Bourbaki

Exercise 9, § 5

Commutative Algebra, II, 148

Bourbaki marks this one as harder than the rest.

(a) Let A be a ring, (fi)i∈I(f_i)_{i \in I} a finite family of elements of A generating the ideal A and M an A-module. Suppose given in each of the AfiA_{f_i}-modules MfiM_{f_i} an element ziz_i such that, for i≠ji \neq j, the canonical images of ziz_i and zjz_j in MfifjM_{f_if_j} are equal, Show that there exists a unique z∈Mz \in M such that, for all ii, the canonical image of zz in MfiM_{f_i} is equal to ziz_i. (Note that, for every integer k>0k > 0, there exists a family (gi)i∈I(g_i)_{i \in I} of elements of A such that ∑igifik=1\sum_i g_i f_i^k = 1.)

(b) Let A be a ring and P a projective A-module of rank 1. Show that the ring End⁡A(P)\operatorname{End}_A(P) is canonically isomorphic to A (use Theorem 3).

(c) Let A be a ring and P a projective A-module of rank nn. For every endomorphism uu of P, there corresponds canonically to AnuA^n u, by (b), an element det⁡(u)\det(u) of A, called the determinant of uu, such that det⁡(u∘v)=(det⁡u)(det⁡v)\det(u \circ v) = (\det u)(\det v) for two endomorphisms u,vu, v of P. The element χu(T)=det⁡(T⋅1−u)\chi_u(T) = \det(T \cdot 1 - u) of A[T]A[T] is called the characteristic polynomial of uu (cf. no. 3, Proposition 4); show that χu\chi_u is a monic polynomial of degree nn and that χu(u)=0\chi_u(u) = 0 (use (a) and § 3, no. 3, Theorem 1). Deduce that, for uu to be bijective, it is necessary and sufficient that det⁡(u)\det(u) be invertible in A. If A is an integral domain, χu(T)=∏i=1n(T−αi)\chi_u(T) = \prod_{i=1}^n (T - \alpha_i) the decomposition into the linear factors of χu\chi_u in an algebraically closed extension of the field of fractions of A, and χq(u)(T)=∏i=1n(T−q(αi))\chi_{q(u)}(T) = \prod_{i=1}^n (T - q(\alpha_i)) for every polynomial q∈A[T]q \in A[T]. Generalize similarly Proposition 14 of Algebra, Chapter VII, § 5, no. 6.

Solution

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Transcribed from the printed volume. Not translated.