Finite Galois descent of morphisms of schemes

For every $k$-algebra $A$, $(A{\otimes }_{k}K{)}^{\mathrm{\Gamma }}=A$: express a given tensor using finitely many $k$-linearly independent coefficients in $A$, then equivariance says its coefficients in $K$ are fixed and hence lie in $k$ by [F1]. The projection $p:X_K\to X$ is finite and surjective: on affine charts $A{\otimes }_{k}K$ is a finite free faithfully flat $A$-module. In each fibre $\mathrm{Spec}(\kappa (x){\otimes }_{k}K)$ the group $\mathrm{\Gamma }$ acts transitively on points. Indeed this tensor product is finite étale over the field $\kappa (x)$ and thus a product of fields; a union of orbits of its factors gives an invariant idempotent. The invariant-ring calculation, with $A=\kappa (x)$, says that only the empty and full unions are possible.

Let $V\subset Z$ be affine. The open subset $W=f^{-1}(V_K)$ of $X_K$ is $\mathrm{\Gamma }$-stable. Step 1.1 shows that each fibre of $p$ is either contained in $W$ or disjoint from it. Since a finite morphism is closed (on an affine chart this follows from lying-over for the integral ring extension, including after passage to quotient ideals), $U=X\setminus p(X_K\setminus W)$ is open and $W=p^{-1}(U)$. As $V$ ranges over an affine cover of $Z$, these opens $U$ cover $X$.

Cover each such $U$ by affine opens $T=\mathrm{Spec}A$. Write $V=\mathrm{Spec}B$. The restriction $f:T_K\to V_K$ corresponds to a $K$-algebra map $B{\otimes }_{k}K\to A{\otimes }_{k}K$. Equivariance and step 1.1 show that its restriction to $B$ takes values in $A$. This gives a $k$-morphism $T\to V$ whose base extension is the restriction of $f$. It is unique, since $A\to A{\otimes }_{k}K$ is injective.

The local morphisms glue. On an overlap their base extensions coincide with $f$; equality can be checked after this faithfully flat scalar extension by covering inverse images of affine target opens and using the injectivity of the corresponding coordinate-ring map, exactly as in step 3.1. They therefore agree on the overlap. The glued $k$-morphism has base extension $f$, is unique by the same argument, and every base extension is semilinearly equivariant by construction. No arbitrary choice is used.