Exercise 10, § 5
Commutative Algebra, II
So let A be a ring, the set of classes of finitely generated projective A-modules; let G be the -module of formal linear combinations of the elements of , N the submodule of G formed by the elements , where are the classes of three finitely generated projective A-modules M, M', M'' such that there exists an exact sequence
(it amounts to the same thing to say that M is isomorphic to the direct sum by virtue of Alg., chap. II, 3rd ed., § 2, no 2). Let be the -module quotient G/N, and for every finitely generated projective A-module M, let (or ) be its class in .
a) Now prove that there exists on one and only one structure of a commutative ring whose addition is that of the -module and whose multiplication is such that for two finitely generated projective A-modules, M, N; is the unit element of .
b) For every finitely generated projective A-module M and every integer denote by the element of . Now prove that one has . Denote by the element in the ring of formal series in one indeterminate over the ring ; now prove that ; deduce a homomorphism again denoted from into the multiplicative group of formal series of whose constant term is equal to 1. The coefficient of in will be denoted by ; one has
for in .
c) If A is a local ring, the ring identifies canonically with , and one has . Deduce that, if A is a direct composition of m local rings, is isomorphic to .
d) Let be a homomorphism of rings. Now prove that there exists one and only one homomorphism of rings such that for every finitely generated projective A-module M; one has for every ; if is a second homomorphism of rings, one has .
e) Let be a homomorphism of rings, for which B is a finitely generated projective A-module; for every finitely generated projective B-module, is then a finitely generated projective A-module; deduce that there exists a homomorphism of -modules such that . Now prove that one has, for and , . If is a second homomorphism of rings for which C is a finitely generated projective B-module, then one has .
Solution
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