Bourbaki

Exercise 19, § 2

Commutative Algebra, III

Bourbaki marks this one as harder than the rest.

a) Let EE be an AA-module which is linearly compact. Now prove that the following conditions are equivalent:

α) For every continuous linear mapping uu of EE into a separated linearly topologized AA-module FF, uu is a strict morphism (Top. gén., chap. III, 2e éd., § 2, no 8) of EE into FF.

β) For every closed submodule FF of EE, the quotient topology on E/FE/F is a minimal topology (Exercise 17).

γ) EE is complete and there is a fundamental system (Uλ)(U_\lambda) of open neighbourhoods of 0 in EE, formed by submodules, such that the E/UλE/U_\lambda are artinian modules.

(To see that γ) implies β), reduce to the case where F=0F = 0, and use Exercises 17 and 18.)

When EE satisfies the equivalent conditions α), β), and γ), one says that EE is strictly linearly compact.

b) Let EE be a separated linearly topologized AA-module, FF a closed submodule of EE. For EE to be strictly linearly compact it is necessary and sufficient that FF and E/FE/F be so.

c) Any projective limit of strictly linearly compact modules is strictly linearly compact.

d) Let EE be a separated linearly topologized AA-module, uu a linear mapping of EE into a strictly linearly compact AA-module FF. Now prove that if the graph of uu is closed in E×FE \times F, uu is continuous (use Exercise 16 c), and the fact that if MM is closed in EE and E/ME/M artinian, MM is open in EE.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.