Scalar extension of an irreducible finite-group representation

By [L4], decompose $E{\otimes }_{F}V$ into irreducibles and choose a constituent $U$. The Galois action permutes its isomorphism classes.

Let $X$ be the sum of the isotypic components in the Galois orbit of $U$. The canonical semilinear $\mathrm{\Gamma }=\mathrm{Gal}(E\mathrm{/}F)$-action on $E{\otimes }_{F}V$ preserves $X$. To make descent explicit, choose trace-dual $F$-bases $(x_i)$ and $(y_i)$ of $E$. For $w\in X$, every $T(y_{i}w)={\sum }_{\sigma \in \mathrm{\Gamma }}\sigma (y_{i}w)$ lies in $X^{\mathrm{\Gamma }}$, and the separability identity ${\sum }_i{x}_i\sigma (y_i)={\delta }_{\sigma ,1}$ gives $w={\sum }_i{x}_{i}T(y_{i}w)$. Hence $X=E{\otimes }_F{X}^{\mathrm{\Gamma }}$. Under $(E{\otimes }_{F}V{)}^{\mathrm{\Gamma }}=V$, the $G$-stable space $X^{\mathrm{\Gamma }}$ is an $F$-subrepresentation of $V$; irreducibility therefore forces $X=E{\otimes }_{F}V$.

By [L3], every member of this orbit has one common multiplicity $m$. The orbit is indexed without repetition by $\mathrm{Gal}(E\mathrm{/}F)\mathrm{/}\mathrm{Stab}(U)$, which gives the formula.

Let $\bar{E}$ be an algebraic closure of $E$. For any irreducible constituent $U$, [L1] and [L2] give ${\mathrm{End}}_G(\bar{E}{\otimes }_{E}U)\cong \bar{E}{\otimes }_E{\mathrm{End}}_G(U)=\bar{E}\mathrm{.}$ The scalar extension is semisimple by [L4]; if it were reducible, projection onto a proper summand would be a nonscalar endomorphism. Hence $U$ is absolutely irreducible. Distinct orbit points are inequivalent by the definition of the stabilizer.