If a finite dimensional algebra $A$ over a field $\mathbb{k}$ is semisimple then any two sided ideal of $A$ is generated (as a left module) by a central idempotent, so its (unique) complement is a two sided ideal again. In general, this observation is wrong as e.g., in the $\mathbb{k}$-algebra of upper triangular ($2 \times 2$)-matrices the (non-central) matrix unit $e_{2,2}$ generates a two sided ideal $Ae_{2,2}$ having a complement $Ae_{1,1}$ which is not two sided.

However, I am not aware of such an example if $A$ is a group algebra.

Question: Let $\mathbb{k}$ be a field, and let $G$ be a finite group such that the algebra $\mathbb{k} G$ is not semisimple. Let $e \in \mathbb{k} G$ be an idempotent generating a two sided ideal $\mathbb{k} G e$. Does this imply that $e$ is central?

Note that since group algebras are self-injective, any left ideal $L$ isomorphic to a two sided ideal $I$ must be equal to $I$. So if $\mathbb{k} G e$ has any two sided complement then it has to be unique, and $e$ must be central.

  • $\begingroup$ In a left self-injective group ring, any left ideal isomorphic to a two-sided ideal is equal to that ideal? How does that come about? $\endgroup$ – rschwieb May 22 '18 at 19:12
  • $\begingroup$ Oh, I think i can see the way now: if $\phi:I\to L$ is an isomoprhism of left modules, then self-injectivity gives you an element $x\in R$ such that $ix=\phi(i)$ for all $i\in I$, so that $L = Im\phi \subseteq I$. Then finite dimensionality should make $Im\phi=I$. That's an interesting facet I hadn't seen before. $\endgroup$ – rschwieb May 22 '18 at 20:15

If $A$ is any finite dimensional algebra and $e$ any idempotent, then $A=Ae\oplus A(1-e)$ as a direct sum of left ideals. The endomorphism ring of $A$ as a left module is $A^{op}$ acting by right multiplication, so the condition that $Ae$ is a two sided ideal is equivalent to $\text{Hom}_A\left(Ae,A(1-e)\right)=0$.

But for a finite group algebra this is equivalent to $\text{Hom}_A\left(A(1-e),Ae\right)=0$, and hence to $A(1-e)$ also being a two sided ideal, since finite group algebras are examples of symmetric algebras, for which $\text{Hom}_A(P,Q)$ is naturally dual to $\text{Hom}_A(Q,P)$ for finitely generated projective modules $P$ and $Q$.

  • $\begingroup$ Thank you very much! Could you provide a little more details (or a reference) on the natural duality you mentioned? I only found that $\dim \mathrm{Hom}(P,Q) = \dim \mathrm{Hom}(Q,P)$ in that situation (Lemma 16.75 in Lam's Lectures on Rings and Modules). $\endgroup$ – Dune May 23 '18 at 10:03
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    $\begingroup$ @Dune In Section 3 of my paper sciencedirect.com/science/article/pii/S0021869302005203 I collect various equivalent conditions to "symmetric". Sorry for the self-reference: nothing in that section is original, but I seem to remember that when I wrote that, I couldn't find a reference that said what I needed in quite the way I wanted. Alternatively, Broue's paper "Higman's criterion revisited" projecteuclid.org/euclid.mmj/1242071686 contains more about symmetric algebras than any human needs to know. Of course, for your question the equality of dimensions is all you need. $\endgroup$ – Jeremy Rickard May 23 '18 at 10:27
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    $\begingroup$ I came (nearly) to the same conclusion this way: since $eR\subseteq Re$, $R\cong \begin{bmatrix}eRe & 0 \\(1-e)Re&(1-e)R(1-e)\end{bmatrix}$, and $eRe=Re$ becomes a summand if $(1-e)Re$ is zero (and that is isomorphic to $Hom(R(1-e), Re)$, I think, if I haven't mixed up sides). I knew group algebras had symmetry but didn't know enough to express it. Thanks for posting that piece I was missing :) $\endgroup$ – rschwieb May 23 '18 at 13:27

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