Timeline for Injective modules
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2022 at 15:52 | comment | added | RumDiary | @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? | |
| Sep 10, 2022 at 9:18 | comment | added | Peter Kropholler | 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 $M=M^{**}\rightarrowtail A^*$. | |
| Sep 10, 2022 at 8:04 | history | edited | YCor | CC BY-SA 4.0 |
added part of the argument
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| Sep 10, 2022 at 8:01 | comment | added | Jeremy Rickard | @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. | |
| Sep 10, 2022 at 7:53 | comment | added | YCor | 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? | |
| Sep 10, 2022 at 7:37 | history | answered | Jeremy Rickard | CC BY-SA 4.0 |