Bourbaki

Exercise 10, § 6

General Topology, III, 317

Bourbaki marks this one as harder than the rest.

a) Let BB be a ring with an identity element, let EE be a free BB-module Bd(I)B_d^{(I)}, and let AA be the ring L(E)\mathscr{L}(E) of endomorphisms of EE. For every finite subset FF of EE, let VFV_F be the set of all u∈Au \in A such that u(x)=0u(x) = 0 for all x∈Fx \in F. Show that the VFV_F from a fundamental system of neighbourhoods of 00 for a Hausdorff topology on AA, which is compatible with the ring structure of AA, and with respect to which AA is totally disconnected.

b) Take I=NI = \mathbf{N} and take BB to be a commutative ring which has no zero-divisors but whose radical R\mathfrak{R} is not {0}\{0\}. Show that the radical of AA is not closed. [Let (en)(e_n) be the canonical basis of EE. Remark on the one hand that each u∈Au \in A, such that u(en)=0u(e_n) = 0 for all but a finite number of indices and such that u(E)⊂R⋅Eu(E) \subset \mathfrak{R} \cdot E, belongs to the radical of AA; consider on the other hand the element u0u_0 of AA such that u0(en)=en+ten+1u_0(e_n) = e_n + te_{n+1} with t≠0t \neq 0 in R\mathfrak{R}, for each nn].

Solution

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Transcribed from the printed volume. Not translated.