_rqy's Blog

0%

Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 7.

## Noetherian Rings

Exercise 1. Let be a non-Noetherian ring and let be the set of ideals in which are not finitely generated. Show that has maximal elements and that the maximal elements of are prime ideals.

Proof. Since is nonempty and satisfies conditions of Zorn’s Lemma, it must have maximal elements.

Let be one of the maximal elements of . Suppose , then is finitely generated. If it’s generated by where , let be the finitely generated ideal contained in , then . Particularly, Since is not finite generated, neither is . But , so by the maximality of in .

Hence for every we have . As a consequnece, is a prime ideal.

Hence a ring in which every prime ideal is finitely generated is Noetherian (I. S. Cohen).

Exercise 2. Let be a Noetherian ring and let . Prove that is nilpotent if and only if each is nilpotent.

Proof. By Chapter 1, Exercise 5, is nilpotent implies that each is nilpotent.

Conversely, if each is nilpotent, let denote the nilradical of , then (i.e. all coefficients of belong to ). By (7.15), is nilpotent, so is .

Exercise 3. Let be an irreducible ideal in a ring . Then the following are equivalent:

 1 is primary; 2 for every multiplicatively closed subset of we have for some . 3 the sequence is stationary, for every .

Proof.

 i) ii): Suppose is -primary.If meets , then and contracts to . In this case, . Select , then .If is disjoint from , then . Choose arbitrary, then , hence . ii) iii): Let , then for some . Clearly for every , so . iii) i): (Similar to (7.12)) Suppose and . If , then , for if , then ; and if then , hence , so . Since is irreducible and , we must have (Consider ). Hence is primary.

Exercise 4. Which of the following rings are Noetherian?

 1 The ring of rational functions of having no pole on the circle . 2 The ring of power series in with a positive radius of convergence. 3 The ring of power series in with an infinite radius of convergence. 4 The ring of polynomials in whose first derivatives vanish at origin ( being a fixed integer). 5 The ring of polynomials in all of whose partial derivatives with respect to vanish for .

In all cases the coefficients are complex numbers.

Proof. We denote the ring described as .

 1 Let be the set of all polynomials, which having no zeros on the circle . Then , hence is Noetherian (cause is Noetherian). 2 If , and , then there exists such that for every . Hence has a positive radius of convergence (at least ). Therefore the only ideals of is ( is a valuation ring). Hence is Noetherian. 3 Let .Clearly are ideals of , , and Hence the increasing sequence is strict, so is not Noetherian. 4 Every elements of is in the form , where and . For every ideal of , assume . Let be a monic polynomial of the minimal degree. For any , if and , then . Proceeding in this way, we can get such that and . For every , choose a polynomial in with degree (if it exists). Then clearly . Hence is finitely generated, and is Noetherian. 5 One can show that in . Hence is not finitely generated, and is not Noetherian.

Exercise 5. Let be a Noetherian ring, a finitely generated -algebra, a finite group of -automorphisms of , and the set of all elements of which are left fixed by every elements of . Show that is a finitely generated -algebra.

Proof. We know that (Chapter 5, Exercise 12) is integral over . Since , is Noetherian and is finite generated over , it follows that is finite generated over .

Exercise 6. If a finitely generated ring is a field, it is a finite field.

Proof. If , then .

Since is finitely generated over , it’s finitely generated over . Hence by (7.9)6, is finitely generated as -module. Then is finitely generated over cause is Noetherian, contradiction.

Therefore , and is finitely generated as -algebra. Apply (7.9) again we know is a finite algebraic extension of , hence finite.

Exercise 7. Let be an affine algebraic variety given by a family of equations (Chapter 1, Exercise 27). Show that there exists a finite subset of such that is given by the equations for .

Proof. Let be the ideal generated by . Let be the set of ideals generated by a finite subset of . Since is Noetherian, has a maximal element, and clearly it must be . Hence can be generated by for a finite subset . Thus is given by the equations for .

Exercise 8. If is Noetherian, is necessarily Noetherian?

Proof. Yes. is a homomorphic image of .

Exercise 9. Let be a ring such that

 1 for each maximal ideal of , the local ring is Noetherian; 2 for each in , the set of maximal ideals which contain is finite.

Show that is Noetherian.

Proof. Let be a ideal of , be the maximal ideals which contain . Choose in arbitrary and let be the maximal ideals which contain . Since do not contain , there exists such that . Since each is Noetherian, the extension of in is finitely generated. Hence there exist in whose images in generate for . Let . Then for every maximal ideal of :

 1 If for some : the images of generate , hence . 2 If for some : then , hence ; and . 3 Otherwise, . so .
Summarize above, for every maximal ideal we have . Hence is finitely generated.

Exercise 10. Let be a Noetherian -module. Show that (Chapter 2, Exercise 6) is a Noetherian -module.

Proof. Suppose is a -submodule. Define to be the set of all leading coefficients of polynomials in ; then is a submodule of , so finitely generated, say by . For each , choose with leading coefficient . Let denote the degree of , and . Let be the -submodule generated by .

For any , if with , write , then satisfies and . Proceeding in this way we get with and .

Let denote the -submodule of generated by , then , hence Noetherian as -module. Hence is finitely generated, say by . Therefore and implies is a linear combination of . Hence is generated by .

Exercise 11. Let be a ring such that each local ring is Noetherian. Is necessarily Noetherian?

Proof. No (in general). If is a Boolean ring, then (the following proof), but need not to be Noetherian.

Suppose is a Boolean ring, be its prime ideal, then . If , then Also if then . Hence .

Exercise 12. Let be a ring and a faithfully flat -algebra (Chapter 3, Exercise 16). If is Noetherian, show that is Noetherian.

Proof. If is faithfully flat, then for every ideal , one has .

Therefore the (order-preserving) map is injective, from the set of ideals in to the set of ideals in . Since satisfies a.c.c., so does . That is, is Noetherian.

Exercise 13. Let be a ring homomorphism of finite type and let be the mapping associated with . Show that the fibers of are Noetherian subspaces of .

Proof. For each prime ideal of , , where is the residue field of . Since is finitely generated -algebra, it is Noetherian. Therefore is Noetherian, so is .

### Nullstellensatz, strong form

Exercise 14. Let be an algebraically closed field, let denote the polynomial ring and let be an ideal in . Let be the variety in defined bt the ideal , so that is the set of all such that for all . Let be the ideal of , i.e. the ideal of all polynomials such that for all . Then .

Proof. If , then for all , so . Hence .

Conversely, if , then there exists a prime ideal containing but not . Let , denote the image of in , let , and let be a maximal ideal of . Since is finitely generated as a -algebra, it follows that . Thus the images in of the generators of define a point . For every , is same as the image of in . The image of in (hence in ) is zero, so . But the image of in is a unit, so it is not zero in . Therefore and , implies that .

Exercise 15. Let be a Noetherian local ring, its maximal ideal and its residue field, and let be a finitely generated -module. Then the following are equivalent:

 1 is free. 2 is flat. 3 the mapping of into is injective. 4 .

Proof.

 i) ii): Clearly is flat. ii) iii): The map is injective, tensor with . iii) iv): We have exact, so by definition . iv) i): Let be elements in whose images in (as a -vector space) forms a -basis. Then generate (If they generate , then ). Let with the canonical basis , and define by . Then is surjective. Let , tensor the exact sequence with , notice that , then we get Since and are vector spaces of the same dimension, it follows that is an isomorphism, hence . Since is a submodule of , and is Noetherian, is finitely generated. By Nakayama’s Lemma (), so is an isomorphism and .

Exercise 16. Let be a Noetherian ring, a finitely generated -module. Then the following are equivalent:

 1 is a flat -module; 2 is a free -module, for all prime ideals ; 3 is a free -module, for all maximal ideals .

In other words, flat = locally free.

Proof. Since each is finitely generated over , by the last Exercise, ii) or iii) is equivalent to that (respectively, ) is flat. Since flatness is a local property, the result follows immediately.

Exercise 17. Let be a ring and a Noetherian -module. Show (by imitating the proofs of (7.11) and (7.12)) that every submodule of has a primary decomposition (Chapter 4, Exercise 20-23).

Proof. Definte a submodule to be irreducible, if for any submodules , implies or .

Similar to (7.11) and (7.12), we have

Proof. Suppose not, then there is a maximal submodule of respecting to this property. Since is reducible, there is with . Hence and is a finite intersection of irreducible submodules, so is . Contradiction.

Proof. Suppose is an irreducible submodule of . Passing to the quotient module , we may assume . Let such that , consider the submodules . Since is Noetherian, the chain is stationary, hence exists such that . It follows that , for if , then , so and . Since is irreducible and , there must be , which implies that .

By this two lemma, clearly every submodule of has a primary decomposition.

Exercise 20. Let be a Noetherian ring, a prime ideal of , and a finitely generated -module. Show that the following are equivalent:

 1 belongs to in ; 2 there exists such that ; 3 there exists a submodule of isomorphic to .

Proof. Suppose is a minimal primary decomposition of in , where is -primary.

 ii) iii): If , then . iii) ii): If and , then is finitely generated since is Noetherian. Let generate , then . Hence for some . ii) i): belongs to . i) ii): Assume . Choose but , then . If generate (since is Noetherian), then , hence for some integer . Hence . Let be the minimal integer such that but , and let . Since , it can not belong to . Then , and if , then . Hence .

Deduce that there exists a chain of submodules such that each quotient is of the form , where is a prime ideal of .

Proof. Suppose not. Let be the set of submodules , so that there is no such chain with and , then . Since is finitely generated over the Noetherian ring , it is a Noetherian module. So has a maximal element, say .

Since , apply the result above to the module , then there exists a submodule such that for some prime ideal . Since is maximal in , , hence there is a chain from to . But , so such a chain extends to a chain from , contradiction.

Exercise 21. Let be an ideal in a Noetherian ring . Let be two minimal decompositions of as intersections of irreducible ideals. Prove that and that (possible after re-indexing the ) for all .

State and prove an analogous result for modules.

Proof. We can regard as -module; then ideals correspond to submodules, irreducible ideals correspond to irreducible submodules. So it’s enough to prove the statement for modules:

If is a ring and is a -module, and is a submodule of . Let be two minimal decompositions of as intersections of irreducible submodules of ; then . If is Noetherian, then for all (maybe after re-indexing the ’s).

Clearly is a irreducible submodule containing if and only if is a irreducible submodule of . So pass to the quotient module, we may assume .

Let denote . First we prove the hint in the book: for each , there exists such that .

Let fixed. We have (since ). Therefore for if , then . Since is irreducible, there exists so that , so . Thus for .

Now we have for some . This is also a irreducible decomposition of , hence apply the result again we have for some . Proceeding in this way, then there exists such that Since the decomposition is minimal, . Hence . For the same reason , hence .

If in addition is Noetherian, then each and is primary. Intersecting all submodules with the same radical, then we get two minimal primary decompositions of . By the uniqueness theorem of primary decompositions, the set of the radical ideals is uniquely determined by . Let denote this set (i.e. the set of prime ideals of ), then we must show that, for each , the number of occurrences of (in the radical of irreducible submodules) is uniquely determined.

Reorder such that implies . Then for each , the set is an isolated set of prime ideals of . Hence the intersection of primary submodules corresponding to these prime ideals is uniquely determined, say . Replacing with , we know that the number of occurrences of is uniquely determined ( for ). Subtracting the case of , then we finish the proof.

Exercise 22. Let be a topological space and let be the smallest collection of subsets of which contains all open subsets of and is closed with respect to the formation of finite intersections and complements.

 1 Show that a subset of belongs to if and only if is a finite union of sets of the form , where is open and is closed. 2 Suppose that is irreducible and let . Show that is dense in (i.e. that ) if and only if contains a non-empty open set in .

Proof.

 1 Let be the set of subsets which is a finite union of sets of the form . Then clearly ; Conversely, if , then Hence . Also the complement of is , hence Hence contains all open subsets and is closed under intersection and complement. Therefore , so . 2 If , then by i), for some open subsets and closed subsets . If , then Since is irreducible, there exists that , so and . Hence , and contanins a non-empty open set in .Conversely, if there is a non-empty open set , then . Hence . If , then .

Exercise 23. Let be a Noetherian topological space (Chapter 6, Exercise 5) and let . Show that if and only if, for each irreducible closed set , either or else contains a non-empty open subset of .

Proof. : If , then for each , let be the collection of subsets in , then clearly . Hence if , by Exercise 20, contains a non-empty open subset of .

: Assume . Let be the collection of closed subsets such that . Then , . Since is Noetherian, has a minimal element, say . If , then , hence , so in particular , show that is irreducible.

Let . If , then . But , contradiction. Hence .

If is a non-empty open subset of such that , then , or else . Let , then is closed in , hence closed in . By the minimality of , we have . There exists open subset such that , so . Hence , contradiction.

Summarize above, , but doesn’t contain a non-empty open subset of , contradict the hypotheses. Thus must in .

Exercise 24. Let be a Noetherian topological space and let be a subset of . Show that is open in if and only if, for each irreducible closed subset in , either or else contains a non-empty open subset of .

Proof. : If , then it is a non-empty open subset of .

: Suppose is not open. Let be the collection of closed subsets such that is not open in . Since , it is not empty.

Let be a minimal element of . If closed, then there exists open subsets of such that . Hence . Therefore , or else is open in . Thus is irreducible.

Now is not open in , in particular . Hence contains a non-empty open subset of , say . Let , then , hence is open. Therefore is open in , contradiction.

Exercise 25. Let be a Noetherian ring, a ring homomorphism of finite type (so that is Noetherian). Let and let be the mapping associated with . Then the image under of a constructible subset of is a constructible subset of .

Proof. Since constructible sets are closed under unions, by Exercise 20 it’s enough to show is constructible, for open and closed. If , then replace with , it reduce to the case where is open in .

Since is Noetherian, is quasi-compact, and therefore a finite union of the basic open sets . Hence it’s sufficient to show each is constructible. Replacing with then it reduce to the case where .

By Exercise 21, to show is constructible, is equivalent to show that if is an irreducible subset of where is dense in , then contains some non-empty open subset of . Suppose for some prime ideal of , then , and . Again replacing with , with . Then we should show that if is a Noetherian integral domain, an -algebra of finite type and injective, then contains some open subset of .

Since is Noetherian, for irreducible closed subsets (irreducible complements of ). We have . Since is irreducible, there is some dense in . Suppose , and let denote the composite mapping . Since is dense in , it follows that , hence is also injective. Replacing with , we may assume are Noetherian integral domains, the inclusion map is of finite type. We have to show that contains some non-empty open subset of ; and it is equivalent to that contains some basic open subset of . That is, there exists such that all prime ideals not containing are contracted ideals.

By Chapter 5, Exercise 20, there exists in such that, if is an algebraically closed field and is a homomorphism for which , then can be extended to a homomorphism . If is a prime ideal of not containing , let be an algebraic closure of and be the composite mapping . Then , so extends to a homomorphism . Hence , and is a prime ideal of . Thus every prime ideal not containing is a contracted ideal.

Exercise 26. With the notation and hypotheses of Exercise 23, is an open mapping has the going-down property (Chapter 5, Exercise 10).

Proof. By Chapter 5, Exercise 10, is an open mapping implies has the going-down property.

Conversely, suppose has the going-down property. To show that is an open mapping, it’s enough to show that is open for every . Clearly has the going-down property, hence so does the composite mapping . Repalcing with then it’s sufficient to show that is open in .

Let , then the going-down property of is equivalent to that if and , then . Hence for any irreducible closed subset of , if , then , hence . By Exercise 23, is constructible; by Exercise 20 ii), contains a non-empty open subset of . Therefore by Exercise 22, is open.

Exercise 27. Let be Noetherian, of finite type and flat (i.e. is flat as an -module). Then is an open mapping.

Proof. By Chapter 5, Exercise 11, is flat implies has the going-down property. By Exercise 24, is an open mapping.

### Grothendieck groups

Exercise 28. Let be a Noetherian ring and let denote the set of all isomorphism classes of finitely generated -modules. Let be the free abelian group generated by . WIth each short exact sequence of finitely generated -modules we associate the element of , where is the isomorphism class of , etc. Let be the subgroup of generated by these elements, for all short exact sequences. The quotient group is called the Grothendieck group of , and is denoted by . If is a finitely generated -modyle, let , or , denote the image of in .

 1 Show that has the following universal property: for each additive function on the class of finitely generated -modules, with values in an abelian group , there exists a unique homomorphism such that for all .Proof. The map induces an homomorphism such that . The hypothesis that is additive is equivalent to that . Hence induces a homomorphism , such that for all . Since is generated by all , is unique. 2 Show that is generated by the elements , where is a prime ideal of .Proof. By Exercise 18, for each finitely generated -module , there exists a chain of submodules , such that each is of the form , where is a prime ideal of . Hence there is short exact sequences , and . Therefore .Since is generated by all , it follows that is generated by the elements . 3 If is a field, or MORE generally if is a principal ideal domain, then .Proof. If is a PID, then for each non-zero prime ideal of , the mapping is a -module isomorphism between and . Hence in . By ii), is generated by .Let denote the field of fractions of , then for each finitely generated -module , is a finite dimensional vector space over . Let denote its dimension. Since is flat, each short exact sequence tensor with , gives a short exact sequence . Hence , so is a additive function. By i) induces a homomorphism . Since , it follows that is surjective. But is generated by a single element , hence . 4 Let be a finite ring homomorphism. Show that the restriction of scalars gives rise to a homomorphism such that for a -module . If is another finite ring homomorphism, show that .Proof. If is an -algebra of finite type, then every finitely generated -module can also be regarded as a finitely generated -module by the restriction of scalars; Thus there is a map .Clearly a -module short exact sequence is also a -module short exact sequence under the restriction of scalars. Hence is additive for -modules. By i), induces a homomorphism , such that for all -module .If is another finite ring homomorphism, then for every -module . Since For every -module , it follows that .

Exercise 29. Let be a Noetherian ring and let be the set of all isomorphism classes of finitely generated flat -modules. Repeating the construction of Exercise 26 we obtain a group . Let denote the image of in .

 1 Show that tensor product of modules over induces a commutative ring structure on , such that . The identity element of this ring is .Proof. Let be the free group generated by ; then tensor product induces a commutative ring structure over with as the identity element, since Let be a short exact sequence where . If , then tensor the exact sequence with , we have exact. Let denote the (additive) subgroup of generated by all elements of the form for short exact sequences , then is also an ideal of the commutative ring .So is also a quotient ring of , such that