Let $\mathsf{K}$ be a field, and $A$ a finite dimensional associative $\mathsf{K}$-algebra, and suppose that $\mathcal{L}$ is a complete collection of representatives of the isomorphism classes of irreducible left $A$-modules. If $A$ is semisimple, then a corollary to the Artin-Wedderburn theorem states that
$$
\dim_{\mathsf{K}} A = \sum_{L \in \mathcal{L}} (\dim_{\mathsf{K}}L)^{2}.
$$
I am interested in whether this corollary has a converse. That is, if this equality holds can we say that $A$ is semisimple? If the answer is yes is there a simple (semisimple?) argument that I'm missing here, or if it is false can you provide a counter-example? If it is false, are there extra suppositions that we can apply to $A$ or $\mathsf{K}$ to make the statement hold? (Perhaps characteristic zero or algebraically closed?)
Best Answer
No: consider $A=\mathbf{C}[t]/(t^2)$ and $K=\mathbf{R}$. $A$ has one irrep, namely $\mathbf{C}$, of dimension $2$ over $\mathbf{R}$, and the dimension of $A$ over $\mathbf{R}$ is $4=2^2$, but $A$ is not semi-simple.
The issue is that there are two competing forces at work here: $A$ is not semi-simple, which tends to decrease the number of irreducibles relative to the semi-simple case, but also not split, which tends to increase the dimensions of the irreducibles.
If you assume that every irreducible representation is absolutely irreducible, then the result it true: the right-hand side is in this case the dimension of the semi-simplification $A/\mathrm{Jrad}(A)$ of $A$ (here $\mathrm{Jrad}(A)$ is the Jacobson radical of $A$); if this is equal to the dimension of $A$ then $\mathrm{Jrad}(A)=0$ as desired.