Consider a symmetric algebra $H$ over a field $k$. By definition, this is a $k$-algebra $H$ with a *symmetrizing trace* $\tau$, which is a $k$-linear map $\tau:H\to k$ such that $\tau(hh')=\tau(h'h)$ for all $h,h'\in H$ and the corresponding bilinear form is non-degenerate. I have been using chapter 7 of Characters of finite Coxeter groups and Iwahori-Hecke algebras by Geck and Pfeiffer as a reference for symmetric algebras.


According to Theorem 7.26 of this reference, if $H$ is split, then $H$ is semisimple if and only if all Schur elements are non-zero. The Schur element $c\_V$ of a split simple $H$ module $V$ can be defined by
$$
\sum\_{b \in \mathcal{B}}\chi\_V(b)\chi\_V(b^\vee)=c\_V \dim\_k V,
$$
where $\mathcal{B}$ is a $k$ basis of $H$, and $\lbrace b^\vee:b \in \mathcal{B}\rbrace$ is the basis dual to $\mathcal{B}$ under $\tau$. 



> 
> If $H$ is split and $k$ has characteristic 0, are the Schur elements always non-zero? Equivalently, is a split symmetric algebra over a field of characteristic 0 always semisimple?
> 
> 
> 


I assume that the answer is no, because I would have seen a result along these lines if it were true, but I can't find a counterexample.



When does a symmetric algebra over a field of characteristic 0 fail to be semisimple?
The answer is no. The reason is that every algebra can be embedded into a symmetric algebra, the so called trivial extension:


If $A$ is a $K$-algebra, then define $D(A):=A\oplus Hom\_K(A,K)$. $I:=Hom\_K(A,K)$ is a $A$-$A$-bimodule via


$a\cdot \phi \cdot b:=x\mapsto \phi(bxa)$


Hence you can define an $K$-algebra structure on $D(A)$ such that $I$ becomes an ideal with $I^2=0$. The multiplication is explicitly given by:


$(a+\phi)(b+\psi) := ab+a\cdot\psi+\phi\cdot b$


The trace form is given by


$(a+\phi,b+\psi) := \psi(a)+\phi(b)$


Now $I$ is a nilpotent ideal and therefore contained in the Jacobsen radical of $D(A)$. In particular do $A$ and $D(A)$ have the same simple modules and $D(A)$ is split iff $A$ is. But because $J(D(A))\neq 0$ the $K$-algebra $D(A)$ is never semisimple regardless of what field $K$ you started with.


