Let $A$ be a finite dimensional $k$-algebra and let $I$ be an injective module。My question is whether $I$ is a direct sum of finite-dimensional injective modules。
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Yes.
Let $M$ be any $A$-module. Then its socle is a direct sum of simple modules: $\operatorname{soc}A=\bigoplus_iS_i$.
$A$ is a finite dimensional algebra, so the dual $\mathrm{Hom}_k(A,k)$ of $A$ is a finite dimensional injective into which every simple module embeds. So each $I(S_i)$ is finite-dimensional.
The direct sum $I=\bigoplus_iI(S_i)$ of the injective envelopes of the simples is injective, so the natural inclusion $\operatorname{soc}A\to I$ extends to a map $M\to I$, which it is easy to see is an essential monomorphism, and so $I$ is the injective envelope of $M$.
If $M$ itself is injective, then $M=I=\bigoplus_iI(S_i)$ is a direct sum of finite dimensional injective modules.
answered Sep 10, 2022 at 7:37
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$\begingroup$ This argument (which I would have rephrased as: if $I$ is injective indecomposable, then its socle is simple) boils down to showing that the injective hull of a simple module is finite-dimensional. How do you see this? $\endgroup$– YCorCommented Sep 10, 2022 at 7:53
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1$\begingroup$ @YCor $A$ is a finite dimensional algebra, so the dual $\operatorname{Hom}_k(A,k)$ of $A$ is a finite dimensional injective into which every simple module embeds. $\endgroup$ Commented Sep 10, 2022 at 8:01
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$\begingroup$ When passing to the dual $A^*:=\hom_k(A,k)$ you switch from left modules to right modules and right modules to left modules. So you maybe want to use the double dual. If $M$ is irreducible then so is $M^*$. Thus there is a surjection $A\twoheadrightarrow M^*$ and thus there is an embedding $M=M^{**}\rightarrowtail A^*$. $\endgroup$ Commented Sep 10, 2022 at 9:18
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$\begingroup$ @JeremyRickard Do you know if the analogous holds for graded algebras? So if, say, $A$ is graded over $\mathbb{Z}$ (or perhaps a torsion free group, or any abelian group) and $I$ is graded injective, then is $I$ a direct sum of finite dimensional graded injectives? $\endgroup$– RumDiaryCommented Nov 4, 2022 at 15:52