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Let $B \subseteq C$ be an inclusion of finite dimensional (associative) algebras over a field $k$. Assume that $C$ is a free $B$-module. Let $\bigoplus_i U_i$ be a decomposition of $B$ into indecomposable $B$-modules. Let $\bigoplus_j V_{ij}$ be a decomposition of the extension $C \otimes_B U_i$ into indecomposable $C$-modules.

Question: Pick $u \in U_i$ nonzero. Is there $\varphi_u \in Hom_B(U_i,V_{ij})$ such that $\varphi_u(u)$ is nonzero?

Remark: By Frobenius reciprocity, $Hom_B(U_i,V_{ij}) \simeq Hom_C(C \otimes_B U_i,V_{ij})$ which is nonzero.
In the semisimple case, Schur's lemma implies the existence of an injective map $\varphi$, which solves the problem for all $u$ nonzero. In the non-semisimple case, we cannot expect the existence of such an injective map, we are just asking about the existence of a map $\varphi_u$ as above with $u \not \in ker(\varphi_u)$.

Assuming a positive answer, some conditions could be useless, but it is exactly the case I need.

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No, not necessarily.

Consider the case $B=kH$, $C=kG$ of finite group algebras over a field $k$, where $H\leq G$. I'll write $\downarrow$ and $\uparrow$ for restriction and induction. This case has a couple of simplifying features. Firstly, induction is both left and right adjoint to restriction. Secondly, $kH$ is selfinjective, and so the projective module $U_i$ is also injective. If the answer to the question were "yes", then there would be an injection from $U_i$ to a direct sum of copies of $V_{ij}\!\downarrow$. But since $U_i$ is injective, this would split, and so, by Krull-Schmidt, $U_i$ would be a direct summand of $V_{ij}\!\downarrow$.

Suppose that $U_i$ is the projective cover of the simple $kH$-module $S$ and $V_{ij}$ the projective cover of the simple $kG$-module $T$.

Then $V_{ij}$ is a direct summand of $U_i\!\uparrow$ if and only if $$\text{Hom}_{kG}(U_i\!\uparrow, T)\cong\text{Hom}_{kH}(U_i,T\!\downarrow)$$ is nonzero, which is the case if and only if $T\!\downarrow$ has a composition factor isomorphic to $S$.

$U_i$ does not occur as a direct summand of $V_{ij}\!\downarrow$ if and only if $$\text{Hom}_{kH}(V_{ij}\!\downarrow,S)\cong\text{Hom}_{kG}(V_{ij},S\!\uparrow)$$ is zero, which is the case if and only if $S\!\uparrow$ does not have a composition factor isomorphic to $T$.

So we just need to find simple modules $S$ for $kH$ and $T$ for $kG$ for which $S$ is a composition factor of $T\!\downarrow$ but $T$ is not a composition factor of $S\!\uparrow$.

This happens quite commonly, but $S$ can't appear in the head or socle of $T\!\downarrow$, or else there would be a map in one direction between $S$ and $T\!\downarrow$, and hence between $S\!\uparrow$ and $T$, and so $T$ would appear in the socle or head of $S\!\uparrow$. So $T$ has to be reasonably large.

One example that I happen to be familiar with is where $k$ is algebraically closed of characteristic two, $G=A_5$ and $H=A_4$. Then $kG$ has a $4$-dimensional simple projective module $T$ and $kH$ has three one-dimensional simple modules: take $S$ to be one of the non-trivial ones. Then $S$ and $T$ have the required properties.

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  • $\begingroup$ Do you expect the existence of a counter-example if the field is algebraically closed of characteristic zero? If yes, what about $k=\mathbb{C}$? $\endgroup$ – Sebastien Palcoux Mar 14 '19 at 12:25
  • $\begingroup$ @SebastienPalcoux Not with group algebras, of course, as they'd be semisimple. But for abstract algebras I wouldn't expect the characteristic of $k$ to matter. It might just be harder to construct the examples because you don't have the groups to play with. $\endgroup$ – Jeremy Rickard Mar 14 '19 at 12:39
  • $\begingroup$ Why the projective cover of a simple $kH$-module $S$ must be indecomposable and direct summand of $kH$? $\endgroup$ – Sebastien Palcoux Mar 14 '19 at 17:07
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    $\begingroup$ @SebastienPalcoux For any finite dimensional algebra, there’s a bijection between (isomorphism classes of) indecomposable projective modules and simple modules, given by taking the quotient by the radical in one direction and taking the projective cover in the other. $\endgroup$ – Jeremy Rickard Mar 14 '19 at 20:35

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