Galois Descent of morphisms for field extensions Let $L/K$ be a finite separable extension of fields and $X,Y$ two finite type schemes defined over $K$. Suppose there is a morphism $f: X_L\to Y_L$ over $L$ such that it is Galois equivariant. That is, for any $\sigma \in Gal(L/K)$, we have $f^\sigma = f$. How do I prove that there exists a $f' : X\to Y$ over $K$ such that $f'_L = f$?
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I know that one can prove this for a general fpqc extension by proving it for quasi coherent sheaves first but I would like a more direct method just for fields.
I also know the following "proof": Prove first that galois equivariant closed sets descent and then, to construct $f'$, consider the graph $\Gamma_f$ of $f$ in $X_L\times_L Y_L = (X\times Y)_L$ and descent this to a closed subscheme of $X\times Y$ and use this to define $f'$.
There are two problems with this approach: First, The graph need not be a closed subscheme (I am not assuming that our morphism is separated). Second, I don't know how to actually get a morphism given a closed subscheme of $X\times Y$ that is an isomorphism under the projection to $X$.
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So I would really appreciate it if someone could construct the morphism quite explicitly. For instance, suppose $X = K[x]/(a)$ and $Y = K[y]/(b)$ and suppose $f$ is defined by sending $y \to p(x)$ in $L[x]$ such that $\sigma(p) \equiv p \pmod{a}$ for all galois conjugates $\sigma$. In this case, what should I define $f'$ to be?
The first try might be to define $y \to \frac{1}{[L:K]}\sum_\sigma p^\sigma(x)$ but what if $[L:K] = 0 \in K$?
 A: One needs $L/K$ Galois (otherwise the theorem as stated is false). I will give a proof only in the case $X=K[x]/(a)$ and $Y=k[y]/(b)$.
As you wrote, a map $f:X_L\rightarrow Y_L$ corresponds to a polynomial $\overline{p}(x)\in L[x]/(a)$. Choose a representative $p\in L[x]$. If moreover $f$ is Galois-invariant, then $u_\sigma=\sigma(p)-p\in aL[x]$.
The polynomial $p$ is not in $K[x]$, but in fact there is a representative which is. Indeed, $(u_\sigma)_\sigma$ forms a cocyle in $aL[x]\simeq L[x]$ (that is $u_{\sigma\tau}=u_\sigma+\sigma(u_\tau)$ for all $\sigma,\tau$). Let us show that it is also a boundary (this is the additive form of Hilbert 90).
Note that, because of the independence of character, there exists an element $l\in L$ of trace 1 : $\sum_\tau \tau(l)=1$. Now put $q=\sum_\tau u_\tau \tau(l)\in aL[x]$. We have 
$$\sigma q=\sum_\tau \sigma(u_\tau)\sigma\tau(l)= \sum_\tau(u_{\sigma\tau}-u_\sigma)\sigma\tau(l)=\sum_\tau u_{\sigma\tau}\sigma\tau(l) -\sum_\tau u_\sigma\sigma\tau(l)=q-u_\sigma$$
where the last equality follows from a change of indices in the sums.
Thus $u_\sigma=q-\sigma(q)$ for some $q\in aL[x]$. But $u_\sigma=\sigma(p)-p$. So $\sigma(p)+\sigma(q)=p+q$. This means that the polynomial $p+q$ is Galois-invariant. From Galois theory, this mean that $p+q\in K[x]$. Since $q\in aL[x]$, $p\equiv p+q\ (\operatorname{mod} a)$.
Finally, the map $f$ came indeed from  $f':X\rightarrow Y$ : on coordinate rings it corresponds to $K[y]/(b)\rightarrow K[x]/(a), y\mapsto p+q\ (\operatorname{mod} a)$ 
