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Problem: find indecomposable injective modules and indecomposable projective modules over $k[x]/ \langle x^n \rangle. (n\geq 2)$.

From this post, I learned that any indecomposable module over $k[x]/ \langle x^n \rangle$ is of the form $k[x]/ \langle x^i \rangle, 1\leq i \leq n$. Hence the problem reduces to: pick out the injective modules and projective modules from $k[x]/ \langle x^i \rangle, 1\leq i \leq n$.

I could determine that $k[x]/ \langle x^n \rangle$ is projective, since it's a free module. And from this post $k[x]/ \langle x^n \rangle$ is also an injective module.

What about the rest? Any help is appreciated.

cxh007
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1 Answers1

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First, every finitely generated indecomposable projective module of a finite-dimensional algebra $A$ is a summand of the regular module $A\in\mathrm{mod}A$. Think about what this means in your case.

Second, if $I$ is an injective module and $f:I\to M$ an injective morphism then, by the definition of an injective module, there exists a morphism $g:M\to I$ such that $gf = \mathrm{id}_I$. You can apply this to the inclusions $M_i\to M_{i+1},\: [1]\mapsto [x]\;$ (where $M_i:=k[x]/\langle x^i\rangle$) to prove non-injectivity of many of the modules $M_i$. Then you should be able to sort out the rest of it for yourself.

Erik D
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  • Could I have a reference about 'every finitely generated indecomposable projective module of a ring $R$ is a summand of the regular module $R\in\mathrm{mod}R$'? Thanks! – cxh007 Oct 26 '21 at 16:54
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    Thinking of it, you most likely need some finiteness condition on the $R$ for this to be true (left Artinian would be sufficient). In your case it is ok, since the ring is a finite-dimensional algebra. The point is that the projective modules are precisely the direct summands of free modules, so a finitely-generated projective $P$ is a direct summand of $R^{\oplus n}$ for some $n$. Then apply the Krull–Schmidt theorem.

    A proof over finite-dimensional algebras can be found, for example, in M. Barot: Introduction to the representation theory of algebras (Springer, Cham, 2015)

     – Erik D Oct 27 '21 at 08:03