Question

Let R be a finite-dimensional k-algebra and let L ⊇ K ⊇ k be fields. Assume that L is a splitting field for R. Show that K is a splitting field for R iff, for every simple left RL-module M, there exists a (simple) left RK-module U such that UL ∼ M.

Answer 1

The "only if" part is clear. For the "if" part, assume every simple (left) $R^L$-module "comes from" a simple $R^K$-module. Let $M1, \ldots, Mn$ be all the distinct simple $R^L$-modules, and let $M_i = U^L_i$, where $U_1, \ldots, U_n$ are (necessarily distinct) simple $R^K$-modules. If there is another simple $R^K$-module $V$ not isomorphic to any $U_i$, then, a composition factor of $V^L$ would give a simple $R^L$-module not isomorphic to any $M_i$, a contradiction. Thus, $\{U_1, \ldots, U_n\}$ give all the distinct simple $R^K$-modules. Each $U_i$ remains irreducible over $L$ and hence over the algebraic closure of $L$. Therefore, each $U_i$ is absolutely irreducible, and we have proved that $K$ is a splitting field for $R$.