Let $A$ be a finite dimensional algebra over an algebraically closed field and let $mod A$ denote the category of finite dimensional A-modules.
Problem Suppose that every simple $A$-module is one-dimensional. I want to show that this implies that $A$ is a basic-algebra (i.e $A/rad(A)$ is a finite direct product of copies of $K$).
Here's what I've tried: since $A/rad(A)$ is semisimple then $A/rad(A) \cong \displaystyle \prod_{i=1}^{l} M_{n_{i}}(K)$ as rings. On the other hand $A/rad(A)=\displaystyle \bigoplus_{i=1}^{l} e_{i}(A/rad(A))$. Then since each of the factors of both decompositions of $A/rad(A)$ are simple rings then one obtains that $M_{n_{i}}(K) \cong e_{i}(A/rad(A))$ as rings. Since $e_{i}(A/rad(A))$ is a simple $A$-module this implies, by assumption, that $e_{i}(A/rad(A))$ is one-dimensional.
I want to claim that this implies that each $M_{n_{i}}(K)$ is one-dimensional because then $n_{i}=1$ and we're done.
Question: is the ring isomorphism $M_{n_{i}}(K) \cong e_{i}(A/rad(A))$ also $K$-linear? this would settle the question because then we would have an isomorphism of $K$-vector spaces.
It's not only true that $A/J(A) \cong \prod M_{n_i}(K)$ as rings; each factor $M_{n_i}(K)$ canonically corresponds to an $n_i$-dimensional simple module of $A$ (you get this from looking at the proof of Artin-Wedderburn more closely). So the hypothesis that each simple module is $1$-dimensional immediately gives that each $n_i = 1$.