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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 9.

## Discrete Valuation Rings and Dedekind Domains

Exercise 1. Let be a Dedekind domain, a multiplicatively closed subset of . Show that is either a Dedekind domain or the field of fractions of .

Proof. is a Noetherian domain of dimension at most one, and is integrally closed. If the dimension of is one then it is a Dedekind domain; otherwise it’s the field of fractions of .

Suppose that , and let be the ideal class groups of and respectively. Show that extension of ideals induces a surjective homomorphism .

Proof. Let denote the groups of ideals of and respectively, and denote the subgroups of principal fractional ideals, and denote the field of fractions of (and ). For every , the extension ; and clearly . Hence the extension of ideals induces a homomorphism . If in , let be the fractional ideal of generated by , then is pricisely the fractional ideal of generated by . Hence the homomorphism is surjective, and maps into . Therefore it induces a surjective homomorphism .

Exercise 2. Let be a Dedekind domain. If is a polynomial with coefficients in , the content of is the ideal in . Prove the Gauss’s Lemma that .

Proof. First assume is a discrete valuation ring with the maximal ideal . If , then such that . So it’s sufficient to show that if then . Passing to the residue field , it’s equivalent to that if are non-zero, then , which is clearly true.

For the case that is a Dedekind domain, for every prime ideal and every polynomial we have (where is the image of in and is the extension of the ideal in ). By the result above, , so for every prime ideal . Hence .

Exercise 3. A valuation ring (other than a field) is Noetherian if and only if it is a discrete valuation ring.

Proof. : Obvious.

: Let be a Noetherian valuation ring, be its valuation. Let . For every , let , then is an ideal of , and if and only if . Since is Noetherian, there is no infinite decreasing chain in . So has a minimal element, named . If , then there exists such that for any natural number . Assume for some . Then there exists such that . Thus , contradicts the choice of . So for every natural number . But in this case we have a infinite decreasing chain also a contradiction. Hence .

Exercise 4. Let be a local domain which is not a field and in which the maximal ideal is principal and . Prove that is a discrete valuation ring.

Proof. Suppose . Since , for every non-zero , there exists such that . Let . Then clearly and . If , then for some ; and since . Hence is a unit, therefore .

Extend to the field of fractions of by . If , then hence . So , and . Since , it follows that for any . Therefore is surjective. Hence is a discrete valuation ring.

Exercise 5. Let be a finitely generated module over a Dedekind domain. Prove that is flat is torsion-free.

Proof. Since torsion-freeness and flatness are local properties, we can passing to the localization, and assmue that is a finitely generated module over a discrete valuation ring.

Suppose is a discrete valuation ring with the maximal ideal , is a finitely generated module over . By Chapter 7, Exercise 15, is flat if and only if the mapping is injective. But as modules, by . Hence . So is flat if and only if , i.e. . If , and , then for some unit , hence Hence is equivalent to that is torsion-free. So is flat if and only if it is torsion-free.

Exercise 6. Let be a finitely-generated torsion module () over a Dedekind domain . Prove that is uniquely representable as a finite direct sum of modules , where are non-zero prime ideals of .

Proof. Suppose is a discrete valuation ring with the maximal ideal . Then is a PID, so by the structure theorem of finitely generated module over PID, is uniquely representable as a finite direct sum of modules , for primary ideals . Since the only primary ideals of is , it follows that uniquely.

In the case where is a Dedekind domain, for each prime ideal , as above. By the exactness of localization, we have . Hence can be written as the finite direct sum of modules uniquely. Now it’s sufficient to show that there are only finite such that is not zero, and that the homomorphism is an isomorphism.

By Chapter 3, Exercise 19, since is finitely generated, we have . Let generate , then for any . Since is prime in , it follows that . Hence is Artinian and hence has finite prime ideals. Therefore is finite.

Let , then it remains to show that is an isomorphism. Since being isomorphism is a local property, it’s sufficient to show each is an isomorphism, which is obvious.

Exercise 7. Let be a Dedekind domain and an ideal in . Show that every ideal in is principal.

Proof. Since , it follows that is a Noetherian ring of dimension , hence an Artin ring. So there are finitely many prime ideals containing , and . Let be an prime ideal of , then is a quotient ring of . Therefore the maximal ideal of is principal for that is a discrete valuation ring.

Let denote the prime ideals of , then . If is generate by in , then it is generate by in . Hence every maximal ideal is principal. For every ideal in , let be a primary decomposition of . Then clearly every is a power of some maximal ideal, and hence is principal.

Deduce that every ideal in can be generated by at most elements.

Proof. Let be an non-zero ideal in . Let be any non-zero ideal in , then is principal by the result above. Let which image generates , then clearly .

Exercise 8. Let be three ideals in a Dedekind domain. Prove that

Proof. By localizing at each prime ideal, then we may assume the ring is a discrete valuation ring, in which every ideal is either or a power of the maximal ideal . We have and . Then the result is obvious.

Exercise 9 ((Chinese Remainder Theorem)). Let be ideals and let be elements in a Dedekind domain . Then the system of congruences has a solution in whenever .

Proof. (According to the hint) The propsition is equivalent to that the sequence of -modules is exact, where

To show the sequence is exact, it is enough to show that it is exact when localized at any . Hence it reduces to the case where is a discrete valuation ring.

If is a discrete valuation ring with the maximal ideal , then for every . In this case for every , either ot . Therefore the result is trivial.