Exercise 10, § 5
Commutative Algebra, X, p. 168
Let be a discrete valuation ring, its normalized valuation, its field of fractions. Let us consider a noetherian -algebra endowed with a homomorphism of -algebras . We denote by the kernel of , and by its annihilator in .
(a) Prove that the following conditions are equivalent: (i) the -module is of finite length ; (ii) the local ring of at the maximal ideal is isomorphic (as a -algebra) to ; (iii) one has .
We now suppose that these conditions are satisfied ; we denote by , or simply , the length of the -module , and by , or simply , that of the -module .
(b) Let be an ideal of contained in ; we consider the -algebra and the homomorphism deduced from . Prove that also satisfies the conditions of (a), and that one has and .
(c) Suppose that the -module is free of finite type. Prove that is a free -module of rank one, a direct factor in , and that one has for all .
(d) Suppose moreover that is a Gorenstein ring ; denote by a basis of the -module (§ 3, exerc. 5), and by a generator of the different ideal of over (exerc. 7, b)). Prove the equality (prove that is equal to , and use (c)).
e) We place ourselves in the situation of b), with ; we suppose that the -algebras and are finite and flat, and that is a Gorenstein ring. Prove that one has (observe that the image of in is the socle of , hence is contained in ).
Solution
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