Solution to Commutative Ring Theory

Anonymous

This is the solution of the homework in the course《Commutative Ring Theory》. Problemscan be founded on http://www.wwli.url.tw/downloads/CommRing-2019-HW.pdf, and I’mglad to any errata to this document.

We suppose A is a commutative ring, p > 0 is a prime number.

1 Ring Theory Revisited

1. Suppose x = 0, then

(1 + x)(1− x+ x2 − · · ·+ (−1)n−1xn−1) = 1

⇒1 + x is a unit.

If a ∈ A×, b ∈ Nil(A), then a−1b ∈ Nil(A),⇒ 1 + a−1b ∈ A× ⇒ a+ b = a(1 + a−1b) ∈ A×

2. (1) (⇐=): a1, . . . , an ∈ Nil(A)

⇒a1, . . . , an ∈ Nil(A[x])

⇒a1x, . . . , anxn ∈ Nil(A[x]) a0 ∈ A× ⊂ (A[x])×

⇒f ∈ (A[x])×

(=⇒): If there exists g = b0 + b1x+ · · ·+ bmxm such that fg = 1, then

(denote ak = 0 when k > n, bk = 0 when k > m)

1

1 RING THEORY REVISITED 2

a0b0 = 1

a0b1 + a1b0 = 0

......

t∑k=0

akbt−k = 0

......

an−1bm + anbm−1 = 0

anbm = 0

⇒ a0, b0 ∈ A×

⇒

am+1n b0 = 0 ⇒ am+1

n = 0 ⇒ an ∈ Nil(A)...

...

a2nbm−1 = 0

anbm = 0

then a0 + a1x + · · · + an−1xn−1 = f − anx

n is nilpotent. Continue the previousprocess, we obtain that a1, . . . , an are nilpotent.

(2) (⇐=): a1, . . . , an ∈ Nil(A)

⇒a0, . . . , an ∈ Nil(A[x])

⇒a0, a1x, . . . , anxn ∈ Nil(A[x])

⇒f = a0 + a1x+ · · ·+ anxn ∈ Nil(A[x])

(=⇒): f ∈ Nil(A) ⇒ fn = 0

⇒ an0 = 0 ⇒ a0 ∈ Nil(A)

⇒ f − a0 ∈ Nil(A)

⇒ a1 + a2x+ · · ·+ anxn−1 ∈ Nil(A)

Continue the previous process, we obtain that a1, . . . , an are nilpotent.

(3) (⇐=):Obvious.(=⇒):If g = b0 + b1x+ · · ·+ bmx

m such that b0 6= 0 and fg = 0, then

1 RING THEORY REVISITED 3

a0b0 = 0

a0b1 + a1b0 = 0

......

t∑k=0

akbt−k = 0

......

an−1bm + anbm−1 = 0

anbm = 0

⇒

a1b20 = 0

a2b30 = 0

......

anbn+10 = 0

⇒ fbn+10 = a0b

n+10 + a1b

n+10 x+ · · ·+ anb

n+10 xn = 0

(4) Denote fg =∑m+n

i=0 cixi

(⇐=):If I = (a0, a1, . . . , an) 6= (1), then

(c0, . . . , cm+n) ⊆ (a0, . . . , an) $ (1)

⇒fg is not primitive, Absurd.

(=⇒):If I = (c0, . . . , cm+n) 6= (1), then we consider the maximal ideal M ⊇ I, andA/M[x], we have:

f g = 0 in (A/M)[x]

⇒∃ a 6= 0in A/M such that af = 0

⇒(A/M is a field)f = 0

⇒(a0, . . . , an) ⊆M 6= (1). Absurd.

3. We prove it by contradiction. Suppose there exists e 6= 0, 1 such that e2 = e.

• Denote the unique maximal ideal by M, then

∀ a /∈ A×, (a) $ A⇒ (a) ⊆M

⇒ArA× ⊆M

⇒ArA× = M is an ideal

• Let e′ := 1− e, then e′ 6= 0, 1 and (e′)2 = e′.

• e, e′ are nonunits and e+ e′ = 1

⇒ e, e′ ∈M while e+ e′ /∈M, contradiction!

2 ZARISKI TOPOLOGY 4

2 Zariski Topology

4. Suppose U ⊆ X is closed, then there exists I ▹ A,

Ar U = V (I) =⋂f∈I

V (f)

⇒U = Ar⋂f∈I

V (f) =⋃f∈I

Ar V (f) =⋃f∈I

Xf

Thus {Xf} form a basis of open sets s for the Zariski topology.

(1)Xf ∩Xg =(Ar V (f)) ∩ (Ar V (g))

=Ar (V (f) ∪ V (g))

=Ar V (fg)

=Xfg

(2)Xf = ∅⇔V (f) = X

⇔f ∈⋂

p∈Spec(A)

p = Nil(A)

⇔f is nilpotent.

(3)Xf = X ⇔V (f) = ∅

⇔f /∈⋃

p∈Spec(A)

p =⋃

M∈Max(A)

M = ArA×

⇔f is a unit.

3 Prime Avoidance

5. (1) DenoteA := k[x, y]/(x, y)2

= {0, 1, x, y, 1 + x, 1 + y, x+ y, 1 + x+ y}

Then(x, y) ={0, x, y, x+ y}

={0, x} ∪ {0, y} ∪ {0, x+ y}

to be the union of 3 properly small ideals.

4 LOCALIZATION OF RINGS AND MODULES 5

(2) Suppose a ∈ J is a homogeneous element of degree k.

• k is odd: each monomial must have y, so a ∈ I2.• If a monomial have y2, then it equals to 0 in k[x, y]/(xy, y2), so a ∈ I1.

Obvious y ∈ J r I1, x2 ∈ J r I2 and I2 is prime, because (k[x, y]/(xy, y2))/I2 ∼= k[x]

is a domain.

4 Localization of rings and modules

6. Suppose 0 /∈ T , then T is the multiplication subset, and

B[S−1] =

{b

f(s)| b ∈ B, s ∈ S

}=

{b

t| b ∈ B, t ∈ T

}as sets. Moreover, A[S−1] acts on B[S−1] and B[T−1] by the exactly same way:

A[S−1]×B[S−1] −→ B[S−1]

(a

s1,

b

f(s2)

)7−→ f(a)b

f(s1s2)

A[S−1]×B[T−1] −→ B[T−1]

(a

s1,b

t

)7−→ f(a)b

f(s1)t

thus B[S−1] and B[T−1] are isomorphic as A[S−1]-modules.

7. (1)⇒(2)⇒(3):Obviously.(3)⇒(1):Fix m ∈M , then

m

1=

0

1in MM ⇔ ∃ s ∈ ArM, sm = 0

⇒s ∈ Ann(m)⇒ Ann(m) * M for any M ∈Max(A)

⇒Ann(m) = R⇒ 0 = 1 ·m = m.

5 Nakayama’s lemma

8. N is a finite generated A-module. So suppose N is generated by n1, . . . , nk, then N/aN

is generated by n1, . . . , nk, suppose ni = u(mi), then ni = π ◦ u(mi).

6 RADICALS 6

So by Nakayama’s lemma,

n1, . . . , nk generate N/aN

⇒u(m1), . . . , u(mk) generateN

⇒u is surjective.

6 Radicals

1. Obviously Nil(A) ⊆ Rad(A).If Nil(A) $ Rad(A), Let x ∈ Rad(A)rNil(A), We obtain

Ax *√0 is an ideal of A

So there exists a ∈ A such that (ax)2 = ax 6= 0⇒ ax(1− ax) = 0

⇒ 1− ax is not invertible, thus lies is one maximal ideal M of A⇒ 1 = 1− ax+ ax ∈M, Contradiction!

7 Noetherian and Artinian rings

2. For any I ▹ A,√I is finitely generated, suppose

√I = 〈a1, . . . , an〉, and arii ∈ I, then

(√I)

∑ri ⊆ I.

3. First, we know

ϕ :M/(N1 ∩N2) −→M/N1 ×M/N2 m 7−→ (m, m)

is a monomorphism,so we can view M/(N1 ∩ N2) as a submodule of M/N1 ×M/N2.then

M/N1&M/N2 are Noetherian(Artinian)

⇒M/N1 ×M/N2 are Noetherian(Artinian)

⇒M/(N1 ∩N2) are Noetherian(Artinian)

8 Support of a Module

4. (1) We have a short exact sequence

0 −→M −→ A −→ k −→ 0

9 SUPPORT, ASSOCIATED PRIMES AND PRIMARY DECOMPOSITIONS 7

thus induce a exact sequence

MM −→M −→ k ⊗A M −→ 0

thus Mk = k ⊗A M ∼=M/MM .

Wrong[0 = k ⊗A M ⊗A N ⊗A k =Mk ⊗A Nk∼=Mk ⊗k Nk]

0 =M ⊗A (k ⊗k k)⊗A N ∼=Mk ⊗k Nk

⇒Mk = 0 or Nk = 0. Suppose Mk = 0

⇒M = MM

Then we use the Nakayama’s lemma, and we obtain

∃ a ∈M, such that (1 + a)M = 0 ⇒ M = 0

(2) We know thatMp ⊗Ap

Np = (Ap ⊗A M)⊗Ap(Ap ⊗A N)

=M ⊗A Ap ⊗A N

= Ap ⊗A M ⊗A N

= (M ⊗A N)p

p ∈ Supp(M ⊗A N)

⇔ 0 6= (M ⊗A N)p =Mp ⊗ApNp

⇔Mp 6= 0 & Np 6= 0

⇔ p ∈ Supp(M) ∩ Supp(N)

9 Support, Associated primes and Primarydecompositions

Facts.

• Spec(Z) = {(p) | p is prime } ∪ {(0)}

• Let p = (p),then

Zp =

{a

b

∣∣∣∣ p - b}and

(Z/nZ)p ∼= Zp ⊗Z Z/nZ =

0, p - n

Z/pα0Z, n = pα0pα11 · · · pαn

n

9 SUPPORT, ASSOCIATED PRIMES AND PRIMARY DECOMPOSITIONS 8

• Let p = (0),then

Zp =

{a

b

∣∣∣∣ b 6= 0

}= Q

and(Z/nZ)p ∼= Q⊗Z Z/nZ = 0

• localization preserves the exactness.

• localization commutes with (infinitely) direct sum.

• Spec(C[T ]) = {(T − a) | a ∈ C} ∪ {(0)}

1. (Another way: Use Supp(M) = V (annR(M)) when M is finitely generated)We know that

Mp∼=⊕n>0

(Z/nZ)p

= 0, p = (0)

6= 0, p = (p)

SoSupp(M) = Spec(Z)r {(0)}

is not a closed subset of Spec(Z).[If V (I) = Spec(Z)r {(0)} ⇒ (p) ⊇ I for any p ⇒ (0) ⊇ I]We also have

V (annZ(M)) = V ({0}) = Spec(Z) = Supp(M)

2. We knowϕ : Z −→

∞∏a=1

Z/pαZ n 7−→ (n, . . . , n, . . .)

is injective, so we have an exact sequence

0 −→ Z −→∞∏a=1

Z/pαZ

and then an exact sequence

0 −→ Zp −→

(∞∏a=1

Z/pαZ

)p

So Supp(

∞∏a=1

Z/pαZ)

= Spec(Z). While

∞⋃a=1

Supp(Z/pαZ) =∞⋃a=1

{(p)} = {(p)}

9 SUPPORT, ASSOCIATED PRIMES AND PRIMARY DECOMPOSITIONS 9

So∞⋃a=1

Supp(Z/pαZ) = {(p)} 6= Spec(Z) = Supp

(∞∏a=1

Z/pαZ

)

3. (1) Naturally⊕∞

a=0 p ∩Ra ⊆ p. The other direction is followed by this method:

∀p =k∑

i=0

pi ∈ p, pm =k∑

i=0

pim = 0

⇒ Use the double induction: first, shows pkm = 0,

to prove it needs the induction of degree of m replace m by pkm.

⇒ pim = 0 for any i ∈ {0, . . . , k}

⇒ pi ∈ annR(m)

⇒ p ⊆⊕∞

a=0p ∩Ra

(2)Suppose p = annR(m),m =∑k

i=0mi,we have

(p ∩Ra)m = 0

⇒ (p ∩Ra)mi = 0

⇒ pmi = 0 for any i ∈ {0, . . . , k}

⇒ p ⊆ annR(mi)

⇒ p ⊆ ∩ki=1annR(mi)

If for any i, p $ annR(mi), then there exists pi ∈ annR(mi)r p, so

p :=n∏

i=0

pi /∈ p

but pm = 0,contradiction!

4. We know (T − 1)e1 = (T − 2)e3 = 0, and dim(Ker(T − a)) = 0 when a /∈ {1, 2}. so

Ass(C3) = {(T − 1), (T − 2)}

5. We have(x3y, xy4) = (x) ∩ (y) ∩ (x3, y4)

They are all primary ideals.((x, y)4 ⊆ (x3, y4) ⊆ (x, y))Method:(x3y, xy4) = (x3, xy4) ∩ (y, xy4) = (x3, x) ∩ (x3, y4) ∩ (y, x) ∩ (y, y4)...

10 INTEGRAL DEPENDENCE, NULLSTELLENSATZ 10

10 Integral dependence, Nullstellensatz

1. Prove that the integral closure of R := C[X,Y ]/ (Y 2 −X2 −X3) in Frac(R) equals C[t]with t := Y /X, where X, Y denote the images of X,Y in R.

证明. Letϕ′ : C[X,Y ] −→ C[T ] X 7−→ T 2 − 1 Y 7−→ T 3 − T

We haveϕ′(Y 2 −X2 −X3) = (T 3 − T )2 − (T 2 − 1)2T 2 = 0

induces the mapϕ′ : C[X,Y ]/(Y 2 −X2 −X3) −→ C[T ]

Suppose f(X,Y ) = a0(X) + a1(X)Y ∈ Ker ϕ where

• a0(X) = 0 or a1(X) = 0, easy to know f(X,Y ) = 0.

• a0, a1 ∈ C[X] has no common nontrivial factors. Then

ϕ(f) =(a0(T

2 − 1))+(a1(T

2 − 1))(T 3 − T )⇒ X|a0, X|a1

Contradiction!

So ϕ is injective, thus we can view R := C[X,Y ]/ (Y 2 −X2 −X3) as a subring ofC[T ],and

XT − Y = 0

⇒ T is integral over R

⇒ R[T ] is integral over R

R[T ] = C[T 2 − 1, T 3 − T, T ] = C[T ] is a UFD, so normal.

2. Consider a Noetherian ring R with K := Frac(R). Show that y ∈ K is integral over R ifand only if there exists u ∈ R such that u 6= 0 and uyn ∈ R for all n.

证明. Suppose R is a domain.(=⇒): y ∈ K is integral over R

⇒ ∃am−1, . . . , a0 ∈ R, ym + am−1xm−1 + · · ·+ a0 = 0

If x = vw

, then there exists u := wm 6= 0, such that for any n ∈ N+,

uxn ∈ u(R+Rx+ · · ·+Rxm−1) ⊆ R+R · · ·+R = R

10 INTEGRAL DEPENDENCE, NULLSTELLENSATZ 11

(⇐=): R is a Noetherian ring

⇒ u−1R is a Noetherian R-module

⇒ R[x] is a finitely generated R-module

⇒ x is integral over R

3. Let R = Q [X1, X2, . . .] (finite or infinitely many variables). Show that nil(R) = rad(R) ={0}.

证明. Claim: For any f ∈ Q[X1, . . . , Xn], there exists (a1, . . . , an) ∈ Q, such thatf(a1, . . . , an) 6= 0.We can prove it by induction on n. Suppose it holds for k < n.WLOG, suppose f =

∑mi=0 gix

βin where gi ∈ Q[X1, . . . , Xn] and gm 6= 0.

Then by induction, there exists (a1, . . . , ak−1) such that gk(a1, . . . , ak−1) 6= 0,now

f(x) := f(a1, . . . , ak−1, x)

has at most m roots. We can choose ak ∈ Q such that f(ak) 6= 0 For any f ∈ R nonzero,there exists n ∈ N, such that f ∈ Q[X1, . . . , Xn].Let

ϕf : R −→ Q g −→ g(a1, . . . , ak, 0, . . .)

It is a surjective homomorphism.

⇒ R/Kerϕf∼= Q

⇒ Kerϕf is a maximal ideal not containingf

⇒ rad(R) ⊆ {0}

⇒ {0} ⊆ nil(R) ⊆ rad(R) ⊆ {0}

⇒ nil(R) = rad(R) = {0}

4. Let R = Q [X1, X2, . . .] (infinitely many variables). Show that R is not a Jacobson ring.

证明. Denote p = (X) in Q[X], R′ = (Q[X])p and an bijection map Ψ : N+ −→ Qr {0}We constrct

ψ : R −→ R′ Xi 7−→1

X −Ψ(i)

10 INTEGRAL DEPENDENCE, NULLSTELLENSATZ 12

which is surjective.Consider all the irreducible polynomials.If R is a Jacobson ring, then so is R′, but

MaxR′ = {p} 6= {p, (0)} = SpecR′

Contradiction!

5. (1) Let A be a subring of an integral domain B, and let C be the integral closure of Ain B. Let f, g be monic polynomials in B[x] such that fg ∈ C[x]. Then f, g are in C[x].

(2) Prove the same result without assuming that B (or A) is an integral domain.

证明. 偷懒抄书，此题不算！[2]Take a field (Frac(B)) containing B in which the polynomials f, g split into linearfactors; say f = Π(x− ξ1) , g = Π(x− ηj) . Each ξi andeach η, is a root of fg, hence isintegral over C. Hence the coefficients of f and g are integral over C.[1]LEMMA (14.7).−Let R ⊂ R′ be a ring extension, X a variable, f ∈ R[X] a monicpolynomial. Suppose f = gh with g, h ∈ R′[X] monic. Then the coefficients of g and h

are integral over R.Proof: Set R1 := R′[X]/〈g〉. Let x1 be the residue of X. Then 1, x1, x

21, . . . form a free

basis of R1 over R′ by (10.25) as g is monic; hence, R′ ⊂ R1 . Now,g (x1) = 0; sog factors as (X − x1) g1 with g1 ∈ R1[X] monic of degree 1 less than g. Repeat thisprocess, extending R1. Continuing, obtain g(X) =

∏(X − xi)and h(X) =

∏(X − yj)

with all xi and yj in an extension of R′. The xi and yi are integral over R as they areroots of f. But the coefficients of q and h are polynomials in the xi and yj ; so they tooare integral over R.

6. Let f : A → B be an injective map, with A Noetherian and B integral over A. Assumethat neither A nor B have zero divisors.(1) Show that if A is a field, then so is B.(2) Deduce that a field k is algebraically closed (i.e., every polynomial has a root) if andonly if for every finite field extension k ⊂ k′ i.e., k′ is f.d. as a k -vector space, we havek = k′.(3) Show that if B is a field, then so is A.

11 FLATNESS 13

证明. (1) For any x ∈ B, there exists a0, . . . , an−1 ∈ A, a0 6= 0, such that

xn + an−1xn−1 + · · ·+ a0 = 0

⇒ x · (− 1

a0)(xn−1 + an−1x

n−1 + · · ·+ a1) = 1

⇒ B is a field.

(2) (=⇒): For any x0 ∈ k′, there exists x1, . . . , xn ∈ k, such that

f(x) = (x− x1) · · · (x− xn) & f(x0) = 0

So x0 ∈ k. (⇐=):If not, there exists an irreducible polynomial f which has no root.Then k[t]/(f(t)) is a finite field extension of degree degf .

(3) For any x ∈ A, there exists y ∈ B such that xy = 1.

⇒ yn + bn−1yn−1 + · · ·+ b0 = 0

⇒ y + bn−1 + · · ·+ b0xn−1 = 0

⇒ y ∈ A

So k is a field.

11 Flatness

Facts.

• k[[t]] is a PID. Its ideal can be written as (tm), while for any f ∈ k[[t]], there existsm ∈ Z>0, g ∈ (k[[t]])∗, such that f = gtm.

• A beautiful diagram:

· · · TorA3 (A,M) TorA3 (K,M) TorA3 (K/A,M)

TorA2 (A,M) TorA2 (K,M) TorA2 (K/A,M)

TorA1 (A,M) TorA1 (K,M) TorA1 (K/A,M)

A⊗A M K ⊗A M K/A⊗A M 0

11 FLATNESS 14

1. For a field k, show that k[[t]][Y, Z]/(Y Z − t) is flat over k[[t]].

证明. We only need to prove k[[t]][Y, Z]/(Y Z − t) has no zero divisors except 0. This iseasy: We claim that every item in k[[t]][Y, Z]/(Y Z − t) can be uniquely written as theform

+∞∑n=0

(fn(Y ) + gn(Z) + an)tn

wherefn ∈ k[Y ], gn ∈ k[Z], an ∈ k, fn(0) = gn(0) = 0

If+∞∑n=0

(fn(Y ) + gn(Z) + an)tn 6= 0

then

tm+∞∑n=0

(fn(Y ) + gn(Z) + an)tn 6= 0

gtm+∞∑n=0

(fn(Y ) + gn(Z) + an)tn 6= 0 g ∈ (k[[t]])∗

2. Let N ′, N,N ′′ be A -modules, and 0→ N ′ → N → N ′′ → 0 be an exact sequence, withN ′′ flat. Prove that N ′ is flat ⇔ N is flat.

证明. If 0 → Mφ−→ M ′ is a A-modular exact sequence, then We have the exact se-

quences:

0 Kerϕ⊗ IdN ′ Kerϕ⊗ IdN 0 · · ·

Tor1(M,N ′′) M ⊗N ′ M ⊗N M ⊗N ′′ 0

Tor1(M′, N ′′) M ′ ⊗N ′ M ′ ⊗N M ′ ⊗N ′′ 0

Because Tor1(M,N ′′) = Tor1(M′, N ′′), the upper line is exact. So N ′ is flat⇔ N is flat.

Better solution:

0 = Tori+1(M,N ′′) −→ Tori(M,N) −→ Tori(M,N ′) −→ Tori(M,N ′′) = 0

11 FLATNESS 15

3. A ring A is absolutely flat if every A -module is flat. Prove that the following areequivalent:

(1) A is absolutely flat.

(2) Every principal ideal is idempotent.

(3) Every finitely generated ideal is a direct summand of A.

证明.

(1)⇒(2): Consider the exact sequence

0 −→ I −→ A −→ A/I −→ 0

Tensoring with I, we obtain the exact sequence (I ⊗A I = I2 because I is principalideal)

0 −→ I2 −→ I −→ I ⊗A A/I −→ 0

So I/I2 = I ⊗A A/I = 0.(I ⊗A A/I = 0 because I ⊗A A/I ↪→ A⊗A A/I ∼= A/I is azero map(using the fact that A/I is flat))Especially, when I = (x), then (x)2 = (x), there exists a ∈ A such that x = ax2,now (x) = (ax) is idempotent.

(2)⇒(3): We just need to prove that every finitely generated ideal is principal ideal.(Then,use the decomposition A = (a)⊕ (1− a))We only need to prove 〈a, b〉 = 〈a+ b − ab〉 when a2 = a, b2 = b.

(3)⇒(1): Suppose any I ▹ A is the direct summand, then A/I is flat, we obtain

Tor1(N,R/I) = 0 for any N, I

. So A is absolutely flat.

4. Prove the following properties of absolutely flat:

(1) Every homomorphic image of an absolutely flat ring is absolutely flat.

(2) If a local ring is absolutely flat, then it is a field.

(3) If a ring A is absolutely flat, then every non-unit in A is a zero-divisor.

12 GOING-UP AND GOING-DOWN 16

证明. (1)A is absolutely flat ⇒〈x〉2 = 〈x〉 in A

⇒〈x〉2 = 〈x〉 in A/I

⇒A/I is absolutely flat

(2) Suppose m is the unique maximal ideal of A, then A/m is a field. If there exists〈x〉 ( A, x 6= 0, then 〈x〉 ⊆ m ( A⇒ x = ax2 ⇒ x(1− ax) = 0.However, 1− ax /∈ m⇒ (ax− 1) ∈ A×, so we obtain x ∈ A×.

(3) If there exists 〈x〉 ( A, x 6= 0, then there exists e ∈ A such that e2 = e, (x) = (e).we get a ∈ A such that x = ae(x 6= a)⇒ x(x− a) = 0.So every non-unit in A is a zero-divisor.

12 Going-up and Going-down

5. Let f : A → B be an integral homomorphism of rings, i.e. B is integra over its subringf(A). Show that f# : Spec(B) → Spec(A) is a closed mapping, i.e. that it maps closedsets to closed sets.

证明. When f(B) 6= B, It may fail. Consider

f : Z =⇒ Z/3Z f#((0)) = (3)

is the counterexample.(Notice that Spec f(A) ∼= V (ker f) is closed set of SpecB)When A ⊆ B, Suppose V (I) ⊆ Spec(B) is the closed set (I / B), then we claim:f#(V (I)) = V (f−1I).

(a) For all p ∈ V (I)⇒ p ⊇ I ⇒ f#(p) = f−1(p) ⊇ f−1(I)⇒ f#(p) ∈ V (f−1(I))

(b) For all q ∈ V (f−1(I)) ⊆ Spec(A), there exists p ∈ Spec(B) such that p ∩ SpecA =

q. Easy to find that p ⊇ I.

6. Let A ⊂ B be an extension of rings, making B integral over A, and let p be a prime idealof A. Suppose there is a unique prime ideal q of B with q ∩A = p. Show that

12 GOING-UP AND GOING-DOWN 17

(a) qBp is the unique maximal ideal of Bp := B [(Ar p)−1]

(b) Bq = Bp

(c) Bq is integral over Ap

证明. (a) We have the following commutative diagram:

A B p q

Ap Bp pAp qBp

Bp is integral over Ap, and pAp is maximal in Ap

⇒ qBp is maximal in Bp

If there exists m ▹ Bp is maximal, then f#p (m) = pAp , thus m = qBp.

(b) p ⊆ q⇒ Bq ⊇ Bp.By the universal property, we only need to show

∀ x ∈ B r q, x is invertible.

If x1= q

b

a1where a1 ∈ A r p, there exists a2 ∈ A r p such that a1a2x = a2bq ⇒

x ∈ q.So x

1/∈ qBp is invertible.

(c) Bq = Bp is integral over Ap.

7. Let the integral extension A ⊂ B and the prime ideal p be as above. Suppose that A isa domain and q, q′ are distinct prime ideals of B, both mapping to p under Spec(B) →Spec(A). Show that Bq is not integral over Ap.

证明. Take y ∈ q′rq. If y−1 ∈ Bq is integral over Ap, then there exists a0, . . . an−1 ∈ Ap

with a0 6= 0 such that a0yn + a1yn−1 + · · · + an−1y = −1. Now there exists s ∈

Ar p, a′0, . . . a′n−1 ∈ A with

a′0yn + a′1y

n−1 + · · ·+ a′n−1y = −s

Now the left is in q′ while −s /∈ q′, contradiction!

Graded rings and modules, Filtrations

12 GOING-UP AND GOING-DOWN 18

1. Many basic operations on ideals, when applied to homogeneous ideals in Z -graded rings,lead to homogeneous ideals. Let I be a homogeneous ideal in a Z-graded ring R. Showthat:

(1) The radical of I is homogeneous, that is, the radical of I is generated by all thehomogeneous elements f such that fn ∈ I for some n.

(2) If I and J are homogeneous ideals of R, then (I : J) := {f ∈ R|fJ ⊂ I} is ahomogeneous ideal.

(3) Suppose that for all f, g homogeneous elements of R such that fg ∈ I one of f andg is in I. Show that I is prime.

证明. We need to show that:

Claim 12.1. For any r =∑

i∈Z ri ∈√I, ri ∈ Ri, we have ri ∈

√I

Consider i0 = min{i | ri 6= 0}, we know that

rni0 = (rn)ni0 ∈ I ⇒ ri0 ∈ Ri0 ∩√I

then consider r − ri0 . By induction, we can show that r ∈ 〈f ∈ Ri | fn ∈ I〉.

We need to show that (the difficult part):

Claim 12.2. For any f =∑

i∈Z fi ∈ (I : J), fi ∈ Ri, we have fi ∈ (I : J)

NowfJ ⊆ I ⇒ fg ⊆ I for any g ∈ Jj

⇒fig ⊆ Ii+j ⊆ I for any g ∈ Jj⇒fiJ ⊆ I

When f =∑

i∈Z fi /∈ I, g =∑

j∈Z gj /∈ I, we need to show that

fg =∑i,j∈Z

figj /∈ I.

Choosei0 = min{i | fi /∈ I} j0 = min{j | fj /∈ I}

then(fg)i0+j0 =

∑i

figi0+j0−i ∈ fi0gj0+I ⊆ A \ I

so fg /∈ I.

12 GOING-UP AND GOING-DOWN 19

⇒

2. Suppose R is a Z -graded ring and 0 6= f ∈ R1

(1) Show that R [f−1] is again a Z-graded ring.

(2) Let S = R [f−1]0 , show that S ∼= R/(f − 1), and R [f−1] ∼= S [x, x−1] where x is anew variable.

证明. (1) R[f−1] is a ring which is graded by

(R[f−1])i =

⟨ai+j

f j

⟩ai+j∈Ri+j

.

(2) We know

(R[f−1])0 =

⟨ajf j

⟩aj∈Rj

.

and we have the surjective ring homomorphism

R −→ (R[f−1])0 ri ∈ Ri 7−→rif i

and the kernel of which is (f − 1).Now the homomorphism

R[f−1

]−→ S

[x, x−1

] ri+j

f j∈ (R[f−1])i 7−→

ri+j

f i+jxi

is an isomorphism.

3. Show that if R is a graded ring with no nonzero homogeneous prime ideals, then R0 is afield and either R = R0 or R = R0 [x, x

−1] .

证明. We first state a lemma which can be proved using Zorn’s lemma:∗

Lemma 12.3. Let I be a homogeneous ideal of a graded ring R, I 6= R, then there existsa homogeneous prime ideal which contains I.

Using the lemma to Ra, we get:∗for details, see: https://math.stackexchange.com/questions/385292/homogeneous-ideals-are-contained-in-homogeneous-

prime-ideals

13 COMPLETIONS 20

Corollary 12.4. If R is a graded ring with no nonzero homogeneous prime ideals, thenany homogeneous item a ∈ R \ {0} is invertible.

Now R0 is a field. if R = R0, then everything was done; Otherwise, Suppose i0 = {i ∈Z>0 | Ri 6= 0} there exists x ∈ Ri0 \ {0}, which is invertible.Now:

• If i0 - r, then Rr = 0 by the euclidean division;

• If i0 | r, a ∈ Rr, then a = ax−r/i0 · x−r/i0 ∈ R0 [x, x−1].

So we’re done.Another point: you can take the homogeneous items from prime ideal p to construct ahomogeneous prime ideal.

4. Taking the associated graded ring can also simplify some features of the structure of R.For example, let k be a field, and let R = k [x1, . . . , xr] ⊂ R1 = k [[x1, . . . , xr]] be therings of polynomials in r variables and formal power series in r variables over k, andwrite I = (x1, . . . , xr) , I

′ for the ideal generated by the variables in either ring. Showthat grI R = grI′ R1

证明. We know thatIk/Ik+1 ∼= I ′k/I ′k+1

sogrI R =

⊕k∈Z

Ik/Ik+1 ∼=⊕k∈Z

I ′k/I ′k+1 = grI′ R1

13 Completions

1. Let A be a local ring, m its maximal ideal. Assume that A is m -adically complete. Forany polynomial f(x) ∈ A[x], let f(x) ∈ (A/m)[x] denote its reduction mod. m. ProveHensel’s lemma: if f(x) is monic of degree n and if there exist coprime monic polynomialsg(x), h(x) ∈ (A/m)[x] of degrees r, n−r with f(x) = g(x)h(x), then we can lift g(x), h(x)back to monic polynomials g(x), h(x) ∈ A[x] such that f(x) = g(x)h(x)

证明. See [Hensel’s Lemma, Theorem 1]† Using induction makes sense.†html:http://therisingsea.org/notes/HenselsLemma.pdf

13 COMPLETIONS 21

2. (a) With the notation of Exercise 1, deduce from Hensel’s lemma that if f(x) has asimple root α ∈ A/M, then f(x) has a simple root a ∈ A such that α = a mod M

(b) Show that 2 is a square in the ring of 7 -adic integers.

(c) Let f(x, y) ∈ k[x, y], where k is a field, and assume that f(0, y) has y = a0 as asimple root. Prove that there exists a formal power series y(x) =

∑∞n=0 anx

n suchthat f(x, y(x)) = 0. This gives the ”analytic branch ” of the curve f = 0 throughthe point (0, a0) .

证明. (a) Suppose f = (x − α)h(x), then by the Hensel’s lemma, there exists g(x) ∈A[x], h(x) ∈ A[x] such that

f(x) = g(x)h(x)

A[x] −→ A/M[x] g 7−→ (x− α) h 7−→ h

As a consequence, there exists a ∈ A, g(x) = x − a, a ≡ α mod M (the root isevidently simple.)

(b) x2 − 2 has a simple root 5 ∈ Z/7Z. By (a), there exists a ∈ A to be a simple rootof x2 − 2 (in Z(7)).

(c) Using (a), let A = k[[x]],M = (x), we get

Corollary 13.1. If f(0, y) ∈ k[y] has a simple root a0, then f(x, y) ∈ k[[x]][y] hasa simple root y(x) =

∞∑n=0

bnxn ∈ k[[x]] such that

f(x, y(x)) = 0 b0 = a0

3. Let A be a Noetherian ring, a an ideal in A, and A the a-adic completion. For anyx ∈ A, let x be the image of x in A. Show that x not a zero-divisor in A implies x not azero-divisor in A. Does this imply that if A is an integral domain then A is an integraldomain?

证明. suppose x is not a zero-divisor in A, then

0 −→ A×x−→ A

⇒0 −→ A×x−→ A

⇒0 is not a zero-divisor in A

14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS22

To take the example where A is an integral domain but A is not an integral domain, wetake

A = Q[x, y]/(y2 − x2 − x3)

which is a domain because (y2 − x2 − x3) is irreducible in Q[x, y],but

A = Q[[x, y]]/(y2 − x2 − x3)

is not a domain because

[y − x(1 + 1

2x− 1

8x2 · · · )][y + x(1 +

1

2x− 1

8x2 · · · )] = 0 in A

4. Let k be a field and consider the quotient of infinite polynomial ring

R :=k [X,Z, Y1, Y2, Y3, . . .]

(X − ZY1, X − Z2Y2, X − Z3Y3, . . .)

Denote by Z the image of Z in R. Show that the ideal I := (Z) of R satisfies⋂

n≥1 In 6=

{0}. Why is this consistent with Krull’s intersection theorem?

证明. For convinience, we omit the bar.

X = ZnYn ∈ In ⇒ X ∈⋂n≥1

In

R is not Noetherian because

(Y1) ( (Y1, Y2) ( (Y1, Y2, Y3) ( · · ·

is a infinite ascending chain in R.

14 Mittag-Leffler systems and Completions fornon-Noetherian rings

1. Consider an inverse system of sets · · · ←− Anφη+1←− An+1 ←− · · · (where n = 1, 2, . . . .

For each j > i, let ϕj,i : Aj −→ Ai be the composition of ϕj , . . . , ϕi. We say thatMittag-Leffler conditions holds for (An, ϕn)n≥1 if for each i, we have

ϕk,i (Ak) = ϕj,i (Aj) whenever j, k � i

14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS23

Show that if (An, ϕn)n is Mittag-Leffler and An 6= ∅ for each n, then the limit

lim←−n

An :=

{(an)n ∈

∏n

An : ∀n, ϕn+1 (an+1) = an

}

is nonempty as well.

证明. Let Dn = ∩∞k=n+1ϕk,n(Ak) be a set which is nonempty (because

ϕn+1,n(An+1) ⊇ ϕn+1,n(An+1) ⊇ · · · ⊇ · · ·

is stable and nonempty). We have the inverse system

· · · ←− Dnφη+1←− Dn+1 ←− · · ·

and each ϕn is surjective. So lim←−nAn is nonempty.(Find the PREIMAGE).

Remark 14.1. Moreover,

ϕD :∏

Dn −→∏

Dn (an)n∈Z 7−→ (an − ϕn+1(an+1))n∈Z

is surjective. Thenlim←−

1Dn := coker ϕD = 0

2. Suppose we are given an inverse system of short exact sequences of abelian groups, i.e.a commutative diagram with exact rows, where n = 1, 2, . . . Show that if (An, ϕn)n isMittag- Leffler, then

0 −→ lim←−Anlim fn−→ lim←−Bn

lim gn−→ lim←−Cn −→ 0

is exact. You only have to show the surjectivity of lim gn.

证明. We have the exact sequence:

0∏

nAn

∏nBn

∏nCn 0

0∏

nAn

∏nBn

∏nCn 0

θ′ θ θ′′

whereθ′ :

∏n

An −→∏n

An (an)n∈Z 7−→ (an − ϕn+1(an+1))n∈Z

14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS24

So by the Snake lemma, we have the exact sequence:

0 −→ lim←−Anlim fn−→ lim←−Bn

lim gn−→ lim←−Cn −→ lim←−1An

We only need to prove that θ′ is surjective. Define Dn as in Problem 1, we get the exactsequence

lim←−Dn −→ lim←−An −→ lim←−An/Dn

from

0∏

nDn

∏nAn

∏nAn/Dn 0

0∏

nDn

∏nAn

∏nAn/Dn 0

τ ′ θ′ τ ′′

By 14.1, lim←−Dn = 0; we will show that lim←−An/Dn = 0, thus lim←−An = 0, thus the proofends.

Remark 14.2. The Mittag-Leffler conditions tell us that

∀ n ∈ Z,∃ m > n s.t. ϕm,n(Am) = Dn

⇒ϕm,n : Am/Dm −→ An/Dn is the zero map.

So lim←−An/Dn = 0 (ONLY ZERO SOLUTION)

3. Let R be a ring (not necessarily Noetherian), I be a proper ideal, and ϕ : M −→ N bea homomorphism of R -modules. Prove the following statements.

(a) If M/IM → N/IN is surjective, then so is ϕ : M → N . Here M = limn≥1M/InM

stands for the I-adic completion.(Hint: First, show M/InM → N/InN is surjective by Nakayama’s lemma. Next,set Kn = Ker [M → N/InN ] to get exact sequences 0

→ Kn/InM →M/InM → N/InN → 0

then try to establish the surjectivity of Kn+1/In+1M → Kn/I

nM in order to applyMittag-Leffler.)

(b) If 0→ K → M → N → 0 is an exact sequence of R -modules and N is flat, then0→ K → M → N → 0 is exact.

(c) If M is finitely generated, then the natural homomorphism M ⊗R R → M given bym⊗ (rn)

∞n=1 7→ (rnm)

∞n=1 is surjective.

14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS25

证明. (a) We’d like to point out the commutative diagram

0 ∗ Kerfn+1 Kerfn

0 InM/In+1M M/In+1M M/InM 0

0 InN/In+1N N/In+1N N/InN 0

0 coker fn+1 coker fn 0

fn+1 fn

Then by induction we know the two surjectivity in the Hint.

(b) We have

0 K/In+1K M/In+1M N/In+1N 0

0 K/InK M/InM N/InN 0

φ

and ϕ is surjective, so 0→ K → M → N → 0 is exact.

(c) Just see the commutative diagram

R⊕N M ⊗R R

M

∃ surj

4. Suppose I is finitely generated. Let M be an R -module. Prove that

InM = Ker[M →M/InM

]= InM

for all n ∈ Z≥1, and M is I -adically complete as an R -module.

证明. (Hint: Fix n and take generators f1, . . . , fr of In. This yields a surjective homo-morphism of R -modules (f1, . . . , fr) :M

⊕r → InM ⊂M Pass to completions and showthat

(f1, . . . , fr)∧: M⊕r → InM = lim

m≥n

InM

ImM' Ker

[M →M/InM

]⊂ M

15 DIMENSION THEORY 26

which is surjective by the previous exercise. The image of (f1, . . . , fr) : M⊕r → M isboth InM and InM to infer that

M/InM ∼= M/InM ∼=M/InM

then we have M∧∧ ∼=M∧, which shows that M∧ is complete.

15 dimension theory

1. Let k be a field and R = k [X0, . . . , Xn] , graded by total degree. Consider the graded R-module S = R/(f) where f is a homogeneous poly-nomial of total degree d ≥ 1. Showthat when m ≥ d,

χ(S,m) := dimk Sm =

(m+ n

n

)−

(m+ n− d

n

)

证明. We have the SES of the graded k-linear spaces:

0 −→ (f) −→ R −→ S −→ 0

Thus

dimk Sm = dimk Rm − dimk(f)m =

(m+ n

n

)−

(m+ n− d

n

)

2. Let R =⊕

nRn be a Z≥0-graded ring, finitely generated over R0. Assume R0 is Artinian(for example, a field) and let M = ⊕nMn be a finitely generated Z≥0−gradedR -module.Define the Hilbert series in the variable T as

HM (T ) :=∑m≥0

χ(M,m)Tm ∈ Z[[T ]]

where χ(M,m) denotes the length of the R0 -module Mm, as usual. In what follows,graded means graded by Z≥0.

(a) Show that if 0→ M ′ → M → M ′′ → 0 is a short exact sequence of graded R

-modules, then HM (T ) = HM ′(T ) +HM ′′(T ).

(b) Relate HM (T ) and HM(k)(T ) for arbitrary k ∈ Z, where M(k)d := Md+k.

15 DIMENSION THEORY 27

(c) Suppose that R is generated as an R0 -algebra by homogeneous elements x1, . . . , xnwith di := degxi ≥ 1. Show that there exists Q ∈ Q[T ] such that

HM (T ) =Q(T )

(1− T d1) · · · (1− T dn)

as elements of Q[T ].(Hint: you may imitate the arguments for the quasi-polynomiality of χ(M,n).)

(d) What can be said about the Zm≥0 -graded case, for general m?

证明. (a) We just need to prove

l(Mn) = l(M ′n) + l(M ′′

n )

this is true even if R0 is not Artinian. ‡

(b)T kHM(k)(T ) =

∑m≥0

χ(M(k),m)Tm+k

=∑m≥0

χ(M,m+ k)Tm+k

=HM (T )−∑

06n<k

χ(M,n)Tn

(c) This is followed by [2]. We prove by induction on n. When n = 0, R = R0 and M isf.g., so Mn = 0 for n >> 0; now suppose n > 0 and the theorem was true for n− 1,we consider the ES induced by the map ×xn:

0 −→ Km −→Mm −→Mm+dn−→ Lm+dn

We haveHK(T )−HM (T ) = HM(dn)(T )−HL(dn)(T )

then easy to find the required form.

(d) Suppose that R is generated as an R0 -algebra by homogeneous elements x1, . . . , xnwith di := degxi 6= (0, 0, . . . , 0). Then there exists Q ∈ Q[T1, . . . , Tm] such that(T := (T1, . . . , Tm))

HM (T ) =Q(T )

(1− T d1) · · · (1− T dn)

as elements of Q[T1, . . . , Tm].

‡see https://math.stackexchange.com/questions/145564.

15 DIMENSION THEORY 28

1. Let Z3 be the 3 -adic completion of the ring Z, so that Z ↪→ Z3 naturally. Evaluate1 + 3 + 32 + 33 + · · · in Z3.

证明.1 + 3 + 32 + 33 + · · · = 1

1− 3= −1

2

2. Let R be a Noetherian local ring. Suppose that there exists a principal prime ideal p inR such that ht (p) ≥ 1. Prove that R is an integral domain.(Hint: Below is one possible approach. Suppose p = (x) for some x ∈ R. Let q ⊂ p bea minimal prime in R. Argue that (i)x /∈ q, (ii) q = xq, and finally (iii) q = {0}.

证明. (i) Otherwise,x ∈ q ⇒ q = p ⇒ ht(p) = 0

(ii) xq ⊆ q is a prime ideal because

xf /∈ xq, xg /∈ xq =⇒ xfxg /∈ xq

Just verify q ⊆ xq is easier and quicker.

(iii) Nakayama lemma applied to (ii).

3. Let k be a field and R = k[[X]] × k[[X]]. Prove that R is a Noetherian semi-local ring,R contains a principal prime ideal of height 1, but R is not an integral domain.(Hint: It is known that k[[X]] is Noetherian local with maximal ideal (X). Argue thatthe ideals in k[[X]] × k[[X]] take the form I × J where I, J are ideals in k[[X]]. Showthat (X)× k[[X]] and k[[X]]× (X) are the only maximal ideals, and both are of height1. )

证明. (a) R is not an integral domain.

(b) k[[X]] is a Dedekind domain, thus Noetherian local with maximal ideal (X).

(c) We know that I = π1(I) × π2(I), where π1(I), π2(I) ▹ k[[X]] and (π1(a), π2(b)) =

(1, 0)a+ (0, 1)b.Obviously(by contradiction) the maximal ideals are (X)×k[[X]] and k[[X]]× (X);

参考文献 29

thus R is semilocal; moreover,(X)× k[[X]] = ((X, 1)) is principal.the prime ideals are

(0)× k[[X]], k[[X]]× (0), (X)× k[[X]], k[[X]]× (X)

So the height of (X)× k[[X]] is 1.

参考文献

[1] Allen Altman and Steven Kleiman. A term of commutative algebra. Worldwide Center ofMathematics, 2013.

[2] Michael Atiyah. Introduction to commutative algebra. CRC Press, 2018.