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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 2.

## Modules

Exercise 1. Show that , if coprime.

Proof. If coprime then is a unit in , so

Hence , ’cause it’s generated by all .

Exercise 2. Let be a ring, an ideal, an -module, show that is isomorphic to .

Proof. Obviously is an exact sequence, so is . But and , and the first arrow is the inclusion map, so .

Exercise 3. Ler be a local ring, and finitely generated -modules. Prove that if , then or .

Proof. Let be the maximal ideal of and be the residue field of . Let denote , then by Nakayama Lemma, . So we have . But and are vector spaces over field , so implies or , hence or .

Exercise 4. Let be any family of -modules, and let be their direct sum. Prove that is flat each is flat.

Proof. is flat for all injective , is injective. And is injective if and only if each is injective, so qed.

Exercise 5. Ler be the ring of polynomials in one indeterminate over a ring . Prove that is a flat -algebra.

Proof. As a -module, , so by Exercise2.4 is flat (since is flat).

Exercise 6. For any -module , let denote the set of all polynomials in with coefficients in . Defining the product of an element of and an element of in the obvious way, show that is an -module.

Show that .

Proof. By define , trivially the module axioms hold here.

Consider a map defined by , then it’s a well-defined -module homomorphism. If we define by , then it’s clearly -bilinear, so induces an -module homomorphism . It’s easy to prove and are inverses, so .

Exercise 7. Let be a prime ideal in , show that is a prime ideal in . If is a maximal ideal in , is a maximal ideal in ?

Proof. Consider map , then . Then is prime since is an integral domain.

If is maximal, then doesn’t have to be a field, so is not maximal in general. For a counterexample, is a maximal ideal in , but is not maximal.

Exercise 8.

 1 If and are flat -modules, then so is . 2 If is a flat -algebra and is a flat -module, then is flat as an -module.

Proof.

 1 If is an exact sequence, then so is , hence .but , qed. 2 Let be an injective -module homomorphism. Since is flat, is injective. Consider as a -module homomorphism, then since is flat, is injective.By associativity of tensor product, and , we have injective, hence is flat as -module.

Exercise 9. Let be an exact sequence of -modules. If and are finitely generated, so is .

Proof. Let and be the maps in the sequence. If generate and generate . For each select an element such that , then since , and is generated by , so is generated by .

Exercise 10. Let be a ring, an ideal contained in the Jacobson radical of ; let be an -module and a finitely generated -module, and let be a homomorphism. If the induced homomorphism is surjective, then is surjective.

Proof. If is surjective, then forall there exists such that . That means, . So by Nakayama Lemma, , i.e. is surjective.

Exercise 11. Let be a ring . Show that .

Proof. let be a maximal ideal of , and an isomorphism, then is an isomorphism between two -vector spaces, hence the dim of two space are same, i.e. .

Exercise 12. Let be a finitely generated -module and a surjective homomorphism. Show that is finitely generated.

Proof. Let be a set of generators of , and such that . Let be the submodule generating by , then clearly , and for all there exists such that , hence . Summarize results above we get . Since is finitely generated, must be finitely generated too.

Exercise 13. Let be a ring homomorphism, and let be a -module. Regarding as an -module by restriction of scalars, form the -module . Show that the homomorphism which maps to is injective and that is a direct summand of .

Proof. Consider the quotient map . Since , we have which maps to .

Now , so is injective. Consider map defined by . ( so the second part of image of is actually ). We will prove is an isomorphism so .

If , i.e. and , obviously , so is injective. For any and , let , then and , hence is injective. All in all we have is an isomorphism so .

### Direct limits

Exercise 14. A partially ordered set is said to be a direct set if for each pair in there exists such that and .

Let be a ring, let be a direct set and let be a family of -modules indexed by . For each pair in such that , let be an -homomorphism, and suppose that the following axioms are satisfied:

 1 is the identity mapping of for all ; 2 whenever .

Then the modules and homomorphisms are said to form a direct system over the directed set .

We shall construct an -module called the direct limit of the direct system . Let be the direct sum of , and identify each module with its canonical image in . Let be the submodule of generated by all elements of the form where and . Let , let be the projection and let be the restriction of to .

The module , or MORE correctly the pair consisting of and the family of homomorphisms is called the direct limit of the direct system , and is written . From the construction it is clear that whenever .

Proof. No exercise here. For the last sentence, , but .

Exercise 15. In the situation of Exercise 14, show that every element of can be written in the form for some and some .

Show also that if then there exists such that in .

Proof. Any element of can be written in form with only finite nonzero. But for any there exists such , so , hence any element of can be written in form for some .

If , i.e. , then we have

where the sum contains only finite nonzero terms, and is projection to . But and the equation above is in a direct sum, so all elements except . Select an index which any appearing here, then

’Cause for any .

Exercise 16. Show that the direct limit is charactered (up to isomorphism) by the following property. Let be an -module and for each let be an -module homomorphism such that whenever . Then there exists a unique homomorphism such that for all .

Proof. First prove constructed here satisfies this condition.

For any and satisfies , define by , then for all and we have , hence , so induce an -homomorphism and for any . Since all elements in can be written in the form , the map is then unique.

If is another system satisfying the condition, let and , there a unique homomorphism such that . In the other direction there also exists a homomorphism such that , so for any . Again let and , there exists a unique homomorphism such that for any . But both and meet the requirement of , so ; and vice versa. Hence and are inverse, and .

Exercise 17. Let be a family of submodules of an -module, such that for each pair of indices in there exists such that . Define to mean and let be the embedding of in . Show that

Proof. is obviously hold since for all there exists some .

We show by show have the universal property in the previous exercise. If given such that for any pair , then for all pair of indices and element there exists , so . Therefore we can define that agrees each over . Since any element belonging to also belongs to some , so the homomorphism here is unique.

By the previous exercise, we have then .

Exercise 18. Let be direct systems of -modules over the same directed set. Let be the direct limits and the associated homomorphisms.

A homomorphism is by definition a family of -module homomorphisms such that whenever . Show that defines a unique homomorphism such that for all .

Proof. Let , then whenever . Hence there exists a unique homomorphism such that for all .

Exercise 19. A sequence of direct systems and homomorphism

is exact if the corresponding sequence of modules and module homomorphisms is exact for each . Show that the sequence of direct limits is then exact.

Proof. Let be the corresponding homomorphisms, be the homomorphisms in system , and the induced homomorphisms.

First we show . For any , there exists and such that , and then

Then if , again for some and , hence . So there exists such that , hence , and there exists such that . Let , we have

Summarize the results above, we have , hence is exact.

### Tensor products commute with direct limits

Exercise 20. Keeping the tame notation as in Exercise 14, Let be any -module, Then is a direct system; let be its direct limit.

For each we have a homomorphism , hence by Exercise 16 a homomorphism . Show that is an isomorphism, so that

Proof. Let be the projection map form to .

Define by , then clearly , so induce a homomorphism . Since every is -linear over , by the construction of it’s easy to show so is . So we have a homomorphism defined by . We will show that is the inverse of . We have:

Since all can be written in form and all can be written in form , it’s clearly . So

Exercise 21. Let be a family of rings indexed by a directed set , and for each pair in let be a ring homomorphism, satisfying conditions i) and ii) of Exercise 14. Regarding each as a -module we can then form the direct limit . Show that inherits a ring structure from the so that the mappings are ring homomorphism. The ring is the direct limit of the system .

If prove that for some .

Proof. Let be the mappings. In every element is some where and . For any and , let be an index , we define . If there are two indices , find an index we can show that the definition does not depend on the choice of . The ring axioms are easy to verify, with the identity element be any (they are all equal).

For the second part, if , select an index , then . By Exercise 15 there exists such that . Since is a ring homomorphism, must be .

Exercise 22. Let be a direct system of rings and let be the nilradical of . Show that is the nilradical of .

If each is an integral domain, then is an integral domain.

Proof. If , then clearly . So is a direct system where is the restriction of .

Let denote the direct limit of . An element is nilpotent iff. , iff. and such that , i.e. exists such that is nilpotent in . That is, an element is nilpotent if and only if it can be written in form where . So the nilradical of is , the proposition holds.

For the second part, if , then there exists such that and , hence and there exists that . Since is an integral domain, either or is zero, so in either or is zero, and is then an integral domain.

Exercise 23. Let be a family of -algebras. For each finite subset of let denote the tensor product (over ) of the for . If is another finite subset of and , there is a canonical -algebra homomorphism . Let denote the direct limit of the rings as runs through all finite subsets of . The ring has a natural -algebra structure for which the homomorphisms are -algebra homomorphisms. The -algebra is the tensor product of the family .

Proof. For and we have a canonical map defined by . Clearly are -algebra homomorphism.

Let denote the canonical homomorphism associated to the direct sum. For any , if we define . Since are -algebra homomorphisms the result is independent of the choice of . Then for any finite , is a ring homomorphism, and by definition of scalar multiplition over it is also a -algebra homomorphism.

### Flatness and Tor

Exercise 24. If is an -module, the following are equivalent:

 1 is flat; 2 for all and all -modules ; 3 for all -modules .

Proof. i) ii): is flat means is exact functor. If is a projective resolution of , then is exact, and .

ii) iii): Trivial.

iii) i): Consider a exact sequence , then there is a long exact sequence:

If then is exact, that is is flat.

Exercise 25. Let be an exact sequence, with flat. Then is flat is flat.

Proof. For all -module there exists a long exact sequence:

If is flat, then . So is flat for all for all .

Exercise 26. Let be an -module, Then is flat for all finitely generated ideals in .

Proof. By (2.19), is flat preserves injective map of finitely generated -module, so for all finitely generated .

Now if is generated by , and let be the submodule generated by . Then there is exact sequence . So by the last exercise, it’s sufficient to show . But for some ideal , so it’s sufficient to show for all .

By (2.19) again, we can only consider finitely generated ideals .

Exercise 27. A ring A is absoutely flat if every -module is flat. Prove that the following are equivalent:

 1 is absoutely flat. 2 Every principal ideal is idempotent. 3 Every finitely generated ideal is a direct summand of A.

Proof. i) ii): Let , then is flat. Let : , tensor with gives an injective map . By the commutative diagram, (the zero map) equals . Since is injective, . So and .

ii) iii): For any , there exists such that , so is idempotent and , i.e. . For any two idempotent , we have cause . So any finitely generated ideal of is idempotent principal ideal , and .

iii) i): , then . For all -module , . But is a direct summand of , hence flat. So for all finitely generated ideal , and by Exercise 2.26 is flat.

Exercise 28. A Boolean ring is absolutely flat. The ring of Exercise 1.7 is absolutely flat. Every homomorphic image of an absolutely flat ring is absolutely flat. If a local ring is absolutely flat, then it is a field.

If is absolutely flat, every non-unit in is a zero-divisor.

Proof. In a Boolean ring every element is idempotent, so Exercise 27 ii) holds, and it’s absolutely flat.

In the ring of Exercise 1.7 (every satisfy for some ), we have , so and Exercise 27 ii) holds.

If is absolutely flat and , then for any , we have . So is also absolutely flat.

If is a absolutely flat local ring with maximal ideal , then there exists , so for some . But then is a ideal not contained in (otherwise ), contradiction. So and is a field.

If is absolutely flat and is not unit, then , where and for any . So is a zero-divisor.