I want to show that every finite-dimensional simple algebra admits a symmetric Frobenius structure.
My attempt:
I know that by Artin–Wedderburn theorem, every simple ring(algebra) is a matrix ring $M_{n\times n}(R)$ over a division ring $R$.If somehow I could manage to make this $R$ to be actually a field (i.e., commutative division ring) then I can give a symmetric Frobenius form using the trace map.
My intuition says I need to take some field extension $\mathbb{k}$ of this division ring $R$, such that I will have symmetric Frobenius form on $M_{n\times n}(\mathbb{k})\to \mathbb{k}$ by using the trace map and the restriction map on $M_{n\times n}(\mathbb{k}) \to \mathbb{k}$ will give me symmetric Frobenius form on $M_{n\times n}(R) \to R$. I am not sure if I make sense or not.
Any solution/hint or a reference will be of grat help!