Let $A$ be a finite-dimensional algebra over a field $F$. A representation $M$ of $A$ is called **absolutely irreducible** if $M\otimes\_FE$ is irreducible as a representation of $A\otimes\_FE$ for all field extensions $E$ over $F$.


A field $F$ is said to be a **splitting field** of $A$ if any irreducible representation of $A$ is absolutely irreducible.


There are some results on splitting fields. For example, suppose that $F$ is a splitting field of $A$. Then any field extension $E$ over $F$ is a splitting field, and moreover, irreducible representations of $A$ are in one-to-one correspondence with irreducible representations of $A\otimes\_FE$ via $S\mapsto S\otimes\_FE$.


Conversely, suppose that $A$ is an $F$-algebra and $E$ (over $F$) is a splitting field of $A$. Then $F$ is a splitting field if and only if irreducible representations of $A$ are in one-to-one correspondence with irreducible representations of $A\otimes\_FE$ via $S\mapsto S\otimes\_FE$.


Now suppose that $A$ is an $F$-algebra and $E$ (over $F$) is a splitting field of $A$. My question is that, if we only know that every irreducible representation of $A\otimes\_FE$ is equal to $S\otimes\_FE$ for some irreducible representation $S$ of $A$, can we conclude that $F$ is also a splitting field of $A$? Notice that we do not assume that $S\otimes\_FE$ is an irreducible representation of $A\otimes\_FE$ for every irreducible representation $S$ of $A$.



Absolutely irreducible representation and splitting field
Yes, $F$ is a splitting field of $A$. Put $A\_E = A\otimes\_F E$ for convenience. Let $S'$ be a simple $A$-module. Then $S'\otimes\_F E$ has a simple $A\_E$-submodule, which by hypothesis is of the form $S\otimes\_F E$ for some simple $A$-module $S$. Note that $$0\neq \hom\_{A\_E}(S\otimes\_F E,S'\otimes\_F E)\cong \hom\_A(S,S'\otimes\_F E)$$. But as an $A$-module, $S'\otimes\_F E$ is isomorphic to a direct sum of $[E:F]$ copies of $S'$ (this can be infinite if the field extension is infinite, but that's ok). So $$0\neq \hom\_A(S,S'\otimes\_F E)\leq \hom\_A(S,S')^{[E:F]}$$ (this is equality if $[E:F]<\infty$). So by Schur's lemma, $S\cong S'$ and therefore $S'\otimes\_F E$ is simple. It now follows that $F$ is a splitting field because $E$ is one.


