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Let $k$ be a field.

Let $A$ be a $k$-algebra and $B \subset A$ be a subalgebra of $A$ which is a semisimple algebra. Note that we do not assume that $A$ is commutative.

We also assume that $A$ is a finite (right or left) $B$-module.

Then, is there any sufficient conditions for $A$ to be also a semisimple algebra?

In particular, the examples I want to treat is the following. $$ A = \begin{pmatrix} R_0 & R_1 & \cdots & R_n \\ R_{-1} & R_0 & \cdots & R_{n-1} \\ & \cdots & \cdots & \\ R_{-n} & R_{-n+1} & \cdots & R_0 \end{pmatrix} \quad (n \in \mathbb{N}),\quad B= R_0, $$ where $R$ is a $\mathbb{Z}$-graded algebra and $R_i$ is the degree $i$ component. We also think of $B$ as the diagonal subalgebra of $A$. (We suppose the above assumptions for $A$ and $B$ hold.)

In addition, we are interested in the case
$R = (k[x_0,x_1,\cdots,x_m]/(x_0^{\text{deg}(x_i)}-a_ix_i^{\text{deg}(x_0)})_{i=1,\cdots m})[x_0^{-1}]$, where $k[x_0,\cdots, x_m]$ is a weighted polynomial ring, $a_i \in k^*$ and $\text{deg}(x_0)(=:n) > 1$.

Any comment welcome. Thank you.

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  • $\begingroup$ What kind of conditions are you looking for? If $B = k$ you're just asking when a finite-dimensional algebra is semisimple. It's semisimple when it's semisimple, it just depends. $\endgroup$ Commented Aug 26, 2022 at 1:36
  • $\begingroup$ Thank you for the comment. I want to know relations between $A$ and $B$ for A to be semisimple. $\endgroup$
    – YkMz
    Commented Aug 26, 2022 at 1:59

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