Let s≥2. L1∩…∩Ls=0⇒φ:M→ΠjM/Lj is injective. For i=1,…,r, let Ii=L1∩…∩
Lj-1∩Ni∩Lj+1…∩Ls. Suppose any Ii≠0. ∃ni∈Ii,∉Lj. φ(ni)=(0,…,0, ni mod Lj,0,…,0).
Ani≅φ(Ani)=(0,…,0,Ani mod Lj,0,…,0)≅(0,…,0, Lj + Ani mod Lj,0,…,0). ∩i Ani=0⇒
(0,…,0,∩i(L j+ Ani mod L j),0,…,0)=0. ⇒∩i(L j+ Ani mod L j) =0. ⇒ ∩i(Lj+ Ani)=Lj
contradicts to that Lj is irreducible. Then L1∩…∩Lj-1∩∃Ni∩Lj+1…∩L s=0. Let
X=L1∩…∩Lj-1∩L j+1…∩L s. X ≠0⇒ ∃x ∈ X, ∉Lj , ∉Ni. (0: x)= (Lj∩ X: x) = (Lj: x) ∩ (X:
66
x)= (Lj: x). As the same, (0: x)= (Ni: x). (Lj: x) is a primary submodule of M whose prime
ideal is rM(Lj) by (ex4.21). (Ni: x) is the same. ⇒ rM(Ni) = rM (Lj). Each Lj is replaced by
some Ni . Then 0 is decomposed by s of Ni.If r>s,0=N1∩…∩Nr is redundant. r =s by
minimality of s.
Ex.7.20
i):
⇒:Let O be an open set of X. O belongs to F by definition. X is a closed set. O = O∩X
is of the form of an ‘open∩closed’.
Let F be a closed set of X. F belongs to F because of a complement of some open set. X
is an open set. F = X ∩ F is of the form of a ‘open∩closed’ .
Let O i∩F i be a member of F (which is obtained by an intersection of members of F).
(O∩F)c=(X−(O∩F ))=(X−O )∪(X−F) is of the form of a finite union of ‘open∩closed’s.
For the case of a finite intersection of form ∪i<n (O i∩F i), it is enough to show the case
of 2 ‘(∪i<n (O i∩F i))’s. (∪i<n (O i∩F i)) ∩ (∪j<n’ (O j∩F j))= ∪i,j ((O i∩F i) ∩ (O j∩F j)) is
of the form of a finite union of ‘open∩closed’s.
(∪j<n(Oj∩Fj))c=∩j<n (X−(Oj∩Fj))=∩j<n ((X−Oj)∪(X−Fj)) is a finite intersection of the
form ∪i<n (Oi∩Fi). This is of the form of a finite union of ‘open∩closed’s by above.
⇐:∪i<n (O j∩F j) = (X−∪i<n (O j∩F j)) c= (∩j<n (X− (O j∩F j)))
c= (∩j<n (O j∩F j)
c) c
=(∩j<n (O j∩(∃U j) c)c) c.
ii):
⇒:Let Oi be an open set of X and Fi be a closed set of X. E=∪1≤i≤n (Oi∩Fi) is dense⇒
XE = . ∪1≤i≤n Fi ⊇ E ⇒ ∪1≤i≤n Fi=X. Let ∩1≤i≤n O i=O. O≠∅, because X is irreducible.
E⊇∪1≤i≤n (O∩F i)=O∩ (∪1≤i≤n F i)=O∩X.
⇐: An open set ∃U(≠∅)⊆X, E⊇U. ⇒ UE ⊇ . If XU ≠ ,V=X−U ≠∅ . V is an open set.
U∩V=∅. It contradicts to that X is irreducible. Then XU = ⇒ XE = .
Ex.7.21
⇒:If E belongs to F, E=∪1≤i≤n (O i∩F i) where O i is an open set of X and Fi is a closed
set of X by (ex.7.20i). Let X0 be any irreducible closed set of X.
If not 0XE∩ ≠X0, 0XE∩ =X0. E∩X0=∪1≤i≤n(Oi∩Fi∩X0). After excluding index such
that Oi∩Fi∩X0=∅ and renumbering, we get E∩X0=∪i’≤n’≤n(O i’ ∩F i’ ∩X0).
X0= 0XFO iini ∩∩∪ ′′′≤′ = 0XFO iini ∩∩∪ ′′′≤′ ⊆∪i’≤n’(Fi’ ∩X0) ⊆X0. ⇒ ∪1≤ i’≤n’F i’⊇X0.
X0 is irreducible. ⇒∃F i’⊇X0. Because if not, there exist open sets of X0 which are
67
disjoint. Then for this i’, O i’∩X0 =O i’∩F i’ ∩X0⊆E∩X0.
⇐: Suppose E does not belong to F. Let Σ=X’ : a closed set of X | (X’∩E) ∉F.
E ∈Σ⇒Σ≠∅. X is a Noetherian space⇒ a minimal element ∃X0 ∈Σ. X0 is irreducible,
because if not, open sets ∃U1,U2 ⊆X, X0∩U1≠∅, X0∩U2≠∅, X0∩U1∩U2=∅. X1=X0∩
(X−U2) and X2=X0∩ (X−U1) are closed sets strictly contained in X0. From minimality of
X0, X1 and X 2 do not belong to Σ. Then (E∩X1) ∈ F and ( E∩X2) ∈ F . (X1∩E) and
(X2∩E) are finite unions of ‘open∩closed’s. Then X0∩E=(X1∩E)∪(X2∩E) is a finite
union of ‘open∩closed’s. It is contradictory.
The case of 0XE ∩ ≠X0: 0XE ∩ ⊂X0. ⇒E∩ 0XE ∩ ⊆E∩X0. On the other hand,
E∩X0⊆E∩ 0XE ∩ always holds. E∩X0= 0XEE ∩∩ ∉F. By minimality of X0,
0XE ∩ ∉Σ. ⇒ E∩ 0XE ∩ ∈F. It is contradictory.
case of 0XE ∩ =X0: an open set ∃ O(≠∅) ⊆ X ,E∩X0⊇X0∩O.
⇒E∩X0=(E∩(X0−O))∪(X0∩O). By minimality of X0, E∩(X0−O)∈F.
Then E∩X0=(E∩ (X0−O))∪(X0∩O)∈F. It is contradictory.
Ex7.22
⇒:Let E be an open set of X. For any irreducible closed set X0 of X, E∩X0=∅ or
E∩X0≠∅. If E∩X0≠∅, E∩X0 is a nonempty open subset of X0.
⇐: Suppose E is not an open set of X. F=X−E is not a closed set. Then F ∩ E≠∅. Let
Σ= X ′ |F⊇X’, X ′ ∩E≠∅. F ∈Σ⇒Σ≠∅. X is a Noetherian topological space⇒
minimal element ∃X0 in Σ.
X0 is irreducible: suppose closed subsets ∃X1,X2 ⊂X0, X0=X1∪X2. X0∩F=(X1∩F) ∪
(X2∩F). X0= FX ∩0 ⇒X0= FX ∩1 ∪ FX ∩2 . Xi⊇ FX i ∩ ⇒ FX i ∩ are proper
closed subsets of X0 (i=1,2). Then FX i ∩ ∩E=∅ by minimality of X0. Then X0∩E=∅
is contradictory.
For the irreducible closed set X0, X0∩E≠∅. By the hypothesis, an open set ∃O, X0∩E⊇
X0∩O(≠∅)⇒ X0∩F⊆X0−O. X0= FX ∩0 ⇒X0⊆X0−O⇒X0∩O =∅. Contradictory.
Ex.7.23
68
Let f : A→B be a homomorphism of finite type where A and B are Noetherian rings. Let
E=∪1≤i≤n(O i∩F i)(⊆Spec(B)).
f*(E) is a constructible subset in Spec(A):
Spec(f(A))≅V(Ker(f))(⊆Spec(A)). If H be a constructible subset in Spec(f(A)), f*(H)
≅H∩V(Ker(f)). f*(H) is a constructible subset in A. Then regard f : A→B as the
embedding .
f*(E)=∪ 1≤i≤nf
*(Oi∩Fi). It is enough to show each f
*(Oi∩Fi) is constructible. Let E=O∩F.
Let F=V(b). Regard B/b as B. Then we can regard E=O as an open set of Spec(B). B is
Noetherian⇒ E is quasi-compact (ex.6.6). E is a union of finite basic open sets Yg of
Spec(B). Bg is a ring whose Spec is Yg(ex.3.23) ⇒Let E=Spec(B). Finaly, It is enough to
show that f*(Spec(B)) is constructible in Spec(A). By (ex.7.21), for any irreducible
closed set X0 of X=Spec(A) such that f*(Spec(B))∩X0 is dense in X0, it is enough to
show that f*(Spec(B))∩X0 contains a non empty open set of X0. f
*(Spec(B))∩X0=
f*(f
*−1(X0)). Let X0=p | p⊇p0. f
*−1(X0)=Spec(A/p0⊗AB)(ex.3.25). Let A be a domain. B
is Noetherian⇒Spec(B)=∪ 1≤i≤kYi where Y i is an irreducible component. Yi corresponds
to Spec(B/q) where q is a minimal prime ideal. Let B be an integral domain. f*(Spec(B))
=∪1≤i≤k f*(Yi). It is enough to show that at least one f
*(Yi) contains a nonempty open set
of Spec(A). Now it is reduced to the case where A and B are integral domain and f:A→B
is injective and finite type.
∃ s(≠0)∈A, Bs is integral over A[y1,…,yn]s where y1,…,yn (∈B) are algebraically
independent over A(ex.5.20). Define a homomorphism h:A[y1,…,yn]s→Ω(an
algebraically closed field including κ(A)) by h(yi)=0 and h(a)=a for a∈A and h(1/s)=1/s.
h can be extended to a homomorphism g : Bs→Ω (ex.5.21). g(Bs) is integral over
h(A[y1,…,yn]s)=As. Let s
h
sn AyyAA →→ ],...,[: 1φ be a homomorphism. φ is a
canonical homomorphism, i.e., A→As and injective. Let )(: s
g
s BgBB →→ψ . ψ is
an extension of φ. A is an integral domain⇒Xs≠∅. For ∀p∈Xs, Asφ(p)=pe is a prime
ideal of As. By (5.10), a prime ideal ∃q of g(Bs), q∩As=pe. ψ is an extension of
φ⇒pec⊆ψ−1
(q)=qc⇒p
ec⊆qc∩A. s∉p⇒pec
=p. On the other hand φ (qc∩A)=ψ (q
c∩A)
⊆q∩As=pe. Since φ is injective and s∉pe
, qc∩A⊆p. Then p=q
c∩A. f*(Spec(B))
includes Xs. f*(Spec(B)) is a constructible subset(ex.7.21).
Ex.7.24
⇒:(ex.5.10ii).
⇐: Let O be an open set of Spec(B) and E= f*(O) . Spec(B) is quasi-compact⇒O is a
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finite union of ∪gYg . Let O=Yg . It is enough to show that f*(O) is open. Spec(Bg)=Yg .
Let O=Spec(B). It is enough to show that E=f*(Spec(B)) is open. E is a constructible set
by (ex.7.23). Let X0 be any irreducible closed set of X=Spec(A). X0=p∈Spec(A) |p⊇p0.
E∩X0≠∅⇒∃p∈E∩X0. ∃q∈Spec(B), p=f*(q). p⊇p0⇒ ∃q0 (⊆q)∈Spec(B),p0=f
*(q0) by
going-down property. ⇒p0∈E. ⇒ 00 XXE =∩ . E∩X0 is dense in X0. E∩X0 contains a
nonempty open set of X0 (ex.7.20ii). As A is Noetherian and E satisfies conditions of
ex.7.22, E is an open set.
Ex.7.25
f is a flat-homomorphism(B is A-flat).
f has going-down property(ex.5.11).
f* is an open mapping(ex.7.24).
Ex.7.26
i): Let λ be an arbitrary additive function on the class F(A) of finitely generated
A-modules with values in an abelian group G. λ can be canonically extended to the
function on the free abelian group C generated by F(A). We call this function also
λ:C→G. Since λ is an additive function, it takes same value on the equivalent class of C
by subgroup D. Then λ is factored as following.
GAKDCC →=→ 0)(/λγ
λ=λ0°γ. Then for any additive function λ, λ0 is uniquely defined on K(A).
ii): A is a Noetherian ring. Let M be any finitely generated A-module. Then M has a
finite chain (ex.7.18) as following.
0=M0⊂M1⊂…⊂Mi-1⊂Mi⊂…⊂Mn =M where Mi/M i-1≅A/pi and pi∈Spec(A).
For each i, 0→Mi-1→Mi→Mi/Mi-1→0 is exact. γ(Mi)=γ(A/pi)+γ(Mi-1). Then γ(M)=
γ(A/p1)+ …+γ(A/pn). K(A) is generated by γ(A/pα) where pα∈Spec(A).
iii): Let A be a principal ideal domain. For any prime ideal p, 0→p→A→A/p→0 is
exact. Then γ(p)−γ(A)+γ(A/p)=0. If p≠0, p≅ (∃x) and 0→A →x (x)→0→0 is exact.
γ(A)−γ(p)=0. Then for nonzero prime ideal ∀p, γ(A/p)=0. K(A) is generated by
γ(A/(0))=γ(A). K(A) is isomorphic to Z.
iv): Since f : A→B is finite, B is a Noetherian ring and finitely generated A-module.
For any finitely generated module N over B, γB (N)∈K(B) is defined. Since N is also an
A-module and B is finite over A, N is a finitely generated module over A. Then
γA(N)∈K(A). Let f! : K(B) →K(A) be defined by f! (γB(N))= γA (N).
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f! is a homomorphism:
For any finitely generated B-modules N’ and N”, an exact sequence ∃0→N’→N→ N”
→0(for an example N=N’⊕N”). γB(N)=γB(N’)+γB (N”). 0→N’→N→N”→0 is exact as
A-modules. γA(N)=γA(N’)+γA(N”). Then
f! (γB (N’)+γB (N”))=f! (γB (N) )=γA (N)=γA (N’)+γA (N”).
Let g : B→C be finite. g°f:A→C is finite. (g°f)! exists. (g°f)!=f! °g! is clear.
Ex.7.27
Let K1(A) be a Grothendieck group of finitely generated flat A-modules and γ1 :
C1→K1(A) be a mapping from C1 which is a free group of isomorphic class of finitely
generated flat A-modules.
i): K1(A) is an abelian group. Multiplication is defined in K1(A) as following. For any
finitely generated A-flat modules M and N, M⊗N is a finitely generated flat A-module.
Then there exists a mapping (γ1 (M), γ1 (N))→γ1 (M⊗N). Denote this by γ1 (M)⋅γ1 (N).
This operator is multiplication:
Associative: followed by (M⊗N)⊗L≅M⊗ (N⊗L).
right distribution: for A-flat modules N’ and N”,0→N’ →N→N” →0 is exact for a flat
A-module N=N’⊕N”. γ1(N)=γ1(N’)+γ1(N”). For any flat A-module M, 0→M⊗N’
→M⊗N→M⊗N” →0 is exact. Then γ1(M⊗N)=γ1(M⊗N’)+γ1(M⊗N”). γ1(M)⋅γ1(N)=
γ1(M)⋅γ1 (N’) +γ1 (M) ⋅γ1 (N”). Then
γ1 (M) ⋅ (γ1 (N’) +γ1 (N”))=γ1 (M) ⋅γ1 (N’) +γ1 (M) ⋅γ1 (N”).
left distribution: same as right distribution.
Commutative: followed by (M⊗N)≅ (N⊗M).
Identity element: A is an A-flat module. For any flat A-module, (A⊗M)≅ M. γ1(A)⋅γ1 (M)
=γ1 (M). γ1 (A) is an identity element.
ii): For γ1(M)∈K1(A) and γ(N)∈K(A), M⊗N is a finitely generated A-module. There
exists a mapping: (γ1 (M), γ (N))→γ(M⊗N). Denote γ (M⊗N) by γ1(M)⋅γ (N) .
K(A) is a K1(A)-module:
K1(A) is a commutative ring :by (i).
Distribution 1: For finitely generated A-modules N’ and N”, 0→N’ →N→N” →0 is
exact for A-module N= N’ ⊕N”. γ (N)=γ (N’) +γ (N”). 0→M⊗N’ →M⊗N→M⊗N” →0
is exact. γ (M⊗N)=γ (M⊗N’) +γ (M⊗N”). γ1(M) ⋅γ (N)=γ1(M) ⋅γ (N’) +γ1(M) ⋅γ (N”).
Then
γ1(M) ⋅ (γ (N’) +γ (N”))=γ1(M) ⋅γ (N’) +γ1(M) ⋅γ (N”).
Distribution 2: For finitely generated flat A-modules M’ and M”, 0→M’ →M→M” →0
is exact for a flat A-module M= M’ ⊕M”. γ1(M)=γ1(M’) +γ1(M”). For any finitely
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generated A-module N, 0→M’ ⊗N→M⊗N→M” ⊗N→0 is exact. γ (M⊗N)=γ (M’ ⊗N)
+γ (M” ⊗N). γ1(M) ⋅γ (N)= γ1(M’) ⋅γ (N)+γ1(M” ) ⋅γ (N). Then
(γ1(M’) +γ1(M”)) ⋅γ (N)=γ1(M’) ⋅γ (N)+γ1(M” ) ⋅γ (N).
associative: For finitely generated flat A-modules M and M’ and a finitely generated
A-module N, (M⊗M’)⊗N≅M⊗(M’⊗N).
(γ1(M)⋅γ1(M’)) ⋅γ(N)=γ1(M’)⋅(γ1(M”)⋅γ(N)).
Identity element: γ1(A) is an identity element. For any finitely generated A-module N,
A⊗N≅N. Then γ1(A) ⋅γ (N)=γ (N).
iii): If A is a Notherian local ring, any finitely generated A-flat module M is a free
module (ex.7.15). M is isomorphic to a finite (n>0) direct sum of A. 0→A→
An→A
n−1→0 is exact. γ1(An)=γ1(A
n-1)+γ1(A). γ1(A
n)=nγ1(A). K1(A) is generated by γ1(A).
K1(A)≅ Z.
iv): Let f: A→B be a ring homomorphism and B be Noetherian. For a finitely generated
A-flat module M, B⊗AM is an Noetherian B-module and B-flat(2.20). There exists a
mapping γA1(M) →γB1(B⊗AM). Denote this mapping f !: K1 (A) →K1 (B).
f ! is a ring homomorphism:
sum: For any finitely generated flat A-flat modules M’ and M”,0→M’ →M→M” →0 is
exact for M=M’⊕M”. γA1(M)=γA1(M’)+γA1(M”). 0→B⊗AM’→B⊗AM→ B⊗AM”→0 is
exact. γB1(B⊗AM) =γB1(B⊗AM’) + γB1(B⊗AM”). Then
f ! (γA1(M) )=f
! (γA1(M’) )+f
! (γA1(M”) ).
Multiplication: Let N be an another flat A-module. B⊗A(M⊗N)≅(B⊗AM) ⊗ (B⊗AN).
γB1(B⊗A(M⊗N))=γB1(B⊗AM)⋅γB1(B⊗AN).
f!(γA1(M)⋅γA1(N))=f
!(γA1(M))⋅f !(γA1(N)).
For any ring homomorphism g:B→C where C is a Northerian ring, C⊗B (B⊗AM)
≅( C⊗B B) ⊗AM≅C⊗AM. Then (g°f)! = g
! °f
!.
v): f!(f !(x)y)=xf! (y) (x∈K1(A),y∈K(B)).
Since B is finite over A, (ex.7.26iv) is applicable.
For a finitely generated A-flat module M and a finitely generated B-module N,
f! ( f !( γA1(M))⋅γB (N)) =f!( γB1 (B⊗AM )⋅γB (N))=f! (γB (B⊗AM⊗BN) ) = f! (γB (M⊗AN) )
=γA (M⊗AN). On the other hand, γA1(M)⋅f!( γB (N)) =γA1(M) ⋅γA (N) =γA (M⊗A N ).
Then, f! ( f !( γA1(M)) ⋅γB (N)) =γA1(M) ⋅f!( γB (N)),i.e., f!(f
!(x)y)=xf! (y).
Chapter 8
Ex.8.1
Let r(qi)=pi. qi and pi are finitely generated. Let yi1,…, yimi be generators of pi. For each
72
yij, ∃sij(>0),yijsij
∈qi. Let ri=Σ1≤j≤misij. y∈pi⇒y=Σ1≤j≤miaijyij (aij∈A) ⇒yri∈qi⇒pi
ri⊆qi.
Let Si=A−pi. Si(piri)⊆Si(qi). Si(qi)=qi (4.8). Si(pi
ri)=pi
(ri). pi
(ri)⊆qi. pi(ri)
is pi-primary
(ex.4.13).
The case that pi is isolated:
pi is a minimal prime ideal. piApi is the unique prime ideal of Api. dim Api =0. Api is
Noetherian⇒Api is Artinian(8.5). Let piApi=mi. ∃r(>0),mir=0 (8.6). Si
−1pi=mi. mi
r =
Si−1pi
r=0. Si (0) = pi
(r). Let 0=∩iqi be a primary decomposition of 0. Si (0)=Si(qi)= qi. For
large r, pi(r)
=qi.
The case that pi is embedded:
A prime ideal ∃p’i(⊂pi). Let mi=Si−1pi. mi is a maximal ideal of Api.
mir=Si
−1pi
r≠mir’(r≠r’). (If mi
r=mi
r+1, mi
r=0. mi
r⊆Si−1p’i. mi⊆Si
−1p’i. It contradicts to
that Si−1p’i⊂Si
−1pi (3.13)). pi
(r)≠pi(r’)
(r≠r’)(4.8ii). For ∀r>r’, pi(r)⊂pi
(r’)⊆qi.
q1∩…∩pi(r)∩…∩qn=0 and q1∩…∩pi
(r’)∩…∩qn=0 are different primary
decompositions of 0 (r≠r’).
Ex.8.2
i)⇒ii): Spec(A) is a finite set of maximal ideals. For ∀x ∈Spec(A), x is a closed set.
∪y(≠x)∈Spec(A)y is a closed set, because of a finite union. x is a complement of
∪y(≠x)∈Spec(A)y⇒x is open. Any subset of Spec(A) is a finite set⇒it is open. Then
Spec(A) is a discrete space.
ii) ⇒iii): trivial.
iii) ⇒i): If p (∈Spec(A)) is not maximal, p⊂∃m(∈Spec(A)). If Spec(A) is discrete, any
subset is open,i.e.,any subset is close. m∈ p ⇒p≠ p . p is not a closed set. It
contradicts to (iii). Then Spec(A) is the set of maximal ideals. dimA=0. A is Noetherian.
A is Artinian.(8.5).
Ex.8.3
Let k be a field and A=k[x1,…,xn].
i)⇒ii): Let mi1≤i≤n be maximal ideals of A. ∃r>0, Π1≤i≤n mir=0. A≅Π1≤i≤n Ai, Ai ≅A/mi
is an Artinian local ring (8.7). The maximal ideal of Ai is mi/mir. Ai/(mi/mi
r) ≅ A/mi is a
finite algebraic extension of k (7.10). Then Ai/(mi/qi) is finite over k.
Let A be an Artinian local ring with the maximal ideal m.
A is finite over k :
A is a Noetherian ring and a finitely generated as an A-module. m is the unique prime
ideal of A. There exists a finite ascending sequence of ideals as following by (ex.7.18).
0=M0⊂M1⊂…..⊂Mn=A where Mi+1/Mi≅A/m for each i.
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Since A/m is an finitely generated k-vector space, A is an finitely generated k-vector
space. Then A is a finite k-algebra.
Since A is a finite product of Artinian local rings, A is a finite k-algebra.
ii)⇒i):Since A is a finitely generated k-module, A is a k-vector-space of finite dimension.
Ideals of A, k-modules, are vector subspace of A. Ideals satisfies d.c.c. .
Ex.8.4
i)⇒ii): Let p be a prime ideal of A. The fiber of p by f* is Spec(B⊗Aκ(p)). B is a finitely
generated algebra over f(A) ⇒B⊗Aκ(p) is a finitely generated algebra over the field κ(p).
⇒B⊗Aκ(p) is Noetherian. B⊗Aκ(p) is finite over f(A)⇒B⊗Aκ(p) is finite over κ(p)⇒
B⊗Aκ(p) is Artinian(ex.8.3) ⇒ Spec(B⊗Aκ(p)) is a discrete subset of Spec(B)(ex.8.2).
ii)⇒iii):Spec(B⊗Aκ(p)) is the fiber of p by f*. B is finitely generated over f(A)⇒
B⊗Aκ(p) is finitely generated over a field κ(p). B⊗Aκ(p) is a Noetherian ring. By
hypotheses, Spec(B⊗Aκ(p)) is discrete. B⊗Aκ (p) is an Artinian ring(ex.8.2). B⊗Aκ(p) is
a finite κ(p)-algebra(ex.8.3).
iii)⇒ii): The fiber of p is Spec(B⊗Aκ(p)). B is finitely generated over f(A) ⇒B⊗Aκ(p) is
a finitely generated κ(p)-algebra⇒B⊗Aκ(p) is Noetherian. By hypotheses, B⊗Aκ(p) is a
finite κ(p)-algebra. B⊗Aκ(p) is an Artinian ring(ex.8.3). Spec(B⊗Aκ(p)) is a discrete
subspace of Spec(B) (ex.8.2).
iii)⇒iv): The fiber of p by f*
is Spec(B⊗Aκ(p)). B⊗Aκ(p) is an Artinian ring by the proof
of iii)⇒ii). Spec(B⊗Aκ(p)), i.e. the fiber of p by f* , is a finite set of maximal ideals.
A counterexample: Let A=Q, B=Q [α1,α2,....] (αi is algebraic over Q) and f:A→B be the
embedding. f is integral. B is not finite over A. A and B are fields. Spec(A)=0.
Spec(B)=0. The fiber of 0 by f*
consists of only 0 of Spec(B).
Ex.8.5
Let k be an algebraically closed field. Let X be an affine algebraic variety of kn with the
coordinate ring A=k[x1,…,xn]. By linear combinations of x1,…,xn, we get
y1,…,yr,yr+1,…,yn such that y1,…,yr are algebraically independent over k and yr+1,…,yn
are integral over B=k[y1,…,yr]. Let L be the r-degree linear subspace of kr which y1,…,yr
generate. A surjective linear transformation ∃ γ : X→L (ex.5.16). L is an algebraic
variety with the coordinate ring B=k[y1,…,yr]. The coordinate ring of X is B’=
B[yr+1,…,yn]. γ is a polynomial mapping and regular⇒γ determines the k-algebra
homomorphism φ: B→B’(ex.1.28). Let φ*:Spec(B’)→Spec(B) and m be a maximal ideal
of B which coresponds to a point P∈ L. The fiber of φ* over m is φ*-1
(m)=
Spec(B’⊗Bκ(m))=Spec(B’⊗BB/m)=Spec(B’⊗Bk), because k is an algebraically closed
74
field (7.10). B’ is a finitely generated B-module (5.2). B’ ⊗B k is a finite k-algebra
⇒B’⊗B k is an Artinian ring (ex.8.3). Spec(B’⊗Bκ(m)) consists of a finite number of
maximal ideals. Since one point corresponds to one maximal ideal (ex.5.17), the number
of points which are lying over P∈L is finite and bounded by the number of maximal
ideals B’ ⊗B k which is an Artinian ring.
Ex.8.6
Let q⊂q1⊂…⊂qn⊂…⊂p be any chain of p-primary ideals.
Regard A/q as A and q as 0: in the one to one correspondence between ideals of A
including q and ideals of A/q, there exists one to one correspondence of primary ideals.
(Let q’(⊇q) be a primary ideal of A. A/q’ is primary in A/q because (A/q)/(q’/q)≅A/q’.
Inverse images of primary ideals by a homomorphism are primary ideals by the
explanation preceding (4.1).)
Localize A by S(=A−p). There exists one to one correspondence between p-primary
ideals and pAp- primary ideals (4.8ii). Ap is an Artinian local ring, because Ap is
Noetherian and dim Ap =0. Every ideal(≠(1)) of an Artinian local ring is primary. Since
an Artinian ring satisfies both a.c.c. and d.c.c., there exists a composition series from the
maximal ideal to zero ideal (6.8). Let s be the length of a composition series. Every
composition series has the same length s (6.7). Returning to A, the length of
q⊂q1⊂…⊂qn⊂…⊂p is finite and the maximum length of chains of p-primary ideals
equals to s.
Chapter 9
Ex.9.1
Let A be a Dedekind domain, K be the field of fraction of A and S be a multiplicatively
closed set of A. A is a Noetherian domain of dim=1 and integrally closed (9.3i). S−1
A is a
Noetherian domain (7.3). There is the one to one correspondence between prime ideals
of S−1
A and prime ideals of A which are disjoint with S (3.11iv). If a prime ideal ∃p(≠0)
of A which is disjoint with S, S−1
A is a Noetherian domain of dim=1 and its quotient
field is K and an integrally closed(5.12). S−1
A is a Dedekind domain (9.3i). If 0 is the
unique prime ideal of A which is disjoint with S, the prime ideal of S−1
A is only 0,i.e.,
S−1
A is a field. A⊆S−1
A ⊆K. K is the smallest field including A. S−1
A=K.
Let S≠A−0. Let I be the ideal group of A, P be the principal fractional ideal group of A
and H be the ideal class group of A and I’,P’,H’ be ones of S−1
A respectively. Let
75
M’=Σi≤nS−1
Ami’ (mi’ = mi/si, mi∈A, si∈ S) be any invertible ideal of S−1
A. M=Σi≤nAmi’ is
an fractional ideal of A because s1…snM⊆A. A is a Dedekind domain⇒M is invertible
by (9.8). M’=S−1
M. S−1
(MN) =S−1
M S−1
N. S−1
is a surjective homomorphism:I→I’. Let J
be its kernel. Following is commutative.
11
111
1
→ →→→
φ↓φ↓↓
→→→→
−
'IIJ
'
*K*K
S
By (2.10),Coker(φ)(=I/Im(φ)=I/P=H) →Coker(φ’)( =I’/Im(φ’) =I’/P’ =H’) is surjective.
Ex.9.2
Since c(fg)⊆c(f)c(g), it is enough to show c(f)c(g)/c(fg)=0. It is enough to show
(c(f)c(g))m=c(fg)m for maximal ideal ∀m of A. Am is a DVR(9.3)⇒mAm=(∃tm). An ideal
of Am is a power of (tm)(9.2). For each coefficient ai of f, ∃ni(>0), (ai)m=(tm ni
).
c(f)=(a1,…, ani)⇒c(f)m= (tm nf
) where nf =minni. For bj, a coefficient of g, let
(bj)m=(tmnj
). c(g)m=(bj)m =(tm ng
) where ng =minnj. c(fg) =Σij Aaibj . c(fg) m=(tmk) where
k is the smallest of ni+nj, i.e. nf+ng. c(fg)m=c(f)mc(g)m=(c(f)c(g))m.
Ex.9.3
⇒: Let A be a valuation ring which is Noetherian and not a field and v be its valuation.
A is a local domain. Let K be a field of fraction of A. A=x∈K | v(x)≥0(ex.5.31). For
any ideal a=(x1,…,xn) (xi≠0) of A, ∃xk, a=(xk) because xixj −1∈ A or xjxi
−1∈A(i≠j). A is not
a field⇒ a maximal ideal ∃m=(t)(≠0). If dimA=1, A is a DVR(9.2).
dimA=1: Let p=(s)(⊂m) be a prime ideal. Suppose s≠0. t/s∈A or s/t∈A. If t/s∈A, it
contradicts to p⊂m. If s/t∈A, then s=ta∈p(a∈A). Since t∉p, a∈p and a=sb(b∈A). v(s)=
v(t)+v(a)=v(t)+v(s)+v(b). v(t)+v(b)=0. Since v(b)≥0, v(t)=0. t is a unit. It is contradictory.
Then s=0. p=0.
⇐: A is a DVR. A has a valuation v to Z. v(x)≥ 0 for x∈A. Let a be any ideal of A. There
exists min v(x)|x∈a (=k ). ∃y∈a, k=v(y). For ∀x∈a, v(x)−v(y) ≥ 0. Then x/y∈A⇒a=(y).
A is Noetherian.
Ex.9.4
Let A be a local domain and not a field. Suppose the maximal ideal is m=(t) and
∩n≥1mn=0.
A is Noetherian :
Let a=(a1,..,an,…) (≠0, ⊆m) be any ideal. Since r(ai)=m=(t), let ri be smallest r such
76
that tr⊆(ai). If t
ri=uai and u is not a unit, ∃s >0, vt
s=u and t
ri−s=vai, because A is a domain.
This contradicts to the minimality of r. So u is unit. Let r= minri. a=(tr).
There is not a prime ideal p(≠0,⊂m) :
p is represented as (ts) for some s(>1) by above. Then p is not a prime ideal.
Because A is a domain, 0 is a prime ideal. Then dimA =1.
A is a DVR by(9.2iii).
Ex.9.5
Let A be a Dedekind domain and M be a finitely generated A-module.
M is flat⇔Mm is an Am-free module for maximal ideal ∀m⊂A(ex.7.16)⇔Mm is
torsion-free, since Am is a domain ⇔M is torsion-free(ex.3.13 ).
Ex.9.6
Let A be a Dedekind domain and M=Ax1+...+Axn and T(M)=M. Ann(xi)≠0(i=1,…,n).
Ann(xi) is a finite product of prime ideals(≠0)(9.4). Let pjj≤m be the set of prime ideals
which appear in Ann(zi)i and pj≠pj’ (j≠j’). Let Ann(xi)=Πj∈∆i⊆1,...,mpje(i,j)
.
A/Ann(xi)≅⊕j∈∆iA/pje(i,j)
(1.10). Axi≅A/Ann(xi)x’i≅(⊕j∈∆iA/pje(i,j)
)x’i. x’i is given by X i mod
Kerφ where Xi is an indeterminate and φ is a homomorphism ΣA/Ann(xi)Xi→
ΣA/Ann(xi)xi. Ann(x’i)=0. M=Σi=1,...,n (⊕j∈∆iA/pje(i,j)
)x’i. We get
M=Σ i=1,…,n (⊕ j∈∆i A/p je(i,j)
) x’i ≅⊕ j=1,…,m Σ i=1,…,n (A/p je(i,j)
) x’i
where e(i,j)=0 if x’i is absence.
Denote pj-component of M by M(pj). We get
M(pj)pj=Σ i=1,…,n Apj/tie(i,j)
x’i
where tj is a generator of the maximal ideal of Apj.
M(pj)pj is a module over Apj (PID). By the structure theorem for modules over PID,
M(pj)pj=⊕ i=1,…,n Apj/tie(i,j)
for some integers i(j) (>0).
Since S-1
(M⊕N)=S-1
M⊕S-1
N and M(pi)pj are 0 (i≠j), M(pj)≅ ⊕ i=1,…,n (A/p j i (j)
). Then M≅
⊕ j=1,…,m ⊕ i=1,…,n (A/p j i (j)
).
Ex.9.7
Let A be a Dedekind domain and dimA=1. A is a Noetherian ring. Let a(≠0) be an ideal
of A. dimA/a=0. A/a is a Noetherian. A/a is Artinian(8.5). Let a=∩1≤i≤nqi be a primary
decomposition and r(qi) =pi. A/a≅Π 1≤ i ≤ n A/qi where A/qi is a local Artinian ring(8.7).
A/qi≅(A/qi)pi/qi≅(A/qi)pi≅Api/qipi(ex.3.4). Api is a DVR. Its maximal ideal is (ti) and others
are powers of (ti) (9.2). Let qipi=(ti)ei. ti
ei=0 mod qipi.
77
Ideals of A/a are finitely generated⇒ it is enough to prove in the case of 2 generators.
Let (x,y) be an ideal. Let φ : A/a→Π1≤i≤nApi/qipi be an isomorphism. φ(x)=(x1,…,xn)
(xi∈Api/qipi). xi is a polynomial fi of ti having units or 0s as coefficients. If fi≠0, let ri(≥0)
be the smallest degree of fi whose coefficient ≠0. Let ui=fiti−ri
. ui is a unit because ui is a
sum of a unit and nilpotent elements. xi=uitiri. If fi=0, xi=ti
ei. φ(x)=(u1t1
r1,…, untn
rn).
(x)≅((t1r1
,…,tnrn
)). For y, (y)≅(t1s1
,…,tnsn
). (x,y)≅((t1min(r1,s1)
,…,tn min(rn,sn)
)). (x,y) is a
principal ideal.
Let b be any ideal of A. If b is not generated by one element, for any b(≠0)∈b b/(b) is a
principal ideal by above, i.e. b/(b)=(a’). Let a be an element such that a’=a mod b.
b=(a,b).
Ex.9.8
The first equation: It is enough to show that (a∩(b+c))p=((a∩b)+(a∩c)) p for any
maximal ideal p of A(3.8)(3.4).
The second equation: it is enough to show that ((a+b)∩ (a+c)) p=(a+(b∩c))p for any
maximal ideal p of A(3.8)(3.4).
Let A be a Dedekind domain. Ap is a DVR for any maximal ideal p of A. Let m=(t) be a
maximal ideal of Ap. ap=(t k) or ap=0 or ap=(1). Let k=0 for ap=(1) and k=∞ for ap=(0).
For the first equation,
(a∩(b+c))p=ap∩(bp+cp)=(t k1
)∩((t k2
)+( t k3
))=t k.
k=max(k1,min(k2,k3)).
((a∩b)+(a∩c))p=(ap∩bp) +(ap∩cp)=(t k1
)∩(t k2
)+(t k1
)∩(t k3
))=t k’
.
k’=min(max(k1,k2),max(k1,k3)).⇒k= k’ ⇒(a∩(b+c))p=((a∩b)+(a∩c))p.
For the second equation,
(a+(b∩c))p=a p+(bp∩cp)=(t k1
)+((t k2
)∩(t k3
))=t k.
k= min(k1,max(k2,k3)).
((a+ b)∩(a+c))p=(ap+bp)∩(ap+cp)=((t k1
)+(t k2
))∩((t k1
)+(t k3
))=t k’
.
k’=max(min(k1,k2),min(k1,k3)).
⇒k=k’ . (a+ (b∩c))p=((a+b)∩(a+c))p.
⇒k= k’ ⇒ (a+ (b∩c))p= ((a + b) ∩ (a + c))p.
Ex.9.9
Lemma1:
Let A be a commutative algebra, and a1,...,an be ideals of A. i)⇔ii).
i) There exists a solution x≡xi (mod ai) (1≤i≤n) ⇔ xi≡xj (mod ai+aj) for any i≠j.
ii) A →φ ⊕i=1,...,nA/ai →ψ ⊕ i<jA/( ai+aj) is exact. φ is defined by φ(x)=(x+ai,...,x+an).
78
and ψ is defined by ψ( x1,..., xn)=(…, xi−xj +ai+aj,…) (i<j).
Proof: i)⇒ii): ψφ=0 is clear.
Imφ⊇Kerψ: let ψ(x1,..., xn)=(0,…0), i.e. xi−xj∈ai+aj(i<j). ∃x0, x0≡xi (mod ai) (1≤i≤n)
by(i). Then φ(x0)=(x0+a1,...,x0+ an)=(x1+a1,...,xn+an).
ii)⇒i):⇐: Suppose xi≡xj (mod ai+aj) for any i≠j and ψ(x1,...,xn)=(0,...,0).
∃x0∈A,φ(x0)=(x1,...,xn) by (ii). x0≡xi (mod ai) (1≤i≤n) is a solution.
⇒: clear.♦
Lemma2:
Let L →φ M →ψ N be a sequence of A-modules and ψφ=0.
L →φ M →ψ N is exact⇔ L p → pφM p → pψ
N p is exact for any maximal ideal p
of A.
Proof:⇒: by(3.3).
⇐: Imφ=Kerψ: it is enough to show (Kerψ/Imφ)=0,e.i., for any maximal ideal p of A,
(Kerψ/Imφ)p=(Kerψ)p/(Imφ)p=0(3.8)(3.4). By hypotheses, Kerψp=Imφp.
Kerψp=(Kerψ)p : 0→Kerψ→M →ψ N is exact⇒0→(Kerψ)p→ pp
p NM →ψ
is exact
(3.3). ⇒ Kerψp=(Kerψ)p
(Imφ) p=Imφp: →φL Imφ→0 is exact⇒ 0)Im( →→ pp
p φφL is exact(3.3).♦
It is enough to show that lemma 2 holds in the case that A is a Dedekind domain. To
prove the sequence A →φ ⊕i=1,...,nA/a i →ψ ⊕ i<jA/( ai+aj) is exact, it needs only to
prove that A p → pφ(⊕i=1,...,nA/a i ) p → pψ
(⊕ i<jA/( ai+aj))p is exact for any
maximal ideal of A by lemma 2, since ψφ=0.
Ap is a DVR. Let (tp) be the maximal ideal of Ap. Let aip=(tp ei
) (9.2).
A p → pφ⊕i=1,...,nAp/(tp
ei) → pψ
⊕ i<jAp/(( tp ei
)+ (tp ej
)).
ψφ=0⇒ψpφp=0. If ψp(x1,...,xn)=(0,...,0), we prove that ∃x∈Ap , φp(x)= (x1,...,xn) by
induction on n.
n=2: let e1≤e2. Suppose x1−x2∈(tp e1
)+ (tpe2
). x1−x2=atpe1
for ∃a∈A. Let x=x2(=x1−atp e1
).
x≡x1 mod ( tp e1
), x≡x2 mod(tp e2
). x is a solution.
n=m: Suppose if n≤m−1, there exists a solution. Re-indexing ei, let em be the greatest
among eii=1,...,m. By assumption of the induction, ∃x0, x0≡xi mod (tpei) for i=1,…,m−1
and xi−xm∈(tpei)+(tp
em) for i=1,…,m−1. Then x0−xm∈(tp
ei)+(tp
em). ⇒x0−xm∈(tp
ei). Assume
em-1 be the greatest among eii=1,...,m-1. x0−xm=btpem-1
for ∃b∈Ap. Let x’=xm=x0−btp em-1
.
x’≡xm mod (tp em
). For i=1,…,m−1, (tpei)⊇ (tp
em-1). x’ ≡x0 mod ( tp
ei)≡xi mod(tp
ei).
φp(x’)=(x1,...,xn).
79
Chapter 10
Ex.10.1
Let A=Z/(p)⊕…⊕Z/(p)⊕… and B=Z/(p)⊕ Z/(p)2⊕...⊕Z/(p)
n⊕... .
Regard A as a Z-module. Let A(p)
be the completion of A by (p)-topology, i.e.
A(p)
= ))/((lim ApA n . ∀n(>0), (p)nA=0 ⇒A
(p)=A.
Let An=α−1((p)
nB), i.e. An=0⊕0⊕…0⊕ Z/(p)⊕Z/(p)⊕…(leading n 0s).
A = nAA /lim .
A/An =Z/(p)⊕…⊕ Z/(p)⊕0⊕0⊕… (leading n Z/(p)s). Then
A =Π∞Z/(p).
Let Bn=(p)nB, i.e. Bn=0⊕0⊕…0⊕p
nZ/(p
n+1)⊕ p
nZ/(p
n+2)⊕ p
nZ/(p
n+3)⊕… .
B/Bn=Z/(p)⊕…⊕ Z/(p n−1
)⊕ Z/(p n
)⊕ Z/(p n
)⊕ Z/(p n
)… .
B = )/(/lim 1
i
in pZBB ≥Π= .
Let F : M→ M be a completion functor on Z-module category by (p)-topology. Let
C=0⊕B and β:B→C be a canonical homomorphism by component. Following is exact.
A →α B →β C→0.
FA → )(αFFB → βF
FC→0, i.e. A → )(αF Π i≥1Z/(pi) → βF Πi≥0Z/(p
i)→0 is not
exact. Because x=(1 mod p,…, pn-1
mod pn,……)∈Ker(F(×p)), but x∉F(α)(FA)
(components of an element of FA are zero except finite components.). F is not a right
exact functor.
Ex.10.2
Let An=0⊕0⊕…0⊕Z/(p)⊕Z/(p)⊕…(leading n 0s). lim A/An=Π∞Z/(p) (ex.10.1).
lim A=A. 0→An→A→A/An→0 is an exact sequence of commutative three inverse
systems. 0→ lim An→A→Π∞ Z/(p) →0 is not exact, because A→Π∞Z/(p) is not be
surjective. lim is not a right exact fuctor.
1lim An:
Let C=Πn>0 An. Define dc:C→C by d
c (an)=an−θn+1(an+1) where θn+1:An+1→An is an
embedding. Coker dc=
1lim An. Ker d
c=0: we calculate x=(x1,…,xn,…)∈C such that
0=dc(x)=Πn>0(xn−θn+1(xn+1)). θn+1 is an embedding for each direct summand j⇒ for each
j, xi+1(j)
= xi(j)
. x (j)
is of the form (x1(j)
,...,x j(j)
,0,0,...)⇒ x (j)
=0.
80
Let D=Πn>0 A=Πi>0(⊕j>0 Z/(p)). Define dd
: D→D by dd (an)=an−θn+1(an+1) where θn+1:
A→A is an identity map. For a=(a1,a2,…,an,…)∈D (an∈An), we calculate
x=(x1,x2,…,xn,…) ∈ D such that a=dc(x)=Πn>0(xn−θn+1(xn+1)). θn+1 is the identity map
for each direct summand j, we can calculate x by j. xi+1(j)
= xi(j)
− ai(j)⇒ we can calculate
x (j)
for any a(j)
. Then dd is surjective. Coker d
d =0. 0=d
d(x)=xn−θn+1(xn+1) means xn=xn+1.
Kerdd≅A=⊕n>0 Z/(p).
Let E=Πn>0A/An. Define de
: E→E by de(an)=an−θn+1(an+1) where θn+1: ⊕Z/(p)
(n+1-tuple)⊕0….→⊕Z/(p) (n-tuple) ⊕0…. is projective. Ker de=Πn>0 Z/(p).
Following exact sequences is commute.
00
00
→→→→
↓↓↓
→→→→
EDC
ddd
EDC
edc
We get the following exact sequence by(2.10).
0→Ker d c →Ker d
d →Ker d
e →Coker d
c →Coker d
d →Coker d
e →0.
0→0→ ⊕n>0 Z/(p) →Πn>0 Z/(p) → 1lim An →0→ Coker d
e →0.
Then 1
lim An= (Πn>0 Z/(p))/(⊕n>0 Z/(p)).
Ex.10.3
Let E=∩1≤n anM =Ker(M→ M ). E is a finitely generated A-module. E is the set of the
elements of M which are annihilated by some element of 1+a(10.17). Let
F=∩m⊇aKer(M→Mm).
F⊆E : Fm=0 for maximal ideal ∀m(⊇a). F=aF(ex.3.14). F is a finitely generated
A-module⇒F is annihilated by some element of 1+a (2.5).
F⊇E : x∈E(⊆M) ⇒ ∃a∈a, (1+a)x=0(10.17). Let m be any maximal ideal containing a.
(1+a)∉m. (x/1)m=0. x ∈F.
⇒: M =0⇒E=Ker(M→ M )=M⇒aE=E (10.17). p ∈Supp(M)⇒apMp=Mp≠0. Suppose
p⊇a. ap⊆pAp=J(Ap). Then Mp=0(Mp is finitely generated). Contradictory. ¬p⊇a⇒
p∉V(a). Supp(M)∩V(a)= ∅.
⇐: Supp(M)∩V(a)= ∅. Let m∈V(a) be any maximal ideal. m∉Supp(M) ⇒Mm=0. Then
F=∩m⊇aKer(M→Mm)=E=Ker(M→ M ). F=M=E. aM=M⇒ M = MM na/lim =0.
81
Ex.10.4
Let A be a-adic completion of A. Let x be a non zero divisor of A. 0→A →x A is
exact. A is A-flat(10.14) and A =A⊗ A ⇒ 0→ A → ⊗x1 A is exact. 1⊗x is the image
of x in A . 1⊗ x is not a zero-divisor.
A counter example: let A=R[x,y], p=(y2−x
2(x+1)). A is Noetherian. Since p is a prime
ideal, M=A/ p is an integral domain. Let m=(x,y), M be m-adic complete of M and A
be m-adic complete of A. A is A-flat(10.14). 0→ p ⊗ A →A⊗ A → M ⊗ A →0 is exact.
By (10.13), M⊗ A ≅ M . By(10.15), p ⊗ A ≅ p A . Then M ≅ A /p A .
But, y2−x
2(x+1)=( y− 1
022
1
)!(2)12(
)!2()1( +∞
=
−
∑ −− j
jj
j
xjj
j)( y+ 1
022
1
)!(2)12(
)!2()1( +∞
=
−
∑ −− j
jj
j
xjj
j) in A .
M is not an integral domain.
Ex.10.5
Let A be a Noetherian ring and M be a finitely generated A-module.
0→bmM→M→M/b
mM→0 is exact.
0→bmM/(a
nM∩b
mM)→M/a
nM→M/(a
n+b
m)M→0 is a commutative exact
sequence of three inverse systems. SincebmM/ (a
nM∩b
mM) n is surjective,
0→n
lim bmM/ (a
nM∩b
mM) →
n
lim M/anM →
n
lim M/(an+b
m)M →0
is exact(10.2).
bmM is finitely generated. a
nb
mM-topology of b
mM coindides with a
nM∩b
mM (10.11).
0→n
lim bmM/
a
nb
mM→M
a→n
lim M/(an+b
m)M →0
is exact.
n
lim bmM/
a
nb
mM = (b
mM )⊗A (A)
a= b
m⊗A M⊗A (A)a=b
m M
a (10.13).
0→ bm M
a→Ma→
n
lim M/(an+b
m)M →0⇒ aa
bba MMMM mmn
n
/)/(lim ≅+ .
⇒ baba )())/(lim(lim MMM mn
nm
≅+ .
≅+ ))/(lim(lim MM mn
nm
ba MM nn
n
)/(lim ba + (lemma in below).
⇒n
lim M/(an+b
n) M= M (M
a)b .
(a+b)2n⊆an
+bn⊆(a+b)
n⇒ (a+b)2n
M⊆(an+b
n)M⊆(a+b)
nM. (a
n+b
n)M and (a+b)
nM have
bounded difference. Then (an+b
n)M-topology coinsides with (a+b)
nM-topology.
82
⇒ bababa +=+≅+ MMMMM n
n
nn
n
)/(lim)/(lim . baba +≅ MM )( .
Lemma ’m
limn
lim M/(an+b
m)M =
n
lim M/(an+b
n) M ’.
Proof: let f :m
limn
lim M/(an+b
m)M→
n
lim M/(an+b
n) M be a map. For n≥n’,m≥m’, let
θn,m:n',m':M/(an+b
m)M→M/(a
n’+b
m’)M be a canonical homomorphism. θn”,m”:n',m'=
θn,m:n',m'°θn”,m”:n,m (n”≥n≥n’, m”≥m≥m’). m
limn
lim M/(an+b
m)M is an inverse limit with
θn,m:n',m'.n
lim M/(an+b
n) M is an inverse limit with θn,n:n',n'. ∀y∈
m
limn
lim M/(an+b
m)M is a
coherent sequence. Let y=(y1,…, ym,…) (ym ∈n
lim M/(an+b
m)M). Each ym is a coherent
sequence. Let ym= (y1,m,…,yn,m,….) (yn,m ∈ M/(an+b
m)M ). θn,m:n-1,m(yn,m)= yn-1,m. (ym) is a
coherent⇒∃θ*,m: *,m-1=(θn,m:n,m-1)n:(y1,m,y2,m,…,yn,m,….) → (y1,m-1,y2,m-1,…,yn,m-1,….) such
that θn,m:n,m-1(yn,m)=yn,m-1. Then yn,m is a double coherent sequence with
homomorphism θn,m:n',m'. Then diagonal of yn,m,(yn,n), is a coherent sequence. f is
defined as a map from m
limn
lim M/(an+b
m)M to
n
lim M/(an+b
n) M.
Let x∈n
lim M/(an+b
n)M be a coherent sequence. x= (x1,1,x2,2,x3,3,….,xn,n,…)( xn,n ∈
M/(an+b
n) M). We can make a double coherent sequence xn,m (xn,m ∈M/(a
n+b
m)M)
from point (n,n) to (n’,m’) by θn,m:n',m' . By θn”,m”:n',m'=θn,m:n',m'°θn”,m”:n,m , xn,m is
uniquely determined. m
limn
lim xn,m∈m
limn
lim M/(an+b
m)M. f is surjective.
For y∈m
limn
lim M/(an+b
m)M, suppose f(y)=(y1,1,y2,2,….,yn,n,…)=(0,0…,0,..)⇒by θn,m:n',m',
∀yn,m=0. y∈m
limn
lim M/(an+b
m)M is (0,0…,0,..)=0. f is injective.
f is clearly a homomorphism.♦
Ex.10.6
⇒: J(A)⊇a. Let ∀m(⊇a) be a maximal ideal. For ∀x ∈A−m, x+an is its any
neighborhood and x+an ⊆ x+m. (x+m)∩m=∅. ⇒x+a
n⊆ (x+m) ⊆A−m. x is an inner
point of A−m. A−m is an open set. Then m is a closed set.
⇐: If¬a⊆J(A), a maximal ideal ∃m, ¬m⊇a. For ∀n(>0), ¬m⊇an. Then a
n+m=(1). For
83
∃y∈an and ∃z∈m ,1=y+z. For ∀x∈A−m, x=x(y+z) =xy+xz. Then x+a
n∋ x−xy = xz∈m.
(x+an)∩ m≠∅. x is not an inner point. A−m is not open. m is not closed.
Ex.10.7
⇒:Let A be a faithfully flat A-algebra. Substitute M in (ex.3.16v) by A. Let
E=Ker(A→ A )=∩n≥0an=0 (10.17). Let m be any maximal ideal of A. Suppose
(x+an)∩m≠∅ for ∃x∈A−m and ∀n(>0). ∩n≥0 ( x+ a
n)=x⊆m. Contradictory. Then ( x+
an )∩m=∅ for ∃n>0. x is an inner point of A−m. m is closed.
⇐: Let Mi be any finitely generated A-module. Consider Mi→Mi⊗A A . Since A is
Noetherian, Mi⊗A A ≅ iM (10.13). Let E=Ker(M i→ iM )=∩n≥0anMi. E is an A-submodule
of M. A is a Zariski ring⇒a⊆J(A). aE=E⇒E=0. Mi→ iM is injective, i.e, Mi→Mi ⊗A A
is injective. For A-module ∀M, Let M i be all finitely generated submodules of M,
I be a directed set by inclusion, M is the direcit limit of M=(Mi,µij) where µij is the
embedding (ex.2.17). Let N=(0,1 ij) and P=( M i⊗A A ,µ ij⊗1) be direct systems on I. N→
M → P is exact sequence of direct systems. 0→M → lim (M i⊗A A ) is exact (ex.2.19).
lim (Mi⊗A A )= M ⊗A A (ex.2.20). A is a faithfully flat A-algebra(ex3.16v).
Ex.10.8
A=f/g | f,g∈C[z1,…,zn],g(0)≠0. B= f ∈ C[[z1,...,zn]] | f converges in some neighbor-
hood of the origin. Let C=C[[z1,...,zn]].
B is a local ring:Let f1, f2∈B be power series which converge on neighborhoods U1, U2
of 0 respectively. f1−f2, f1 f2 are power series which converge on neighborhoods
0∈U1∩U2. B is a ring. Let a power series f ∈ B converge on 0∈∃U . Since f is analytic ,
1/f is analytic on U except z=( z1,...,zn) | f(z)=0. Then if f(0)≠ 0,1/f can be expanded to
a power series and converges on 0∈∃U’. f such that f(0)≠ 0 is a unit of B. m=f∈B|
f(0)=0 is an ideal of all non units. m is the unique maximal ideal.
Let f ∈ C(=C[[z1,...,zn]]). There exists a Cauchy sequence (fn) of B which converges to f
by m-toplogy, where fn be a sum of monomials upto degree n of f. Then C= B .
B is A-flat: If C is B-faithfully flat and C is A-flat, B is A-flat(ex.3.17). Assuming B is
Noetherian, B is a Neotherian local ring. B is a Zariski ring because m ⊆J(B). C is
B-faithfully flat(ex.10.7). C is a ring of formal power serises of D=C[z1,…,zn]. Since C
is a completion of D by p=(z1,…,zn), C is D-flat(10.14). An A-module is a D-module.
84
For any exact sequence 0→M’ →M of A-modules, 0→M’ ⊗DC →M ⊗DC is exact. Let
S=D−p.
CSMSCSMS DD
11110 −−−− ⊗→⊗′→pp
is exact(3.7). Since Dp=A and S−1
C =C,
CMSCMS AA ⊗→⊗′→ −− 110
is exact. S−1
M=M and S−1
M’=M’ .
0→M’ CA⊗ →M CA⊗
is exact. Then C is A-flat. Then B is A-flat.
Ex.10.9
Let (A,m) be m-adic complete. Let f(x) be a monic polynomial of degree n and
)()()( xhxgxf = where )(xg , )(xh ∈(A/m)[x] are coprime monic polynomials of
degree r and n−r respectively.
We prove ∃gk(x),hk(x)∈A[x], f(x)−gk(x)hk(x) ∈mkA[x] for ∀k>0 by induction.
Case of k=1:
By the hypothesis on f(x), coprime monic polynomials ∃ )(xg (deg=r),
)(xh ( deg=n−r)∈(A/m)[x], )(xf = )(xg )(xh . Let g1(x) (deg=r) and h1(x) (deg=n−r) be
monic polynomials which are inverse images of )(xg and )(xh in A[x] respectively.
Then f1(x)=f(x)−g1(x)h1(x)∈mA[x].
Case of k>1:
Suppose ∃gk-1(x) (deg=r), ∃hk-1(x) (deg=n−r)∈A[x], fk-1(x)=f(x)−gk-1(x)hk-1(x)∈mk-1A[x]
and )(xg =gk-1(x) mod m and )(xh =hk-1(x) mod m.
Then
fk-1(x)=Σ0≤p≤ncpxp
(cp∈mk-1
).
For 0≤p≤n, ∃ )(xa p (deg≤n−r),∃ )( xb p(deg≤r)∈(A/m)[x], x
p= )(xa p )(1 xg k − +
)(xb p )(1 xhk − (lemma in below). Let ap(x),bp(x)∈A[x] be inverse images of
)(xa p (deg≤n−r) and )(xb p (deg≤r) respectively. Then
ap(x) gk-1(x)+ bp(x) hk-1(x)=xp+rp (x) where rp (x)∈mA[x].
Let gk(x)=gk-1(x)+ Σ0≤p≤ncpbp(x) and hk(x)=hk-1(x)+ Σ0≤p≤ncpap(x). )(xg =gk(x) mod m
and )(xh =hk (x) mod m.
fk(x) =f(x) – gk(x)hk(x)= − Σ0≤p≤ncprp(x) −Σ0≤p≤ncpap(x) Σ0≤p≤ncpbp(x).
85
ap(x)gk-1(x)+bp (x)h k-1 (x) =xp+rp (x) (rp(x)∈mA[x] ) .
⇒fk(x) ∈ mkA[x]+m
2k-2A[x]⊆mk
A[x], since k>1.
Coefficients of (gk(x)hk(x))k are Cauchy sequences in the mA[x]-topology and converge
to coefficients of f(x). Coefficients of (gk(x))k and (hk(x))k are Cauchy sequences by
m-topology. Let g(x) and h(x) be their limits respectively. g(x),h(x) ∈A[x] (m-complete) .
Let α=1+Σ1≤nαn (αn∈mn) be the coefficient of highest degree of g(x). A is m-adic
complete ⇒m =m⇒α a unit⇒g(x)/α is a monic polynomial⇒rename it g(x). As the
same for h(x). We get f(x)=g(x)h(x).
Lemma’ let g0(x)(deg n0),g1(x) (deg n1) be coprime monic polynomials on a field and
n0≥ n1. ∃a(x)(deg≤n1), ∃b(x)(deg≤n0),f(x)( deg ≤n0+n1)=g0(x)a(x)+g1(x)b(x).’
Proof: Coprime⇒g0(x)a0(x)+g1(x)b0(x)=1. Using Euclid algorithm,
g0(x)=g1(x)q1(x)+g2(x) (deg(g2)=n2<n1) (deg(q1)=n0−n1) .
g1(x)=g2(x)q2(x)+g3(x) (deg(g3)=n 3<n2)( deg(q2))=n1−n 2).
g2(x)=g3(x)q3(x)+g4(x) (deg(g4))=n4<n3)( deg(q3)=n2−n3).
∃ν≥0,
gv-1(x)=gv(x)qv(x)+k(k≠0 is an element of a field)(deg(qv))=n v-1−n v).
We calculate the degree of a(x) and b(x).
To get degree of polynomial which degree of polynomial which
following polynomials multiply g0(x) multiply g1(x)
g2(x) =g0(x) −g1(x)q1(x) 0 n 0−n 1
g3(x) = g1(x) −g2(x)q2(x) n 1−n 2 n 0−n 1+ n 1−n 2
g4(x)=g2(x) −g3(x)q3(x) n 1−n 2+ n 2−n 3 n 0−n 1+ n 1−n 2+ n 2−n 3
…………………………….
gi (x) n1−n i-1 n0−n i-1
…………………………….
k n 1−n v n 0−n v
deg(a0(x))= n1−n v,deg(b0(x)) =n0−n v.
To get a(x) and b(x) such that f(x)( deg ≤n0+n1)=g0(x)a(x)+g1(x)b(x) by induction, let
deg(f(x))= m≤n0+n1. If m≥n0,f(x)=g0(x)h0(x)+f1(x). deg(h0)≤m−n0=n1. deg(f1)<n0. The
degree of polynomial which are multiplied to g0(x) is ≤ n1. For f1(x), let gi(x) be one of
the maximum degree among gi(x) i≤v and equal or less than deg(f1).
f1(x) =gi(x)hi(x)+f2(x) (deg(f2)) < deg(gi)).
The degree of polynomial which is multiplied to g0(x) for getting f2(x) is
(n1−ni-1+deg(hi)) and is ≤n1−ni . The one for g1(x) is (n 0−n i-1+deg(hi(x))) and is ≤ n 0−n i.
Let f2(x) =gi’(x)hi’(x)+f3(x). f(x) =g0(x)h0(x)+gi(x)hi(x)+gi’(x)hi’(x)+f3(x).
86
If gj(x)|fl(x),end. If not, the degree of the residual fu(x) is less than nv. fu(x)
=g0(x)fu(x)a0(x)+g1(x) fu(x)b0(x) by fu(x)a0(x)(deg<n1) and fu(x)b0(x) (deg< n 0).
Let a(x) be sum of polynomials which are multiplied to g0(x). The degree of a(x) is
equal or less than n1. Let b(x) be a sum of polynomials which are multiplied to g1(x).
The degree of b(x) is equal or less than n0. g0(x)a(x)+g1(x)b(x) =f (x). ♦
Ex.10.10
i): a monic polynomial ∃g(x)∈(A/m)[x], )(xf =(x−α)g(x) where α∈(A/m). α is a simple
root of )(xf ⇒ g(α)≠0. (x−α) and g(x) are coprime, because (A/m)[x] is a Euclidian ring.
monic polynomials ∃u(x),v(x) ∈A[x], f(x)=u(x)v(x), )(xu =g(x), )(xv =(x−α) by
(ex.10.9). v(x) is a monic polynomial of degree 1. ∃a∈A,v(x)=x−a,i.e. )(xv =v(x) mod m.
Then α = a mod m. a is a simple root of f(x)=0. If not, α is a multiple root
of )(xf .Contradictory.
ii): p=(p), p is a prime. Let Zp =Σi≥0 ai pi |0≤ ai <p be a p-adic integer ring. Zp is the
lim Z/pkZ where θ k+1 :Z/p
k+1Z→→→→ Z/p
kZ is a projection.
Zp is a complete local ring with the maximal ideal m=Σi≥1 ai pi |0≤ ai <p :
p=Zpp(10.15i)⇒ p=m. Z/(p) ≅≅≅≅Zp/ p (10.15iii)⇒ p= m is a maximal ideal. p⊆J(Zp)
(10.15iv)⇒ p= m is the unique maximal ideal.
Let A = Z7 and f(x)=(x2−2)∈ A[x]. )(xf =(x−4)(x−3) on (A/m)[x]=(Z/(7))[x]. A is a
complete local ring. 4 and 3 are single roots. by (i) ∃a,b ∈A such that 4≡ a mod m, 3≡ b
mod m and roots of f(x)=0. Then squares of a,b equal to 2. a=3+7α1+…+7nαn+… .
From (3+7α1) 2−2≡0 mod 7
2,α1=1. Consequently , we can obtain αn-1 from f(a) ≡0 mod
7n
(n=2,3,….),i.e., f(a) is in 7n-neighborhood(n=2,3,….). As the same,b =4+
7×5+72×4+… .
iii):Regard f(x,y)∈k[x,y](k is a field) as a polynomial of the indeterminate y having
coefficients in k[x]. Let n=(x) be a maximal ideal of k[x]. Let A=k[[x]]. A is the
completion of k[x] by n and a local ring with maximal ideal m=(x) (ex.1.5i). f(x,y)∈A[y].
f(0,y)= ),( yxf ( ),( yxf =f(x,y) mod m). ⇒ f(0,y) ∈ (A/m) [y]. y=a0 is a single root of
f(0,y)=0,i.e., ),( yxf has a single root a0∈A/m. f(x,y) has a root y∈A such that y≡a0
(mod m) by (i). y(x) =Σn≥0a n xn∈A.
Ex.10.11
Let X be a metric topological space of R . The germ of C∞
functions at x=0 (x∈X) is
defined as the equivalent class of the relation ‘f(x) and g(x), C∞ functions, are
equivalent ⇔f(x)=g(x) on some neighborhood of x=0’. These equivalent classes form a
ring, say A. Let m=(x). g(x)∉m⇒g(x)≠ 0 on some neighborhood of the origin ⇒1/g(x)
87
is define on some neighborhood of the origin ⇒1/g(x)∈C∞⇒g(x) is a unit of A⇒m is
the unique maximal ideal A⇒ A is a local ring.
Let a=∩n>0(x)n. a is an ideal of f(x)∈ m, all of whose derivatives vanish at the origin. If
A is Noetherian, a =0 from a= ma . But 2/1 xe− ∈ a. Then A is not Noetherian.
f∈A ⇒ f mod a is a series of x⇒A/a ⊆R[[x]]. By Borel’s theorem, A ⊇ R[[x]]. R[[x]]∩
a=0⇒(R[[x]]+ a) /a= R[[x]] ⇒ A/a⊇ R[[x]] ⇒ A/a= R[[x]]. Following is exact.
0→ a →A→ R[[x]] →0.
We take m-adic completion of the above sequence. a⊆(x)n .a/(x)
nn is surjective.
A = lim A/ (x)n .Since a⊆mn
, R[[x]]/(R[[x]]∩mn)=R[[x]]/(x)
n and R[[x]] is (x)-adic
complete.
0→0→ A → R[[x]] →0
is exact. Then ]][[xA R= . R[[x]] is Noetherian(10.27). A is Noetherian.
Ex.10.12
Since A is a Noetherian ring, B=A[x1,…,xn] is a Noetherian ring by (7.6). Because
C=A[[x1,…,xn]] is (x1,…,xn)-adic completion of B, C is Noetherian(10.26). B is an A-flat
algebra (ex2.5). C is a B-flat algebra (10.14). C is an A-flat algebra (ex2.8ii).
Spec(C)→Spec(A) is surjective (ex1.5v). C is faithfully flat over A by (ex3.16ii).
Chapter 11
Ex.11.1
Let m be the maximal ideal of A corresponding a point P of variety f(x)=0. Let q⊇(f) be
the maximal ideal of k[x1,…,xn] corresponding to m. q is of the form (x1−a1,…,xn−an)
because k is algebraically closed field (ex.5.17). Let q=(x1,…,xn). No loss of generality.
f(0)=0. Am=(k[x1,…,xn]/(f))q=k[x1,…,xn]q/(f)q (ex.3.4). Poincare series of Gq (k[x1,
…,xn]q) is (1− t)-n⇒dim k[x1,…,xn]q=n. f is an irreducible polynomial⇒ f/1(∈qk[x1,
…,xn]q) is a non-zerodivizor of k[x1,…,xn]q. dimAm=k[x1,…,xn]q−1(11.18). Then
dimAm=n−1.
P is a nonsingular point⇔not all partial derivatives ∂f/∂xi≠0 at P⇔there exist some
monomials of degree 1 in f⇔f∉(x1,…,xn)2. Am is a regular local ring⇔dimk(m/m
2)=n−1
(11.22). It is enough to show dimk(m/m2)=n−1⇔f∉(x1,…,xn)
2.
Let φ : k[x1,…,xn]→k[x1,…,xn]/(f) be a canonical homomorhism. Kerφ= (f) .
m=q/Kerφ=(x1,…,xn)/(f).
88
m2= (q+Kerφ)
2 : q∈∈∈∈q /Kerφ=(x1,…,xn)
2+ (f) /(f).
m/m2= (x1,…,xn) /((x1,…,xn)
2+(f)).
If f∈(x1,…,xn)2, m/m
2≅≅≅≅(x1,…,xn)/(x1,…,xn)2 . Then dimk(m/m
2)=n.
If f∉(x1,...,xn)2, ((f)+(x1,…,xn)
2)=((f ’)+(x1,…,xn)
2) where f ’ is a sum of monomials of
degree 1 of f. Then (f ’) defines a hyper-space in V=kx1+…+kxn. Then dimension of
(x1,…,xn)/((x1,…,xn)2+(f ’)) equals n−1.
Ex.11.2
Let A be m-adic complete and k ⊂A and k≅A/m. Let x1,…,xd be system parameters and
q be the m-primary ideal which x1,…,xd generate. x1,…,xd are algebraically
independent (11.21). The mapping φ : k[[t1,…,td]]→A such that φ(k’ti)=k’xi for 1≤i≤d
and k’∈k is definable because A is m-adic complete.
φ is injective:
If not, ∃f (≠0)∈k[[t1,…,td]] , φ(f)=0, i.e. f(x1,…,xd)=0. Regard x1,…,xd as indeterminates.
k[[x1,…,xi-1]][[xi]]=k[[x1,…,xi]] by (ex.10.5). Regard f(x1,…,xd) as a series of xd having
coeffients in k[[x1,…,xd-1]]. Because f(x1,…,xd) can be regarded as nilpotent, coefficients
which are elements of k[[x1,…,xd-1]] are all nilpotent (ex.1.5). Consequently, elements in
k[[x1]] must be nilpotent. It contradicts to that k is a field.
Apply k[[t1,…,td]] to A in (10.24) and A of this exercise to M in (10.24). Let
f∈k[[t1,…,td]] and a∈A. A is a k[[t1,…,td]]-module, because φ(f)⋅a∈A. Let a=(t1,…,td).
k[[t1,…,td]] is a-complete. φ(a)=q. Let An=qn. (An) is a-filtration. ∩An=∩q
n=f (∩ a
n)=0.
So A is Hausdorff.
G(A)=A/q⊕q/q2⊕…is finitely generated over G(k[[t1,…,td]]):
G(k[[t1,…,td]])≅G(k[t1,…,td])=k⊕a/a2⊕a
2/a
3⊕…(10.22). φ(a/a2⊕a
2/a
3⊕…)=q/q2⊕q
2/q
3
⊕…⇒it is enough show that A/q is finite over k. A/q is Artinian. Let r(q)=m. ∃n>0,
mn⊆q. Consider m
n⊂…⊂ mi+1 ⊂ mi⊂… ⊂A. m
i/m
i+1 is A/m(≅k)-module and m
i is
finitely generated⇒ mi/m
i+1 is k-vector space of finite dimension. A/q is finite over k
(A/q=ky1+…+kyr). G(A)=ky1+…+kyr⊕q/q2⊕…⇒G(A) is finitely generated G(k[[t1,
…,td]]) - module (y1,…,yr are generators).
A is a finitely generated k[[t1,...,td ]]-module(10.24).
Ex.11.3
’For an irreducible variety V on a field k, the local dimension at any point P on V equals
to dimV.’
Proof: Let V be an irreducible variety on affine space kn. I(V)=p is a prime ideal of
k[x1,…,xn]. Let k (⊇k) be an algebraically closed field. k [x1,…,xn] is integral over
89
k[x1,…,xn]. A prime ideal ∃q of k [x1,…,xn], q∩k[x1,…,xn]=p(5.10). Let U be the
irreducible variety of an affine space nk which q defines. A(U)= k [x1,…,xn]/q is
integral over A(V)=k[x1,…,xn]/p (5.6).
Local dimension of any point of V=dimU:the height of the prime ideal of Noetherian
ring is finite(11.12). Let m be a maximal ideal of A(V) and its heights be m and 0=p0⊂
p1...⊂ pm =m be strict chain of prime ideals. By (5.10) and (5.11), there exists a chain of
prime ideals in A(U), say, 0= q0⊂q1...⊂ pm =n such that A(V)∩qi =pi. By (5.8), n is
maximal. Suppose there exists another chain of prime ideals 0= q’0⊂q’1...⊂ q’m’ =n
such that m’> m, the chain 0=(q’0∩ A(V))⊂ (q’1∩ A(V))...⊂ (q’m’ ∩ A(V)) = m is of
length m’ (5.9). This is contradicory to definition of height of m. Then the height of n is
m. By (11.25), the height of n equals to dimU.
dimV=dimU:since A(V)⊆A(U) is integral, k(V)⊆k(U) is an algebraic extention. k⊆k(V)
is a transcendental extention of degree dim V. Consider k⊆ k ⊆k(U). k ⊆k(U) is a
transcendental extention. k⊆ k is an algebraic extention. The transcendental degree of
k(U) over k is dimU. Then dimV=dimU.
Then local dimension at any point of V equals to dimV.
Ex.11.5
Let λ be an additive function which takes values in Z. Let A=A0⊕ A1⊕…⊕… be a
Noetherian graded ring. A is a finitely generated algebra over A0. We may take its
generators from homogeneous elements, say x1,…,xs of degree k1,…,ks respectively. Let
M be a finitely generated graded A-module and Mn be its homogeneous part of degree n.
Let mj (of degree rj) be homogeneous generators of M. Any element of Mn is of the form
Σjgjmj (gj∈jrnA − ) in which gj takes coefficients in A0. Then Mn is a finitely generated
A0-module. Let γ(Mn) be the image of Mn in Grothendieck group K(A0). There exists the
induced homomorphism λ0 : K(A0)→ Z such that λ (Mn)=λ0(γ(Mn)) by (ex.7.26).
Poincare series of M on K(A0) is P(M,t)= Σn≥0λ0 (γ(Mn))tn.
And (11.1) is P(M,t) =f(t)/Πi=1,…s(1−tki) f(t) ∈Z[t] .
Ex.11.6
1+dimA≤dim A[x]:
p[x] is a prime ideal of A[x] for a prime ideal p of A (ex.4.7). (p,x)(⊃p[x]) is a prime
ideal of A[x](A[x]/(p,x)≅A/p is a domain). For a chain p0⊂p1⊂…⊂pd of prime ideals of
A, p0[x] ⊂p1[x] ⊂…⊂pd[x] ⊂ (pd,x) is a chain of prime ideals of A[x].
dimA[x] ≤1+2dimA:
90
Let dimA=d. Let f : A→A[x] be the embedding and f*: Spec(A[x])→Spec(A). The fiber
of p∈Spec(A) is Spec(κ(A/p)⊗AA[x])=Spec(κ(A/p)[x]). Dim κ(A/p)[x]=1. There exists
some chain of 2 prime ideals of A[x] which correspond to p. For a chain of d+1 prime
ideals in A, if all fibers make a chain, the number of member in its chain is at most
2(d+1) . Then number of links is at most 2d+1.
Ex.11.7
Let dimA=m. Let p be a maximal ideal of A whose height is m. There exists a chain of
prime ideal p0⊂ p1⊂ p2.... ⊂ pm=p which can not be inserted. A is Noetherian⇒ideals
are decomposable. Let p0,1,...., p0,n(0) be prime ideals belong to 0. p0 is equals to one of
p0,1,...., p0,n(0). If ∪ jp0,j⊇ p1,∃p0,j⊇p1(1.11). It contradict to height(p1)=1. Then ¬∪ jp0,j⊇
p1. ∃ a1∈p1, a1∉∪ jp0,j. Consider A/p0. 0 is the minimul prime ideal. (a1)⊆ p1. Let
p1,1,...., p1,n(1) be prime ideals belong to (a1). Since r((a1))= p1,1∩....∩ p1,n(1) ⊆ p1,∃ p1,j⊆
p1. Since height(p1)=1, p1,j= p1. p1 is a prime ideal belong to (a1). If ∪ jp1,j⊇ p2,∃p1,j⊇
p2(1.11). It contradicts to Height(p2)=2. Then ¬∪ jp1,j⊇ p2. ∃a2∈p2, a2∉∪ jp1,j.
Consequently,we can get ai such that pi is a prime ideal belong to ai=(a1,a2,...,ai). pm is a
prime ideal belong to (a1,a2,...,am). pi[x] is a prime ideal belong to ai[x] (ex.4.7) . ai[x] is
an ideal of A[x] which a1,a2,...,ai generate.⇒height(pi[x])≤i(11.16). On the other hand,
p0[x]⊂p1[x]⊂p2[x]....⊂pi[x] is a chain of prime ideals⇒height(pi[x])≥i. Then
height(pi[x]) = i = height(pi).
Let m be any maximal ideal of A. hight(q=m∩A)≤m. By the above, height (q[x])=
height(q). Since m=m[x],m⊇q[x]. The number of fiber of q is 2(ex.11.6). (q,x) is a fiber
of q⇒m=(q,x). height(m)=height(q[x])+1≤m+1. dimA[x]≤m+1=dimA+1. From
(ex.11.6) , 1+ dimA≤dimA[x].⇒ dimA[x]= dim A +1. Inductively, dim A[x1,x2,...,xn] =
dim A +n.