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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 5.

## Integral Dependence and Valuations

Exercise 1. Let be an integral homomorphism of rings. Show that is a closed mapping, i.e. that it maps closed sets to closed sets. (This is a geometrical equivalent of (5.10)).

Proof. For the case when is injective, i.e. is a subring of : For any and prime ideals, if , then by (5.11), there exists a prime ideal such that . i.e. . Conversely if then . Hence , so is closed.

In the general case, is the composition of and , where . Obviously is closed, and is closed since is integral over . Thus is closed.

Exercise 2. Let be a subring of a ring such that is integral over , and let be a homomorphism of into an algebraically closed field . Show that can be extended to a homomorphism of into .

Proof. Let . Since is a integral domain, is prime. By (5.10), there exists prime such that . Let and , then is integral over ; and induces an embedding of into . Hence we can assume are integral domains and is injective.

Let be the set such that is a subring, and is an embedding, where . Define a partial order over by if and . Clear the conditions of Zorn’s lemma are satisfied, and therefore has a maximal element, denote by . We must show .

Suppose for a contradiction, then there exists . Since is integral over and , is integral over . Let be the minimal polynomial of . Since is algebraically closed, the polynomial has a root in . Consider the homomorphism defined by . Let be the fraction field of , then can be considered as subfield of ; and clearly is also the minimal polynomial of over . So , and induce an embedding from to which restriction to on . But , contradict the maximality of .

Exercise 3. Let be a homomorphism of -algebras, and let be an -algebra. If is integral, prove that is integral.

Proof. For any , if Then So is integral over . Since generates , we have integral.

Exercise 4. Let be a subring of a ring such that is integral over . Let be a maximal ideal of and let be the corresponding maximal ideal of . Is necessarily integral over ?

Proof. No. Let and (where is a field), and let . Then . Since , is integral over . Hence is integral over .

If is integral over , then for some . Let be the common denominator of , and . Then In particular, . But , contradiction. So is not integral over .

Exercise 5. Let be rings, integral over .

 1 If is a unit in then it is a unit in . 2 The Jacobson radical of is the contraction of the Jacobson radical of .

Proof.

 1 If is a unit in , then is integral over . Suppose where , then So , i.e. is a unit in . 2 Since integral over , for any maximal ideal of , is a maximal ideal of . Hence Conversely, if , then for any , is a unit in , hence by i) a unit in . So . Thus .

Exercise 6. Let be integral -algebras. Show that is an integral -algebra.

Proof. For every and every , is integral of , since is integral over . But every element of is a finite product of such elements. So is integral over .

Exercise 7. Let be a subring of a ring , such that the set is closed under multiplication. Show that is integrally closed in .

Proof. If is integral over , choose a monic polynomial of minimal degree such that . Suppose , then We have , and by the choice of , ; since is closed under multiplication, there must be . Hence is integrally closed.

Exercise 8.

 1 Let be a subring of an integral domain , and let be the integral closure of in . Let be monic polynomials in such that . Then are in . 2 Prove the same result without assuming that (or ) is an integral domain.

Proof.

 1 Let be the field of fraction of , and be the splitting field of over . Then split into linear factors in . Suppose , Then each is a root of the monic polynomial , hence integral over . Since the coefficients of is polynomial of ’s, so integral over . But is integral closed in , so . Similarly we have . 2 Similarly to i), it’s sufficient to show that there exists an extension of in which splits into linear factors. Inductive on the degree of . If then there is nothing to prove. If , let , then can embed into , and is a root of ; hence for some . By inductive hypothesis, there exists an extenstion of such that splits into linear factors, so is .

Exercise 9. Let be a subring of a ring and let be the integral closure of in . Prove that is the integral of in .

Proof. If is integral over , then clearly is integral over for any . Hence is integral over . So it’s sufficient to show: if is integral over , then .

Suppose where . Let be an integer larger than and the degrees of . Let , then or say And is a monic polynomial. Let , then , and , are both monic polynomials, so is . Since , by Exercise 8, . So .

Exercise 10. A ring homomorphism is said to have the going-up property (resp. the going-down property) if the conclusion of the going-up theorem (5.11)1 (resp. the going-down theorem (5.16)2) holds for and its subring .

Let be the mapping associated with .

1.

Consider the following three statements:

 1 is a closed mapping. 2 has the going-up property. 3 Let be any prime ideal of and let . Then is surjective.

Proof.

 (a) (b): If is closed, then for every prime ideal of , is closed. Let , then , so . Clearly this is equivalent to (b). (b) (c): has the going-up property for every and for every , there exists such that for every , the map is surjective.

2.

Consider the following three statements:

 1 is a open mapping. 2 has the going-down property. 3 Let be any prime ideal of , if . Then is surjective.

Proof.

 (a’) (c’): Let be a prime ideal of , and . By Chapter 3, Exercise 23, So by Chapter 3, Exercise 26, Since is an open neighberhood of , and is open, each is an open neighberhood of , thus contains . Hence . (b’) (c’): has the going-down property for every and for every , there exists such that for every , the map is surjective.

Exercise 11. Let be a flat homomorphism of rings. Then has the going-down property.

Proof. By Chapter 3, Exercise 18, for every prime ideal of , let , then is surjective. Hence by Exercise 10, has the going-down propery.

Exercise 12. Let be a finite group of automorphisms of the ring , and let denote the subring of -invariants, that is of all such that for all . Prove that is integral over .

Proof. For every , let . Then , and Hence . So is integral over .

Let be a multiplicative closed subset of such that for all , and let . Show that the action of on extends to an action on , and that .

Proof. For every , define . If , i.e. exists such that , then , so . Hence this extension is well-defined.

For every , clearly ; and this induces a homomorphism . It’s easy to see is injective. Conversely, if , let , and , then , and . So , and hence is surjective. Thus is isomorphic to .

Exercise 13. In the situation of Exercise 12, let be a prime ideal of , and let be the set of prime ideals of whose contraction is . Show that acts transitively on . In particular, is finite.

Proof. Clear if a prime ideal of satisfies , then for any , . Hence acts on .

For any two prime ideals , and for any , we have . In particular, , hence there exists such that ; that is, . So . Thus, for some .

By Cor 5.9, since is integral over , and , we have . Hence acts transitively on .

Exercise 14. Let be an integrally closed domain, its field of fractions and a finite normal separable extension of . Let be the Galois group of over and let be the integral closure of in . Show that for all , and that .

Proof. If , then there exists monic polynomial such that . Suppose , then for any , we have Thus . Conversely , i.e. . Hence .

For the second proposition, we have .

Exercise 15. Let be as in Exercise 14, let be any finite extension field of , and let be the integral closure of in . Show that, if is any prime ideal of , then the set of prime ideals of which contract to is finite (in other words, that has finite fibers).

Proof. Since every field extension can decompose to a separable extension and a purely inseparable extension, it reduce to this two cases.

If is separable, then can embed in a Galois extension of . Let denote the integral closure of in , then by Exercise 14 we have . By Exercise 13, for any , there exists finite prime ideals of , which contract to . Since is integral over , every prime ideal of is the contraction of some prime ideal of . Thus there is only finite prime ideals of which contract to .

If is purely inseparable, is a prime ideal of , is a prime ideal of such that , then for every , since is purely inseparable, there exists such that , so . Conversely if , then must belong to . Hence is unique determined by . In particular, is bijective.

### Noether’s normalization lemma

Exercise 16. Let be a field and let be a finitely generated -algebra. Then there exists elements which are algebraically independent over and such that is integral over .

We shall assumen that is infinite. (The result is still true if is finite, but a different proof is needed.) Let generate as -algebra. We can renumber so that are algebraically independent over and each of is algebraic over . Now proceedby induction on . If there is nothing to do, so suppose and the result true for generators. The generator is algebraic over , hence there exists a polynomial in variables such that . Let be the homogeneous part of highest degree in . Since is infinite, there exists such that . Put (). Show that is integral over the ring , and hence that is integral over . Then apply the inductive hypothesis to to complete the proof.

Proof. Let . Consider the leading coefficient of , clearly it’s . Hence is a monic polynomial in , and So is algebraic over . Thus is algebraic over . Then apply the inductive hypothesis to .

From the proof it follows that may be chosen to be linear combinations of . This has the following geometrical interpretation: if is algebraically closed and is an affine algebraic variety in with coordinate ring , then there exists a linear subspace of dimension in and a linear mapping of onto which maps onto .

Proof. Let denote the image of in , then is generated by . by the proposition above, since is algebraically closed (so infinite), there exists algebraically independent , such that is linear combinations of ; and is integral over .

Let , let , then define as Clearly restricts to on .

To show that maps onto , let be a given point of . Since is algebraically independent, there exists a ring homomorphism such that . Since is integral over , by Exercise 2 can extends to a homomorphism , such that . Let , then clearly .

### Nullstellensatz (weak form)

Exercise 17. Let be an affine algebraic variety in , where is an algebraically closed field, and let be the ideal of in the polynomial ring (Chapter 1, Exercise 27). If then is not empty.

Proof. If then the coordinate ring is not trivial. By Exercise 16, there exists and a surjective map . Hence is nonempty.

Deduce that every maximal ideal in the ring is of the form where .

Proof. Let be a maximal ideal of , and let . Then is a field. By Exercise 16, there exists algebraically independent , such that is integral over . Since is an integral domain, and is field, it follows that is a field, thus . So is integral over , i.e. algebraic over . Hence , cause is algebraic closed. Let denote the image of in . Then clearly

Exercise 18. Let be a field and let be a finitely generated -algebra. Suppose that is a field. Then is a finite algebraic extension of . (This is another version of Hilbert’s Nullstellensatz. The following proof is due to Zariski. For other proofs, see (5.24) (7.9).)

Let generate as a -algebra. The proof is by induction on . If the result is clearly true, so assume . Let and let be the field of fractions of . By induction hypothesis, is a finite algebraic extension of , hence each satisfies a monic polynomial equation with coefficients in , i.e. coefficients of the form where and are in . If is the product of the denominatoes of all these coefficients, then each of is integral over . Hence and therefore is integral over .

Suppose is transcendental over . Then is integrally closed, because it is a unique factorization domain. Hence is integrally closed (5.12), and therefore , which is clearly absurd. Hence is algebraic over , hence (and therefore ) is a finite extension of .

Proof. If is transcendental over , then is isomorphic to the polynomial ring over , hence is a Euclid domain, hence a UFD.

If is a UFD, is its field of fractions, and is integral over , then , where . Assume and is coprime (i.e. ), then , hence . But is coprime with , so must be a unit, hence . Thus is integrally closed.

If is a field, let be the unique factorization of . Since has infinite many prime elements, there exists a prime element distinct from . Let be the invert of in , then . This contradict the fact that is UFD.

Exercise 19. Deduce the result of Exercise 17 from Exercise 18.

Proof. If is a maximal ideal of , then is a finitely generated -algebra. By Exercise 18, is a finite algebraic extension of . But is algebraically closed, hence . Let be the image of in , then , and clearly .

Exercise 20. Let be a subring of an integral domain such that is finitely generated over . Show that there exists in and elements in , algebraically independent over and such that is integral over , where .

Proof. Let and be the field of fractions of . Then is finitely generated -algebra, and therefore by Exercise 16, there exists in algebraically independent over , and such that is integral over .

Let generate as -algebra, then each of satisfies a monic polynomial with coefficients in . Suppose where . Since and , there exists a common denominator of and , such that , and . Since ’s are algebraically independent over , must also be ’s. By , we know that . Hence each is integral over . Since generate as -algebra, they must generate as -algebra. So is integral over .

Exercise 21. Let be as in Exercise 20. Show that there exists in such that, if is an algebraically closed field and is a homomorphism for which , then can be extended to a homomorphism .

Proof. By Exercise 20, there exists algebraically independent in and in , such that is integral over , where . For any such that , first extend to (Since is algebraically independent, we can, for example, let ). Then since , can be extended to . By Exercise 2, is integral over , so can be extended to , hence to .

Exercise 22. Let be as in Exercise 20. If the Jacobson radical of is zero, then so is the Jacobson radical of .

Proof. Let be an element of , then we need to show there is a maximal ideal of which does not contain . Applying Exercise 21 to with its subring , we obtain in . Let be a maximal ideal of which does not contain , and . Then the canonical mapping extends to a homomorphism , where is an algebraic closure of . Its restriction satisfies , so . It’s sufficient to show is maximal. Since , and is -algebra, it follows is -algebra. But is algebraic over , hence is a field. So is maximal.

Exercise 23. Let be a ring. Show that the following are equivalent:

 1 Every prime ideal in is an intersection of maximal ideals. 2 In every homomorphic image of the nilradical is equal to the Jacobson radical. 3 Every prime ideal in which is not maximal is equal to the intersection of the prime ideals which contain it strictly.

Proof.

 i) ii): It’s sufficient to show in every , ii) holds. In this case, every prime ideal containing is an intersection of maximal ideals which containing . Hence So in the nilradical is equal to the Jacobson radical. ii) i): For every prime ideal of , by ii), in the nilradical is equal to the Jacobson radical. This means is equal to the intersection of all maximal ideals of which contains . i) iii): Obvious. iii) i): If i) doesn’t hold, then there exists prime ideal of which is not the intersection of all maximal ideals containing it. Clearly if iii) holds for then it holds for every quotient of . So passing to the quotient ring, we may assume is an integral domain, and the Jacobson radical is not zero.Suppose in the Jacobson radical of , then since is a domain. Let be a maximal ideal in , and be its contraction in . Then , and is maximal with respect to this property. Since is in the Jacobson radical, can’t be maximal. But any prime ideal which contains strictly must contain , hence So is not equal to the intersection of the prime ideals which contain it strictly.

A ring with the three equivalent properties above is called a Jacobson ring.

Exercise 24. Let be a Jacobson ring (Exercise 23) and an -algebra. Show that if is either (i) integral over or (ii) finitely generated as an -algebra, then is Jacobson.

In particular, every finitely generated ring, and every finitely generated algebra over a field, is a Jacobson ring.

Proof.

 1 For any prime ideal of , we must show the Jacobson radical in is zero. Let denote , and the image of in . Then is an integral domain, and since is a Jacobson ring, the Jacobson radical of is zero. Let be the Jacobson radical of , then by Exercise 5, is equal the Jacobson radical of , so zero. If , suppose where , and we may assume since is an integral domain. Then , contradiction. So , and is a Jacobson ring. 2 For any prime ideal of , let and be as in (i). Then is an integral domain, so is . Hence in the Jacobson radical is zero. By Exercise 22, since is finitely generated over , so the Jacobson radical of is also zero.

Exercise 25. Let be a ring. Show that the following are equivalent:

 1 is a Jacobson ring; 2 Every finitely generated -algebra which is a field is finite over .

Proof.

 i) ii): By consider the image of in , we may assume that is a subring of . Let be in Exercise 21, then there exists a maximal ideal of which doesn’t contain . Let , and extend the canonical map to by Exercise 21, where is an algebraically closure of . Since is a field and , is injective. But , so , and is a field. Since can be embeded into the algebraic closure of , it is algebraic over . Thus is finite over . ii) i): If is a prime ideal of which is not maximal, let . Let , then is finitely generated (by ) over . If it’s a field, then it’s finite over , hence integral over , therefore is a field3, contradiction. So is not a field. Let be a maximal ideal of , and . Then is not zero and . Hence in , the zero ideal is the intersection of all non-zero prime ideals. Thus is Jacobson.

Exercise 26. Let be a topological space. A subset of is locally closed if it is the intersection of an open set and a closed set, or equivalently if it is open in its closure.

Proof. If , where open and closed, let denote its closure, then . Also, , hence . So . So is open in .

Conversely, if is a subset such that is open in its closure , then by definition there exists an open subset of such that .

The following conditions on a subset of are equivalent:

 1 Every non-empty locally closed subset of meets ; 2 For every closed set in we have ; 3 The mapping of the collection of open sets of onto the collection of open sets of is bijective.

A subset satisfying these conditions is said to be very dense in .

Proof.

 (1) (2): If is a closed set and , then for each open neighborhood of , meets , thus meets . Hence , so . Obviously , hence . (2) (3): If , let , then . Hence , so . (3) (1): Let be subsets of open or closed respectively. If , then where denotes . Hence , thus .

If is a ring, show that the following are equivalent:

 1 is a Jacobson ring; 2 The set of maximal ideals of is very dense in ; 3 Every locally closed subset of consisting of a single point is closed.

Proof.

 i) ii): For every closed, we will show In the nilradical is equal to the Jacobson radical. Let , then . Clearly corresponds to the Jacobson radical in , and corresponds to the nilradical in . Hence the nilradical and the Jacobson radical in are equal. i) iii): If , write as then there exists such that . Hence iii) is equivalent to: if where closed, i.e. is maximal in , then . This means if is maximal along all prime ideals which containing , then is maximal. But there exists such if and only if is not the intersection of all prime ideals containing it strictly. Hence iii) is equivalent to i).

### Valuation rings and valuations

Exercise 27. Let be two local rings. is said to be dominate if is a subring of and the maximal ideal of is contained in the maximal ideal of (or, equivalently, if ). Let be a field and let be the set of all local subrings of . If is ordered by the relation of denomination, show that has maximal elements and that is maximal if and only if is a valuation ring of .

Proof. To show has maximal elements, we should show it satisfies the conditions of Zorn’s lemma.

Clearly (with maximal ideal ). Hence is not empty. Let be a chain in with maximal ideals . Then is a subring of , and clearly is an ideal of . So it’s sufficient to show is the unique maximal ideal of .

Let be the residue field of , then the inclusion map (when ) induces a embedding (since ). Let be the direct limit of the system , then by definition is a field.

Since for any , we have the commutative diagram between the short exact sequence

[TikZ 编译错误]

By Chapter 2, Exercise 18, 19, there is a exact sequence Hence is a maximal ideal of . If is a ideal of , then for any , hence . So is the unique maximal ideal of , and is an upperbound of the chain.

If is a maximal element of , let be its maximal ideal and its residue field. Let be an algebraic closure of . Then there is a canonical map . We finish this part with Theorem 5.214. Let denote the poset as in Theorem 5.21, and an element in such that , that is, is a subring and . let , then is a maximal ideal of , and . Hence dominate , and by the maximality of we know . Thus is a maximal element in , hence a valuation ring of .

On the other hand, if is a valuation ring of , then it’s local. Let denote the maximal ideal of . If is dominated by , Then is also a valuation ring of . We must show that . Let and . Then and (Here ). Since we must have . But , hence .

Exercise 28. Let be an integral domain, its field of fractions. Show that the following are equivalent:

 1 is a valuation ring of ; 2 If are any two ideals of , then either or .

Deduce that if is a valuation ring and is a prime ideal of , then and are valuation rings of their fields of fractions.

Proof.

 (1) (2): If and , then there exists and . Hence and , contradiction. (2) (1): For any such that , we have or , hence or . So is a valuation ring of .
If is a valuation ring, then it is a valuation ring of , so it satisfies (2). Clearly and also satisfies (2), hence are valuation rings of their fields of fractions.

Exercise 29. Let be a valuation ring of a field . Show that every subring of which contains is a local ring of .

Proof. If is a subring of , then let as We will show .

If , then . Hence , so . Conversely, if , then , and , thus . Hence .

Exercise 30. Let be a valuation ring of a field . The group of units of is a subgroup of the multiplicative group of .

Let . If are represented by , define to mean . Show that this defines a total ordering on which is compatible with the group structure (i.e.