Solutions to Atiyah-Macdonald, Chapter 6

Dave Karpuk

May 19, 2010

Exercise 1. (i) Let M be a Noetherian A-module and u : M → M a module homomorphism. Ifu is surjective, then u is an isomorphism.

(ii) If M is Artinian and u is injective, then again u is an isomorphism.

Proof. (i) The chain. . . ker(un) ⊆ ker(un+1) ⊆ ker(un+2) . . .

of submodules of M must terminate by the Noetherian assumption. So pick n such that ker(un) =ker(un+1). Let m ∈ ker(u). The maps uk are all surjective by the surjectivity of u. Pick x ∈ Msuch that un(x) = m. Applying u, we get the equation

un+1(x) = u(m) = 0

which says that x ∈ ker(un+1) = ker(un). This implies that m = 0, and hence u is injective.

(ii) The chain. . . im(un) ⊇ im(un+1) ⊇ im(un+2) . . .

of submodules must terminate by the Artinian condition. So Pick n such that im(un) = im(un+1).Let m ∈ M . We can find an x ∈ M such that un(m) = un+1(x). Then un(m − u(x)) = 0, but un

is injective because u is injective. Hence m = u(x), and it follows that u is surjective and thus anisomorphism.

Exercise 2. Let M be an A-module. If every non-empty set of finitely generated submodules ofM has a maximal element, then M is Noetherian.

Proof. Let N be a submodule of M , and let S = {Ni} be the set of all finitely generated submodulesof N . By assumption, S has a maximal element, call it N ′. Then for any Ni ∈ S, we have Ni ⊆ N ′.If this were not the case, then we could pick x ∈ Ni, x 6∈ N ′. The submodule N ′ + Ax wouldbe a finitely generated submodule properly containing N ′, contradicting the maximality of N ′.Therefore,

N =⋃i

Ni ⊆ N ′ ⊆ N

which says that N = N ′. We conclude that every submodule of M is finitely generated, which saysthat M is Noetherian.

Exercise 3. Let M be an A-module and let N1, N2 be submodules of M . If M/N1 and M/N2 areNoetherian, so is M/(N1 ∩N2). Similarly with Artinian in place of Noetherian.

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Proof. We may replace M/(N1 ∩ N2) with M and assume that N1 ∩ N2 = 0. So we must provethat for submodules N1, N2 of M such that N1 ∩N2 = 0, M/N1 and M/N2 Noetherian implies MNoetherian. We have an exact sequence

0 → N1 → M → M/N1 → 0

of A-modules. Because N1 ∩N2 = 0, a copy of N1 sits inside M/N2. Thus we have another exactsequence

0 → N1 → M/N2 → M/(N1 + N2) → 0

which by Proposition 6.3(i) proves that N1 is Noetherian. Again using Proposition 6.3(i) andthe first exact sequence, we see that M is Noetherian. The proof for Artinian is the same, usingProposition 6.3(ii).

Exercise 4. Let M be a Noetherian A-module and let a be the annihilator of M in A. Prove thatA/a is a Noetherian ring.

If we replace “Noetherian” by “Artinian” in this result, is it still true?

Proof. Because M is Noetherian, it is finitely generated, say by m1, . . . ,mn. Let ai = Ann(mi).By the in-text Exercise 2.2, we have a =

⋂ni=1 ai. We have an inclusion A/ai ↪→ M of A-modules,

induced by the map 1 7→ mi. Thus A/ai is a Noetherian A-module for all i. By Exercise 3 andinduction, we see that A/a is a Noetherian A-module and thus a Noetherian A/a-module, whichproves that it is a Noetherian ring.

To see that the result fails if we replace Noetherian with Artinian, consider A = Z and M =Qp/Zp. The Z-module M is Artinian by the argument of Example (3) in the text, but its annihilatoris the zero ideal of Z. As Z = Z/(0) is not an Artinian ring, we see that the result fails.

Exercise 5. A topological space X is said to be Noetherian if the open subsets of X satisfythe ascending chain condition (or, equivalently, the maximal condition). Since closed subsets arecomplements of open subsets, it comes to the same thing to say that the closed subsets of Xsatisfy the descending chain condition (or, equivalently, the minimal condition). Show that, if Xis Noetherian, then every subspace of X is Noetherian, and that X is quasi-compact.

Proof. Let Y ⊆ X be a subspace, and suppose that (Un)n≥0 is a chain of open subsets of Y . ThenUn = Vn ∩ Y for some open Vn ⊆ X. We are free to replace Vn with Vn−1 ∪ Vn, and so we mayassume that Vn ⊆ Vn+1 for all n. Because X is Noetherian, there exists N such that Vn = Vn+1

for all n ≥ N . Intersecting with Y , we see that the chain (Un) is eventually stationary, thus Y isNoetherian.

To see that X is quasi-compact, let X =⋃

i Ui be an open cover. Let Σ be the collection ofopen subsets of X which are finite unions of the Ui’s. The set Σ is partially ordered with respectto inclusion, and because X is Noetherian it has a maximal element, say U = U1 ∪ · · · ∪ Un.

If U 6= X, pick x ∈ X − U . Because the Ui’s cover X, there exists Ui such that x ∈ Ui. ThenU is properly contained in U ∪ Ui, contradicting the maximality of U . Therefore U = X and X isquasi-compact.

Exercise 6. Prove that the following are equivalent:(i) X is Noetherian.(ii) Every open subspace of X is quasi-compact.(iii) Every subspace of X is quasi-compact.

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Proof. (i) ⇒ (iii) If Y is a subspace of X, then Y is Noetherian by the previous exercise, and thusquasi-compact (also by the previous exercise).

(iii) ⇒ (ii) Obvious.

(ii) ⇒ (i) Let (Un)n≥0 be a chain of open subsets of X, and let U =⋃

i Ui. The Ui form anopen cover of the quasi-compact open set U , thus there is a finite subcover. Hence

⋃i Ui =

⋃Ni=0 Ui

for some N and thus the chain is eventually stationary, proving that X is Noetherian.

Exercise 7. A Noetherian space is a finite union of irreducible closed subsets. Hence the set ofirreducible components of a Noetherian space is finite.

Proof. Let Σ be the set of non-empty closed subsets of X which are not finite unions of irreducibleclosed subspaces. Suppose that Σ 6= ∅. Because X is Noetherian, Σ has a minimal element C.Then C itself is not irreducible, hence C = D1 ∪D2 for some proper non-empty closed subsets Di

of C. By minimality of C, we can express each Di as a finite union of irreducible subspaces of X,but then C is a finite union of irreducible subspaces of X, which is a contradiction.

Since Σ = ∅, we have that X itself is a finite union of irreducible closed subsets. Each ofthese irreducible closed subsets is contained in a unique irreducible component of X. Conversely,every irreducible component contains one of these irreducible closed subsets, thus the number ofirreducible components of X is finite.

Exercise 8. If A is a Noetherian ring then Spec(A) is a Noetherian topological space. Is theconverse true?

Proof. Let V (a0) ⊇ V (a1) ⊇ · · · be a descreasing chain of closed subsets of Spec(A). Replacingeach ai with its radical does not change the sets V (ai), thus we may assume that all of the ai areradical ideals. Thus they induce a sequence a0 ⊆ a1 ⊆ · · · of ideals of A which must terminatebecause A is Noetherian. Thus an = an+1 for sufficiently large n and because the ai are radical,this implies that V (an) = V (an+1) for sufficiently large n. Therefore Spec(A) is Noetherian.

The converse is false. Let k be a field and consider the ring A = k[x1, x2, . . .]/(x21, x

22, . . .). Let p

be a prime ideal in this ring. Then p corresponds to some prime ideal q of the ring k[x1, x2, . . .] whichcontains x2

i for all i. Because q is prime, we must have xi ∈ q for all i and thus q = (x1, x2, . . .).Therefore p is the prime ideal of A generated by the images of all of the xi. The quotient A/p = k isa field and thus p is maximal. Therefore p is the only prime ideal of A and Spec(A) is the one-pointtopological space which is clearly Noetherian.

However, A is not a Noetherian ring. It contains the strictly increasing chain (x1) ( (x1, x2) ((x1, x2, x3) ( · · · .

Exercise 9. Deduce from Exercise 8 that the set of minimal prime ideals in a Noetherian ring isfinite.

Proof. By Exercise 8, Spec(A) is a Noetherian topological space. By Exercise 7, Spec(A) is thusa union of finitely many irreducible components. By Chapter 1, Exercise 20(iv), the irreduciblecomponents of Spec(A) correspond bijectively to the minimal prime ideals of A. Thus there areonly finitely many minimal prime ideals of A.

Exercise 10. If M is a Noetherian module (over an arbitrary ring A) then Supp(M) is a closedNoetherian subspace of Spec(A).

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Proof. By Chapter 3, Exercise 19, we have Supp(M) = V (Ann(M)) = Spec(A/Ann(M)). ByExercise 4, A/Ann(M) is a Noetherian ring and so by Exercise 8 Spec(A/Ann(M)) is a Noetheriantopological space.

Exercise 11. Let f : A → B be a ring homomorphism and suppose that Spec(B) is a Noetherianspace (Exercise 5). Prove that f∗ : Spec(B) → Spec(A) is a closed mapping if and only if f hasthe going-up property.

Proof. The fact that f∗ closed implies that f has the going-up property is contained in Exercise10 of Chapter 5. To prove the converse, we must prove that f∗(Spec(B/b)) is closed in Spec(A)for all b ⊂ B. Replacing B by B/b (whose spectrum is still Noetherian) and f by the compositeA → B → B/b (which still has the going-up property), we are reduced to proving that f∗(Spec(B))is closed in Spec(A).

As Spec(B) is Noetherian, we can write Spec(B) =⋃n

i=1 Spec(B/qi) as the finite union ofirreducible closed subspaces Spec(B/qi). So it suffices to prove that f∗(Spec(B/qi)) is closedfor all i. That is, we may assume that B is a domain and that Spec(B) is irreducible. Thusf∗(Spec(B)) = f∗(V (0)), so by Chapter 1, Exercise 21(iii) we have

f∗(Spec(B)) = V (ker(f)).

As B is a domain, we have ker(f) = p for some prime p of A. To prove f∗(Spec(B)) =f∗(Spec(B)), we must prove that every prime ideal containing p is contracted. But if p ⊆ p′,then there exists a prime q′ of B such that q′c = p′ by the fact that f has the going-up property.Therefore f∗(Spec(B)) is equal to its closure and hence closed.

Exercise 12. Let A be a ring such that Spec(A) is a Noetherian space. Show that the set of primeideals of A satisfies the ascending chain condition. Is the converse true?

Proof. Let p1 ⊆ p2 ⊆ · · · be a chain of prime ideals of A. Applying V (−) we obtain a descendingchain V (p1) ⊇ V (p2) ⊇ · · · of closed subsets of Spec(A), which must terminate because Spec(A) isNoetherian by assumption. Thus there exists n such that V (pn) = V (pn+1) = · · · , but for primeideals we have V (pn) = V (pn+1) if and only if pn = pn+1. Therefore the chain of prime idealsterminates.

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