Bourbaki

Exercise 8, § 3

Algebra, VII, p. 61

Bourbaki marks this one as harder than the rest.

Let A be a principal ideal domain which is not a field. a) Show that the product A-module A'' admits no countable system of generators. (If A is countable, consider Card(A^N); otherwise notice that, if for all x \in A we let v_x denote the element (xn)n≥0(x^n)_{n \geq 0} of A'', then the vxv_x form a free system). b) Let π\pi be an irreducible element of A. Let S denote the submodule of A'' consisting of those sequences z=(zn)n∈Nz = (z_n)_{n \in \mathbf{N}} for which there exists a sequence (k(n))n∈N(k(n))_{n \in \mathbf{N}} of integers tending to +∞+\infty (depending on z), such that zn∈Aπk(n)z_n \in A \pi^{k(n)} for all n. Show that the (A/π)(A/\pi)-vector space S/πSS/\pi S has a countable basis. c) Deduce from a) and b) that ANA^N is not a free A-module. (Otherwise, S would be a free A-module with a countable basis; but (x,(x, (xnπn)(x_n \pi^n) is an injective A-linear map from ANA^N into S.)

Solution

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Transcribed from the printed volume. Not translated.