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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 3.

## Rings and Modules of Fractions

Exercise 1. Let be a multiplicatively closed subset of a ring , and let be a finitely generated -module. Prove that if and onle if there exists such that .

Proof. : if generate , then , i.e. exists such that . Let , then for all .

: if and , then for all we have .

Exercise 2. Let be an ideal of , and let . Show that is contained in the Jacobson radical of .

Proof. If , then . Therefore , and , hence is unit, i.e. only contains units. So is contained in the Jacobson radical of .

Use this result and Nakayama’s lemma to give a proof of (2.5) which does not depend on determinants. (2.5: Let be a finitely generated -module and let be an ideal of such that . Then there exists such that )
Proof. If , let , then . Since is contained in the Jacobson radical, by Nakayama we have , so by Exercise 3.1 there exists such that .

Exercise 3. Let be a ring, let and be two multiplicatively closed subsets of , and let be the image of in . Show that the rings and are isomorphic.

Proof. Define by , then for all we have is a unit (if and , then ), so induces a map satisfying: forall .

In the other direction, consider the canonical map defined by , then are units, so induces a map and . It’s trivial to show are inverses, so .

Exercise 4. Let be a homomorphism of rings and let be a multiplicatively closed subset of . Let , show that and are isomorphic as -modules.

Proof. Define by . By definition the action of on is , so two elements is equal in if and only if there exists such that ; this is equivalent to the equality of , so induces a bijection . Since is ring homomorphism, it’s easy to show is a group isomorphism. Finally, if and , then

So is a -module isomorphism.

Exercise 5. Let be a ring. Suppose that for each prime ideal , the local ring has no nilpotent element . Show that has no nilpotent element . If each is an integral domain, is necessarily an integral domain?

Proof. Let denote the nilradical of and , then . So for all , , hence .

For the second part, if is product of fields , then is not a integral domain. But the only prime ideals of is ; and is direct sum of as a -module, so . Now if because they meet, and , hence is a field (so an integral domain).

Exercise 6. Let be a ring and let be the set of all multiplicatively closed subsets of such that . Show that has maximal elements, and is maximal if and only if is a minimal prime ideal of .

Proof. Since , it is nonempty. By Zorn’s lemma, it’s sufficient to show every chain in has a upper bound. If is an ordered chain that , then consider the onion of them, denoted by . Clearly , and if , then , so is multiplicatively closed, and is an upperbound of .

If is a maximal element of , let . We will show is an ideal. If , then the minimal multiplicatively closed set containing and is . It’s a multiplicatively closed set strictly containing , so by hyposthesis it must contains , i.e. such that . Conversely if these exist then clearly .

Now if , then there exists and so that . Consider , by binomial expanding it’s a sum of some , where or , so equals ; hence . Together with we have is an additive subgroup of . If , then for some , so and , and so is an ideal of . If , that is, , then , and is an prime ideal. If is some prime ideal, then is a multiplicatively closed set containing , so equals to , an . This means is minimal.

Conversely, if is minimal prime ideal, clearly , and is maximal.

Exercise 7. A multiplicatively closed subset of a ring is said to be saturated if

Show that

 1 is saturated is a union of prime ideals. 2 If is any multiplicatively closed subset of , there is a unique smallest saturated multiplicatively closed subset containing , and that is the complement in of the union of the prime ideals which do not meet . ( is called the saturation of ).

If , where is an ideal of , find .

Proof.

 1 : If , then for any : so is saturated.: If is saturated and , then is disjoint from . Let be all ideals containing and disjoint from , then , and all increasing chains are closed in , so by Zorn’s Lemma, there is a maximal element . If , then strictly contain , so not in , i.e. they meet with . Let , then , so , hence is prime. Every element in is contained in a prime ideal disjoint from , so is union of prime ideals. 2 Let be the complement of the union of all prime ideals disjoint from : , then by (1), is saturated. If is any saturated multiplicatively closed subset containing , then is the complement of union of some prime ideals disjoint from , hence .
If , then is disjoint if and only if there exists so that , that is, . So is the union of all prime ideals not coprime with . Since iff. there is a maximal ideal contains both and , so it is sufficient to union all maximal ideals which contains : .

Exercise 8. Let be multiplicatively closed subsets of , such that . Let be the homomorphism which maps each to considered as an element of . Show that the following statements are equivalent:

 1 is bijective. 2 For each , is a unit in . 3 For each there exists such that . 4 is contained in the saturation of . 5 Every prime ideal which meets also meets .

Proof.

 i)ii): If is bijective, then since is a unit in , it must be a unit in . ii)i): Consider defined by , clearly and are inverses, so is bijective. ii)iii): is a unit there exists such that (). Let , then . iii)iv): Let denote the saturation of , then , so . iv)v): If , then . Since is disjoint from any prime ideal which doesn’t meet , so there must be . v)ii): Since every prime ideal of is for some not meet , so: is unit in Any prime ideal in doesn’t contain for any disjoint from , we have .By v), every prime ideal disjoint from also disjoint from , so for all , is a unit in .

Exercise 9. The set of all non-zero-divisors in is a saturated multiplicatively closed subset of . Hence the set of zero-divisors in is a union of prime ideals (see Chapter 1, Exercise 14). Show that every minimal prime ideal of is contained in .

Proof. If , then and must not be zero-divisors, so and . Conversely and implies , so is a saturated multiplicatively closed subset.

If is a minimal prime ideal, then is a maximal multiplicatively closed subset not containing . Consider , it’s multiplicatively closed, and clearly . Since is maximal, there must be , that is, , hence .

The ring is called the total ring of fractions of . Prove that
 1 is the largest multiplicatively closed for which the homomorphism is injective. 2 Every element in is either a zero-divisor or a unit. 3 Every ring in which every non-unit is a zero-divisor is equal to its total ring of fractions (i.e. is bijective).

Proof.

 1 the kernel of if and only if there exists such that , i.e. . Hence is injective if and only if only contains non-zero-divisors, so is the largest of them. 2 If , then is a unit.On the other hand if , then is a zero-divisor in , hence there is a non-zero element so that . Since is injective, . Then and is a zero-divisor in . 3 If satisfies the condition, then only contains unit. Then for any , . So is surjective, and by i) it is injective, hence bijective.

Exercise 10. Let be a ring.

 1 If is absolutely flat, and is any multiplicatively closed subset of , then is absolutely flat. 2 is absolutely flat is a field for each maximal ideal .

Proof.

 1 For any , there exists such that ; so and . 2 : is a absolutely flat local ring, hence a field.: For any -module , is a vector space of , hence flat. Since flatness is a local property, is flat.

Exercise 11. Let be a ring. Prove that the following are equivalent:

 1 is absolutely flat ( being the nilradical of ). 2 Every prime ideal of is maximal. 3 is a -space (i.e. every subset consisting of a single point is closed). 4 is Hausdorff.

If these conditions are satisfied, show that is compact and totally disconnected (i.e. the only connected subsets of are those consisting of a single point).

Proof. Cause ii) iii) iv) holds for if and only if it holds for , without loss of generality we may assume .

i) ii): If is a prime ideal and contained in the maximal ideal , then is a prime ideal of , which is a field. Hence is maximal. Conversely if ii) holds, then has only as it’s prime ideal. But it’s nilradical is , so and is a field.

ii)iv): Let be distinct prime (hence maximal) ideals of , and let . Consider , it’s a open subset of , and , so it’s sufficient to show is close:

Proof. Every prime ideal which does not contains correspends to a prime ideal of , and every prime ideal which contains correspends to a prime ideal of . So it’s sufficient to show maps prime ideal to prime ideal (the preimage of a prime ideal is automaticly prime).

If is a prime (also maximal) ideal of , then is also injective, hence is not trivial (except , in which case is prime) and have a prime ideal, so it correspends to a prime ideal of which preimage is contained in . Cause every prime ideal is maximal, we have ; this shows every prime ideal maps to a prime ideal too.

iv)iii): Trivial.

iii)ii): If a prime ideal is closed in then there exists so that but not contained in any other prime ideal. This shows is maximal.

Exercise 13. Let be an integral domain and an -module. An element is a torsion element of if , that is if is killed by some non-zero element of . Show that the torsion elements of form a submodule of .

Proof. If , then , and since is an integral domain, so the torsion elements are closed under addition and scalar-multiplication, and so it contains , hence a submodule of .

This submodule is called the torsion submodule of and is denoted by . If , the module is said to be torsion-free. Show that
 1 If is any -module, then is torsion-free. 2 If is a module homomorphism, then . 3 If is an exact sequence, then the sequence is exact. 4 If is any -module, then is the kernel of the mapping of into , where is the field of fractions of .

Proof.

 1 If is a torsion element in , then there exists such that , so there exists another such that , hence and . 2 If such that , then , so . 3 If is exact, let and denote the induced map, then clearly , and is injective.If , then there exists such that . Since , there exists such that . Since is injective, and , that is, . 4 Let , then and by , hence . Then , so .

Exercise 14. Let be a multiplicatively closed subset of an integral domain . In the notation of Exercise 12, show that . Deduce that the following are equivalent:

 1 is torsion-free. 2 is torsion-free for all prime ideal . 3 is torsion-free for all maximal ideal .

Proof. Let , then exists such that , so for all , , and . Conversely, if , then there exists such that . Since is integral, , and , so .

For the second part, is torsion-free . So the statement is equivalent to: i) ; ii) For all prime ideal , ; iii) For all maximal ideal , . Clearly they are equivalent.

Exercise 15. Let be an -module and an ideal of . Suppose that for all maximal ideals . Prove that .

Proof. Consider -module , then any maximal ideal in is in the form for some maximal ideal . For any maximal ideal , we have . For any , we have , so it is isomorphic to .

Now we have , and is the same as , so is zero for any maximal ideal ; that is, it’s zero under any localization by maximal ideal in , hence itself is zero. So , then .

Exercise 16. Let be a ring, and let be the -module . Show that every set of generators of is a basis of .

Proof. Let be the canonical basis of , and be a set of generators of . Define by , then it is surjective, and we have to prove it’s a isomorphism. By Propsition 3.9, injective is a local property, so it’s sufficient to assume is a local ring. Let and is the residue field of . Since is exact, tensoring with we have exact. Since is a vector space, and is a surjective automorphism over it, there must be injective, and , i.e. . Also by Chapter 2, Exercise 12, is finitely generated; since is the unique maximal ideal, by Nakayama’s Lemma we have .

Deduce that every set of generators of has at least elements.
Proof. If generate for some , extend it by random non-zero elements . By the results above, is a basis of ; but for some coefficient , contradiction.

Exercise 17. Let be a flat -algebra. Then the following conditions are equivalent:

 1 for all ideals of A. 2 is surjective. 3 For every maximal of we have . 4 If is any non-zero -module, then . 5 For every -module , the mapping of into is injective.

Proof.

i) ii): By (3.16), is contraction if and only if .

ii) iii): Trivial.

iii) iv): Since is flat, then implies . So it’s sufficient to consider only for some , and in this case we have for some ideal of . Then , and for some maximal ideal of . Hence .

iv) v): Let , then is exact. By flatness of we have exact. By Exercise 2.13, is injective, hence , so .

v) i): Let , then is injective, so .

is said to be faithfully flat over .

Exercise 18. Let be ring homomorphisms. If is flat and is faithfully flat, then is flat.

Proof. If is a injective -module homomorphism, we want to prove is injective. By Exercise 16 v), is injective. But . By the following commutative diagram:

[TikZ 编译错误]

And since is flat over , we know is injective.

Exercise 19. Let be a flat homomorphism of rings, let be a prime ideal of and let . Then is surjective.

Proof. Cause flatness is local we have is flat over . Also is a localization of , so flat over . So is flat over .

Also, the only maximal ideal of is , and it’s extension in is contains in , hence not . By Exercise 16 iii)i) we have surjective.

Exercise 20. Let be a ring, an -module. The support of is defined to be the set of prime ideals of such that . Prove the following results:

 1 . 2 . 3 If is an exact sequence, then . 4 If , then . 5 If is finitely generated, then (and is therefore a closed subset of . 6 If are finitely generated, then . 7 If is finitely generated and is an ideal of , then . 8 If is a ring homomorphism and is finitely generated -module, then .

Proof.

 1 for all prime ideal , . 2 , So . 3 For every prime ideal we have exact sequence , so . 4 If , then , so at least one . 5 If , then , and each is isomorphic to some . By ii), , so . 6 If , then either or . So . 7 If is finitely generated then also is . So . Now if , and generate , then for some . By linear algebra we know , that is, . so and hence . By we know . 8 If then . So and . Cause , it’s sufficient to show when for some ideal of .If then , and .

Exercise 21. Let be a ring homomorphism, the associated mapping. Show that

 1 Every prime ideal of is a contracted ideal is surjective. 2 Every prime ideal of is an extend ideal is injective.

Is the converse of ii) true?

Proof.

 1 If is contracted ideal then . By (3.16) is contraction of some prime ideal, i.e. is surjective. The converse is trivial. 2 Let be distinct prime ideal of , then , so .
The converse is not true in genereal. For example, Let be a integral domain with a unique non-zero prime ideal , and where is the fraction field of ; and define where is the image of in . Then and . Hence is injective, but , so it’s not a extend ideal.

Exercise 22.

 1 Let be a ring, a multiplicatively closed subset of , and the canonical homomorphism. Show that is a homeomorphism of onto its image in . Let this image be denote by .In particular, if , the image of in is the basic open set . 2 Let be a ring homomorphism. Let and , and let be the mapping associated with . Identifying with its canonical image in , and with its canonical iamge in , show that is the restriction of to , and that . 3 Let be an ideal of and let be its extension in . Let be the homomorphism induced by . If is identified with its canonical image in , and with its image in , show that is the restriction of to . 4 Let be a prime ideal of . Take in ii) and then reduce mod as in iii). Deduce that the subspace of is naturally homeomorphic to , where is the residue field of the local ring .

is called the fiber of over .

Proof.

 1 If in is the contraction of some prime ideal in then it must be . So is injective. Consider and a prime ideal of , then . So is also continuous, hence is a homomorphism between and it’s image in .If , then a prime ideal of is the contraction of some prime ideal of , i.e. . 2 To show is a restriction of , it’s equivalent to say that the right diagram below commutes. Cause the left diagram commutes, so is the induced maps. [TikZ 编译错误]If , i.e. is disjoint from , then is disjoint from , so . Hence . 3 Like ii), the map is the same as , so the induced maps are the same too, i.e. is the restriction of . Also we have . 4 is identified with in (since it’s maximal in ), and is identified with , hence in , so it is . And . But image of in is zero, so .

Exercise 23. Let be a ring and a prime ideal of , then the canonical image of in is equal to the intersection of all the open neighborhoods of in .

Proof. The intersection of all the open neighborhoods of is , i.e. all prime ideals contained in . This is equivalent to be a contraction ideal of .

Exercise 24. Let be a ring, let and let be a basic open set in (i.e. for some ).

 1 If , show that the ring depends only on and not on . 2 Let be another basic open set such that . Show that there is an equation of the form for some integer and some , and use this to define a homomorphism (i.e. ) by mapping to . Show that depends only on and . This homomorphism is called the restriction homomorphism. 3 If , then is the identity map. 4 If are basic open sets in , show that the diagram [TikZ 编译错误](in which the arrowsare restriction homomorphism) is commutative. 5 Let be a point of . Show that

The assignment of the ring to each basic open set of , and the restriction homomorphism , satisfying the condition iii) and iv) above, constitutes a presheaf of rings on the basis of open sets . v) says that the stalk of this presheaf at is the correspending local ring .

Proof. If then . So there exists such that . So we can define a map by . If there are another pair of s.t. , then because . So only depends on and .

If there are , and , then . So . So .

 1 If , then And . So we can identity with . 2 If then is a map from to . Also if , then . By the definition in i), and are identity map, so this map only depends on . 3 Trivial. 4 Trivial if we select . 5 Prove by the universal property of direct limit and fractions. If there is another ring and a family of maps whenever , and these maps are conpatible; that is, a map with all invertible for , then it induces a unique map on , also induces a unique map on . Hence they have the same universal property, so must be isomorphic.

Exercise 25. Show that the presheaf of Exercise 23 has the following propety. Let be a covering of by basic open sets. For each let be such that, for each pair of indices , then images of and in are equal. Then there exists a unique whose image in for all . (This essentially implies that the presheaf is a sheaf).

Proof. First, we can choose a finitely covering of , so without loss of generality (cause is unique) we assume that .

By replace by (Cause ), we can assume . If the image and are same in , i.e. , then there exists such that . Choose a large such that this equation is satisfied for every pair of , and rewrite it as

Since generate , so do . So we have for some . Let , then