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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 6.

## Chain Conditions

Exercise 1.

 1 Let be a Noetherian -module and a module homomorphism. If is surjective, then is an isomorphism. 2 If is Artinian and is injective, then again is an isomorphism.

Proof.

 1 Let , then ; and . Since is surjective, . But is Noetherian, so for some . Therefore ; repeating the progress we have , i.e. . Hence is an isomorphism. 2 Let , then ; and . Since is injective, . is Noetherian, so for some . Hence ; so , in particular, , and is an isomorphism.

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

Proof. Assume is not Noetherian, then there exists a strictly ascending chain of submodules . Choose arbitrary, and let be the submodules generated by . Then , hence . So there exists a strictly ascending chain of finitely generated submodules , contradiction. Hence is Noetherian.

Exercise 3. Let be an -module and let be submodules of . If and are Noetherian, so is . Similarly with Artinian in place of Noetherian.

Proof. Passing to the quotient , one may assume . Therefore . If is Noetherian (respectively, Artinian), then so is . Hence if and are both Noetherian (respectively, Artinian), then so is .

Exercise 4. Let be a Noetherian -module and let be the annihilator of in . Prove that is a Noetherian ring.

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

Proof. Since is Noetherian, it must be finitely generated. Let be its generator, and . Then as -module. Hence is Noetherian. Since , by the previous Exercise, is also Noetherian.

If we replace “Noetherian” by “Artinian”, then the result does not hold. For example, let and , then (since ). is Artin, but isn’t.

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

Proof. If is a subspace of , and be an increasing sequence of open subsets of ; then by definition, there exists open subsets such that . Assume (consider ), then the chain is stationary, so is . Hence is Noetherian.

Suppose is an open cover of . Let be all finite union of , i.e. where is a finite subset of . Since open subsets of satisfy the maximal condition, has a maximal element, say . If , then there exists such that , contradiction. Therefore and there exists a finite subcover. Hence is quasi-compact.

Exercise 6. Prove that the following are equivalent:

 1 is Noetherian. 2 Every open subspace of is quasi-compact. 3 Every subspace of is quasi-compact.

Proof.

 i) iii): By the last Exercise, every subspace is also Noetherian, hence quasi-compact; iii) ii): trivial. ii) i): Let be a increasing sequence of open subsets of . Let , then is an open subspace. Therefore is quasi-compact, so the open cover has a finite subcover. So for some , and . Hence is Noetherian.

Exercise 7. A Noetherian space is a finite union of irreducible closed subspaces.

Proof. Let be the set of closed subsets of , which are not finite unions of irreducible closed subspaces.

If is not empty, then it has a minimal element (since is Noetherian), say . By definition itself is not irreducible, hence there exists proper closed subsets and of such that . Since is minimal, it follows , i.e. and can be written as finite union of irreducible closed subspaces. So , contradiction.

Hence . In particular, is a finite union of irreducible closed subspaces.

Hence the set of irreducible components of a Noetherian space is finite.

Exercise 8. If is a Noetherian ring then is a Noetherian topological space. Is the converse true?

Proof. Suppose be an decreasing sequence of closed subsets of . Let . Then , and . Since is Noetherian, there exists such that . So , and is Noetherian.

The converse is not true. Let be an arbitrary field, and . Then the unique prime ideal of is the ideal generated by . Hence is a singleton, clearly Noetherian. But for every , there is an ideal generated by , and . So is not Noetherian.

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

Proof. For every closed subset, suppose (where is a radical ideal). Then is not irreducible iff. there exists proper subset such that iff. there exists radical ideals such that iff. is not a prime ideal.

Hence prime ideal of corresponds to irreducible closed subspace of ; and minimal prime ideal of corresponds to irreducible components of . If is Noetherian, then by Exercise 8, is Noetherian. So has finite irreducible components by Exercise 7. Hence has finite minimal prime ideals.

Exercise 10. If is a Noetherian module (over an arbitrary ring ) then (Chapter 3, Exercise 19) is a closed Noetherian subspace of .

Proof. Since is Noetherian, it’s finitely generated. Hence closed. By Exercise 4, is Noetherian, so is Noetherian.

Exercise 11. Let be a ring homomorphism and suppose that is a Noetherian space. Prove that is a closed mapping if and only if has the going-up property (Chapter 5, Exercise 10).

Proof. By Chapter 5, Exericise 10, is a closed mapping has the going-up property.

Suppose has the going-up propery, and be a closed subset. We should show that is closed. Since has the going-up property, the result holds when where is a prime ideal (Chapter 5, Exercise 10), i.e. when is irreducible. In the general case, is Noetherian, so is . By Exercise 7, is a finite union of irreducible closed subspaces. Hence is also closed.

Exercise 12. Let be a ring such that is a Noetherian space. Show that the set of prime ideals of satisfies the ascending chain condition. Is the converse true?

Proof. Clearly is a Noetherian space if and only if the set of radical ideals of satisfies the ascending chain condition. In particular, the set of prime ideals satisfies a.c.c.

The converse is not true. Let , then is Boolean, so every prime ideal is maximal (Chapter 1, Exercise 11). In particular, prime ideals of satisfy a.c.c. But for every , let Then is a radical ideal of (In every element is idempotent, hence every ideal is a radical ideal). Therefore is a strictly increasing sequence of radical ideals. So is not Noetherian.