Bourbaki

Exercise 22, § 4

Commutative Algebra, VII, 577

Bourbaki marks this one as harder than the rest.

Let A be a local ring, m its maximal ideal and E a finitely generated A-module.

(a) If r=γ(E)r = \gamma(E) is the least of the cardinals of a system of generators of E, r is also the least integer h such that the determinantal ideal dh(E)d_h(E) is equal to A and also the least of the integers k such that ⋀k+1E=0\bigwedge^{k+1} E = 0 (note that r is the rank of E/mE over A/m).

(b) If we write e(E)=dr−1(E)e(E) = d_{r-1}(E), show that ⋀rE\bigwedge^r E is isomorphic to A/e(E)A/e(E). For an ideal a of A to contain e(E), it is necessary and sufficient that E/aE be a free (A/a)-module (reduce it to the case where a=0a = 0).

(c) If e(E) is a principal ideal AαA\alpha, show that E contains a direct factor isomorphic to A/AαA/A\alpha (write E as a quotient of ArA^r by a submodule R and note that there is in R an element of the form αy\alpha y, where y is an element of a basis of ArA^r).

(d) Let dh′(E)d'_h(E) be the annihilator of ⋀h+1E\bigwedge^{h+1} E for 0≤h≤r−10 \leq h \leq r-1. Show that the following conditions are equivalent: (α) The ideals dh(E)d_h(E) (0≤h≤r−10 \leq h \leq r-1) are principal. (β) The ideals dh′(E)d'_h(E) (0≤h≤r−10 \leq h \leq r-1) are principal. (γ) E is the direct sum of r modules isomorphic to A/AλnA/A\lambda_n, where λn+1\lambda_{n+1} divides λn\lambda_n for 0≤h≤r−10 \leq h \leq r-1. Moreover, if these conditions hold, then dh′(E)=Aλhd'_h(E) = A\lambda_h for 0≤h≤r−10 \leq h \leq r-1 and dh(E)=λnλn+1…λr−1Ad_h(E) = \lambda_n \lambda_{n+1} \ldots \lambda_{r-1} A for 0≤h≤r−10 \leq h \leq r-1. (Proceed by induction on r, using (c).)

(e) Suppose further that A is an integral domain. Show that, if p is a prime ideal of A such that E/pEE/pE is a free (A/p)-module and E, a free A,-module, then E is a free A-module (using (b), show that γ(Ep)=γ(E)\gamma(E_p) = \gamma(E) and that e(Ep)=e(E)e(E_p) = e(E)).

Solution

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Transcribed from the printed volume. Not translated.