Bourbaki

Exercise 7, § 10

Tag 00CN, permanent. Cite it as [Bourbaki, Tag 00CN].

Algebra, VIII, p. 179

Bourbaki marks this one as harder than the rest.

Let A be an Artinian ring, r\mathfrak{r} its radical, and A‾\overline{A} the ring A/rA/\mathfrak{r}. Let l\mathfrak{l} be a nonnilpotent indecomposable left ideal; we have l=Ae\mathfrak{l}= Ae, where ee is an idempotent in A.

a) Prove that the ideal n=l∩r\mathfrak{n}=\mathfrak{l}\cap \mathfrak{r} is the unique maximal submodule of l\mathfrak{l}. The A-module l/n\mathfrak{l}/\mathfrak{n} is simple and isomorphic to Ae‾A\overline{e}, where e‾\overline{e} is the image of ee in A.

b) Let M be an A-module. Then (Ae)M(Ae)M is a submodule of M, and e((Ae)M)=eMe((Ae)M)=eM is an eAeeAe-module. If N is an A-submodule of M, then the eAeeAe-modules e(M/N)e(M/N) and eM/eNeM/eN are isomorphic. If the A-module M is simple, then either we have eM=0eM = 0, or M is isomorphic to Ae‾A\overline{e} and eMeM is a simple eAeeAe-module.

c) Prove that the length of the eAeeAe-module eMeM is equal to number of quotients of a Jordan–Hölder series of the A-module M isomorphic to Ae‾A\overline{e}.

d) Suppose that M has finite length and admits a unique maximal submodule N and that M/NM/N is isomorphic to l/n\mathfrak{l}/\mathfrak{n}. Prove that M is isomorphic to a quotient of AeAe (observe that an element xx of eMeM that does not belong to N generates M).

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Transcribed from the printed volume. Not translated.