Bourbaki

Exercise 7, § 2

Commutative Algebra, I, 42

(a) With the notation of Exercise 6, show that, if M is a projective module, then λ(M)=−1\lambda(M) = -1 or λ(M)=+∞\lambda(M) = +\infty. If A is a left Noetherian ring, then, for every A-module M, λ(M)=−1\lambda(M) = -1 or λ(M)=+∞\lambda(M) = +\infty.

(b) If aa is a left ideal of a ring AA, which is not finitely generated, As/aA_s/a is a monogenous AA-module which does not admit a finite presentation, in other words λ(As/a)=0\lambda(A_s/a) = 0 (no. 8, Lemma 9).

(c) Give an example of a monogenous left ideal aa of a ring AA such that As/aA_s/a (which has a finite presentation) admits a dual which is not a finitely generated right AA-module.

(d) Let KK be a commutative field, EE the vector space K(N)K^{(\mathbf{N})}, (en)(e_n) the canonical basis of EE and TT the tensor algebra of EE, which therefore has a basis consisting of the finite products ei1ei2…eike_{i_1}e_{i_2}\ldots e_{i_k} (k⩾0,ij∈Nk \geqslant 0, i_j \in \mathbf{N} for all jj). For a given integer nn, let bb be the two-sided ideal of TT generated by the products

e1e0,e2e1,…,enen−1 e_1e_0, e_2e_1, \ldots, e_ne_{n-1}

and en+kene_{n+k}e_n for all k⩾1k \geqslant 1; let AA be the quotient ring T/bT/b, and for every integer mm, let ama_m be the canonical image of eme_m in AA. Show that, if M=As/Aa0M = A_s/Aa_0, then λ(M)=n\lambda(M) = n (observethat, for m⩽n−1m \leqslant n - 1, the left annihilator of aa, is Aam+Aa_{m+}, and use Exercise 6(b)).

Solution

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Transcribed from the printed volume. Not translated.