Galois fixed points recover finite-dimensional scalar extensions

Put $n=[E:F]=\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }\ge 1$, choose an $F$-basis $a_1,\dots ,a_n$ of $E$, and form its trace Gram matrix $H_{ij}=\mathrm{Tr}(a_i{a}_j)$. Separability and nondegeneracy make $H$ invertible. Set $b_j={\sum }_k(H^{-1}{)}_{kj}{a}_k$. Then $\mathrm{Tr}(a_i{b}_j)={\delta }_{ij}$. The invertible coefficient matrix also shows that $(b_j)$ is a basis.

Any finite $F$-independent list $v_1,\dots ,v_r$ in $W^{\mathrm{\Gamma }}$ is $E$-independent. Indeed, if a relation exists, select one with the least positive number of nonzero coefficients and normalize one of those coefficients to $1$. Applying $T_{\sigma }$ and subtracting produces a relation with that coefficient zero; minimality forces every other coefficient to be fixed by every $\sigma$. They all lie in $F$, contradicting $F$-independence. A one-term relation is already impossible since a nonzero scalar cannot annihilate a nonzero vector.

For the canonical action, choose a finite $F$-basis $(v_j)$ of $V$. Each tensor has a unique expression ${\sum }_j{e}_j\otimes v_j$: the product $F$-basis in F6, regrouped by the $v_j$, proves existence and uniqueness of its coefficients in $E$. Such a tensor is fixed exactly when $\sigma (e_j)=e_j$ for every $\sigma ,j$, equivalently each $e_j\in F$. It then equals $1\otimes {\sum }_j{e}_j{v}_j$. Conversely every $1\otimes v$ is fixed. This also proves that $v\mapsto 1\otimes v$ is injective, including the empty-basis case.

Let $X_{\sigma i}=\sigma (a_i)$ and $Y_{\sigma j}=\sigma (b_j)$. A relation among the rows of $X$ vanishes on the $a_i$, hence by $F$-linearity on every $x\in E$; restricting to $E^{\times }$ and using independence of the distinct automorphisms as multiplicative characters makes every coefficient zero. Thus $X$ is invertible. The trace formula gives $X^{\mathsf{T}}Y=I$, so $X{Y}^{\mathsf{T}}=I$. Its row at $1\in \mathrm{\Gamma }$ yields ${\sum }_i{a}_i\sigma (b_i)={\delta }_{1,\sigma }$.

Any tensor in the kernel is a finite sum ${\sum }_j{e}_j\otimes w_j$. By eliminating dependent members of the finite list $(w_j)$, rewrite it as ${\sum }_{k=1}^r{f}_k\otimes v_k$ with the $v_k$ $F$-independent and invariant. Its image is ${\sum }_k{f}_k{v}_k=0$, so step 1.2 forces every $f_k=0$. Hence the tensor is zero and $\mu$ is injective. For $W=0$ the domain and codomain are zero and the same argument uses the empty list.

For $w\in W$ define $P_i(w)={\sum }_{\sigma \in \mathrm{\Gamma }}\sigma (b_i)T_{\sigma }(w)$. For $\tau \in \mathrm{\Gamma }$, semilinearity gives $T_{\tau }(P_i(w))={\sum }_{\sigma }(\tau \sigma )(b_i)T_{\tau \sigma }(w)=P_i(w)$, since $\sigma \mapsto \tau \sigma$ permutes the finite group. Moreover ${\sum }_i{a}_i{P}_i(w)={\sum }_{\sigma }{\delta }_{1,\sigma }{T}_{\sigma }(w)=w$. Thus $P_i(w)\in W^{\mathrm{\Gamma }}$, and $\mu$ is onto. The formula $\mu (e\otimes w)=ew$ is $F$-balanced and $E$-linear.

If $W$ is an algebra, its fixed space is an $F$-subalgebra containing $1$. For invariant $u,v$ and $e,f\in E$, $\mu ((e\otimes u)(f\otimes v))=efuv=(eu)(fv)$ and $\mu (1\otimes 1)=1$. Distributing over finite sums proves multiplicativity on all tensors. Together with bijectivity, this proves both algebra assertions. If $E=F$, $\mathrm{\Gamma }=\{1\}$ and the map is the usual multiplication $F{\otimes }_{F}W\cong W$; the proof divides by no group order in any characteristic. [F1, step 3.1, step 2.2, step 1.3, algebra] QED