Exercise 24, § 2

On a commutative ring A let \mathscr{T}_m(A) denote the linear topology which has a fundamental system of neighbourhoods of 0 the products (equal to the intersections) of a finite number of powers of maximal ideals (cf. no. 13, Proposition 17); if every finite intersection of ideals \neq 0 of A is \neq 0, let \mathscr{T}_u(A) denote the linear topology for which a fundamental system of neighbourhoods of 0 consists of all the ideals \neq \{0\} of A.

(a) Show that \mathscr{T}_c(A) (Exercise 23) is coarser than \mathscr{T}_m(A) (observe that the Jacobson radical of an Artinian ring is nilpotent); give an example where \mathscr{T}_c(A) \neq \mathscr{T}_m(A) (cf. Exercise 22 (c)). \* If A is the ring of a non-discrete valuation of height 1 (Chapter VI), then \mathscr{T}_c(A) = \mathscr{T}_m(A) and \mathscr{T}_u(A) \neq \mathscr{T}_m(A). \*

(b) The topology \mathscr{T}_m(A) is the coarsest linear topology on A under which the maximal ideals of A are closed and every power of an open ideal is an open ideal (note that under a linear topology on A, under which a maximal ideal m is closed, m is necessarily open).

(c) If A is a Noetherian ring, then \mathscr{T}_m(A) = \mathscr{T}_c(A) (observe that, if m is a maximal ideal of A, A/m^k is an Artinian ring for every integer k \geq 1, noting that m^h/m^{h+1} is an (A/m)-module which is necessarily of finite length). Give an example of a Noetherian local ring A for which \mathscr{T}_m(A) \neq \mathscr{T}_u(A) (consider a local ring of a polynomial ring over a field).