This is a continuation of the discussion in the mathoverflow, [Pushforward of semi-stable sheaves](https://mathoverflow.net/questions/279034/pushforward-of-semi-stable-sheaves-under-finite-field-extension). Let $X$ be a smooth projective variety over a field $k$ and $L$ be a finite field extension of $k$. Denote by $p:X\_L \to X$ the natural morphism. From the comments on the mathoverflow post, it seems that given a coherent sheaf $E$ on $X\_L$, $p^\*p\_\*E$ need not be a finite number of copies of $E$. My questions is, under what conditions on $X$ or $E$ will $p^\*p\_\*E$ be a finite number of copies of $E$? I am mainly interested in the case $X$ is an unirational variety and $E$ is locally-free.


More generally, I am looking for examples of varieties of dimension at least $2$ such that for *any* locally free sheaf $E$ on $X\_L$, there exists a morphism of sheaves from $E$ to $p^\*p\_\*E$.


Any reference which deals with similar topics/questions will be most welcome.



Pushforward of coherent sheaves and field extensions
There are at least two things that can go wrong. To explain this, let's carefully prove the following lemma.



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> > **Lemma.** Let $X$ be a finite type $k$-scheme, let $k \to \ell$ be a finite extension, and let $\mathscr F$ be a coherent sheaf on $X\_\ell$. Let $p \colon X\_\ell \to X$ and $q \colon X\_{k^{\operatorname{sep}}} \to X$ be the natural maps. Then
> >  $$q^\* p\_\* \mathscr F \cong \bigoplus\_{\sigma \in \operatorname{Hom}\_k(\ell,k^{\operatorname{sep}})} \sigma^\* \mathscr F.$$
> >  If $k \to \ell$ is Galois, then
> >  $$p^\*p\_\* \mathscr F \cong \bigoplus\_{\sigma \in \operatorname{Gal}(\ell/k)} \mathscr F^{\sigma},$$
> >  where $\mathscr F^{\sigma}$ is shorthand for $\sigma^\* \mathscr F$.
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*Proof.* By flat base change, we may replace $q^\* p\_\*$ by $p'\_\* q'^\*$, where $p'$ and $q'$ are the maps
$$\begin{array}{ccc}X\_\ell \underset{X}\times X\_{k^{\operatorname{sep}}} & \stackrel{p'}\to & \ X\_{k^{\operatorname{sep}}} \\ {\scriptsize q'}\downarrow\ \ \ \ & & \downarrow {\scriptsize q} \\ X\_\ell & \stackrel p\to & X. \ \end{array}$$
But $X\_\ell \times\_X X\_{k^{\operatorname{sep}}}$ is none other than $X\_{\ell \otimes\_k k^{\operatorname{sep}}}$. We have a natural isomorphism
\begin{align\*}
\ell \otimes\_k k^{\operatorname{sep}} &\stackrel\sim\longrightarrow \prod\_{\sigma \in \operatorname{Hom}\_k(\ell, k^{\operatorname{sep}})} k^{\operatorname{sep}}\\
x \otimes y &\longmapsto \sigma(x)y.
\end{align\*}
On the level of spectra, this means that $\operatorname{Spec}(\ell \otimes\_k k^{\operatorname{sep}}) = \coprod\_\sigma \operatorname{Spec} k^{\operatorname{sep}}$, with the natural map
$$\coprod\_\sigma \operatorname{Spec} k^{\operatorname{sep}} = \operatorname{Spec}(\ell \otimes\_k k^{\operatorname{sep}}) \to \operatorname{Spec} \ell$$
given on the component corresponding to $\sigma$ by $\sigma \colon \ell \to k^{\operatorname{sep}}$. Thus, $q'^\* \mathscr F$ is given on the component corresponding to $\sigma$ by $\sigma^\* \mathscr F$. This proves the first statement. The proof of the second statement is similar (but we should be careful about the two maps $p\_1', p\_2' \colon X\_{\ell \otimes\_k \ell} \to X\_\ell$). $\square$


**Remark.** If $k \to \ell$ is a separable extension but possibly no longer Galois, then the primitive element theorem gives $\ell \cong k[x]/(f)$ for some irreducible polynomial $f \in k[x]$. Then we get
$$\ell \otimes\_k \ell \cong \ell[x]/(f) \cong \ell \times \ell[x]/(g),$$
where $f = (x-\alpha)g$ for $\alpha \in \ell$ a root of $f$ and $g \in \ell[x]$. Thus, we always split off at least one factor $\ell$, which gives a summand of $p^\*p\_\*\mathscr F$ isomorphic to $\mathscr F$ corresponding to the component $X\_\ell$ of $X\_{\ell \otimes\_k \ell}$. So this does give you a map $\mathscr F \to p^\*p\_\*\mathscr F$ splitting the counit $p^\*p\_\*\mathscr F \to \mathscr F$ of the adjunction $p^\* \vdash p\_\*$. This answers positively your second weaker question.


**Remark.** So what can go wrong for $p^\* p\_\* \mathscr F$ to not be a power of $\mathscr F$?


Firstly, we can have a Galois extension $k \to \ell$ and a sheaf $\mathscr F$ such that $\mathscr F^\sigma$ is not isomorphic to $\mathscr F$ for some $\sigma \in \operatorname{Gal}(\ell/k)$. This is for example the case if $X = E$ is an elliptic curve over $\mathbb R$, and $\mathscr F = \mathcal O\_{E\_\mathbb C}(P)$ for a non-real point $P \in E(\mathbb C)$. But this is not rationally connected, so we will have to do better.


Another thing that can go wrong is that the splitting
$$q^\* p\_\* \mathscr F \cong \bigoplus\_{\sigma \in \operatorname{Hom}\_k(\ell,k^{\operatorname{sep}})} \sigma^\* \mathscr F.$$
is not defined over $\ell$ if $k \to \ell$ is not Galois.


It seems pretty hard to (provably) construct an example of the second type. Therefore, we will construct one of the first type, with $X\_{\bar k}$ rationally connected.


**Example.** Let $X \subseteq \mathbb P^3\_{\mathbb R}$ be the smooth quadric given by $x\_0^2+x\_1^2+x\_2^2-x\_3^2 = 0$. We have an isomorphism
\begin{align}
X\_\mathbb C &\stackrel\sim\longrightarrow V(x\_0x\_3-x\_1x\_2) \subseteq \mathbb P^3\_\mathbb C\label{Eq form}\tag{1}\\
[x\_0:x\_1:x\_2:x\_3] &\longmapsto [x\_0+ix\_1:x\_2+x\_3:x\_3-x\_2:x\_0-ix\_1].\nonumber
\end{align}
We also have an isomorphism (over any field)
\begin{align\*}
\mathbb P^1 \times \mathbb P^1 &\stackrel\sim\longrightarrow V(x\_0x\_3-x\_1x\_2) \subseteq \mathbb P^3\\
([x\_0:x\_1],[y\_0:y\_1]) &\longmapsto [x\_0y\_0:x\_0y\_1:x\_1y\_0:x\_1y\_1].
\end{align\*}
Under this isomorphism, the divisor $V(x\_0) \in |\mathcal O\_{\mathbb P^1 \times \mathbb P^1}(1,0)|$ corresponds to $V(x\_0,x\_1) \subseteq \mathbb P^3$. Under Galois conjugation in (\ref{Eq form}), this becomes $V(x\_1,x\_3)$, corresponding to $V(y\_1) \in |\mathcal O(0,1)|$. Therefore, we conclude that $\sigma^\* \mathcal O(1,0) = \mathcal O(0,1)$, where $\sigma \colon X\_\mathbb C \to X\_\mathbb C$ is the complex conjugation, and $X\_\mathbb C$ is identified with $\mathbb P^1 \times \mathbb P^1$. Thus, by the analysis above, we have
$$p^\*p\_\* \mathcal O(1,0) = \mathcal O(1,0) \oplus \mathcal O(0,1).$$
Thus, we have constructed a counterexample with $X\_\bar k$ rationally connected and $\mathscr F$ locally free.



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> > **Closing remark.** It seems that in general, it is hardly ever true that $p^\*p\_\* \mathscr F \cong \mathscr F^{[\ell:k]}$, even if $X\_{\bar k}$ is rationally connected and $\mathscr F$ is locally free.
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