Let $k$ be a field of characteristic zero and $X$ be a nonsingular rationally connected variety over $k$. Let $L$ be a finite field extension of $k$. This induces a proper morphism $p:X_L \to X_k$. Is it true that for any torsionfree semistable sheaf $E$ on $X_L$, the coherent sheaf $p_*E$ is semistable on $X_k$?

1$\begingroup$ What definition of semistability are you using? If $p_*E$ is destabilized by $F$, i.e., some appropriate slope of $F$ is greater than the slope of $p_*F$, then presumably the same holds for $p^*F$ and $p^*p_*E$. Since $p^*p_*E$ is $E\otimes_L(L\otimes_k L) \cong E^{\oplus n}$ for $n=[L:k]$, this would seem to contradict semistability of $E$. $\endgroup$ – Jason Starr Aug 18 '17 at 17:31

$\begingroup$ @JasonStarr: your comment and my answer are virtually identical. However, $L \otimes_k L \cong L^{[L:k]}$ only holds if $k \subseteq L$ is Galois. Otherwise, you first have to pass to the Galois closure of $L/k$ (or, as I do, all the way to $\bar k$). $\endgroup$ – R. van Dobben de Bruyn Aug 18 '17 at 17:42

$\begingroup$ @R.vanDobbendeBruyn. I absolutely agree that as a (left or right) $L$algebra, the tensor product $L\otimes_k L$ is not always isomorphic to the product algebra $L\times \dots \times L$. However, as a (left or right) $L$module, the tensor product is a vector space of dimension $n$. I believe that is what is relevant here. Are you saying that the $L$module structure gets shifted through $L\otimes_k L$, so that the algebra structure is relevant? $\endgroup$ – Jason Starr Aug 18 '17 at 17:59

$\begingroup$ @JasonStarr: Sorry, I misread your argument and gave the wrong objection. I think $p^* p_* E$ is not $E \otimes_L (L \otimes_k L)$, but rather $E_k \otimes_k L$ (where $E_k$ forgets the $L$structure). I see no way of relating the two (over $L$), and in fact I don't think that they are the same. For example, if $X$ is an elliptic curve over $\mathbb R$ and $E = \mathcal O(P)$ for a nonreal point $P$, then $p^* p_* E$ should be $\mathcal O(P) \oplus \mathcal O(\bar P)$, which is not isomorphic to $E^2$. $\endgroup$ – R. van Dobben de Bruyn Aug 19 '17 at 15:33

$\begingroup$ @R.vanDobbendeBruyn. For a Cartesian square of schemes, say $f:X\to Z$, $g:Y\to Z$ and $f':W\to Y$ and $g':W\to X$, if $f$ is quasicompact and quasiseparated, and if $g$ is flat, then for every quasicoherent $\mathcal{O}_X$module $E$, the natural map $g^*f_*E \to (f')_*(g')^*E$ is an isomorphism. For $f$ and $g$ both equal to $\text{Spec}(L)\to \text{Spec}(k)$, it follows that $g^*f_*E$ is the same as the pushforward of the pullback of $E$. Thus, for every scheme $T$ over $\text{Spec}(L)$, for every quasicoherent sheaf $E$ on $T$, $g^*f_*E$ equals $(L\otimes_k L)\otimes_L E$. $\endgroup$ – Jason Starr Aug 19 '17 at 17:25
Assuming what you mean is that $X$ is projective and $\mathcal E$ is Giesekersemistable, the answer is yes.
Lemma. Let $X$ be a projective $k$scheme, and let $\mathcal E$ be a semistable sheaf on $X$. Then $\mathcal E_{\bar k}$ is semistable on $X_{\bar k}$.
Proof. Let $\mathcal F \subseteq \mathcal E_{\bar k}$ be a proper nonzero subsheaf, and assume that $\mu(\mathcal F) > \mu(\mathcal E)$. Then $\mathcal F$ is defined over some finite extension $k \subseteq \ell$. Let $p \colon X_\ell \to X$ be the projection. Then $p_* \mathcal F \subseteq p_* \mathcal E_\ell$ is a subsheaf, with $\mu(p_*\mathcal F) > \mu(p_* \mathcal E_\ell)$. But $p_* \mathcal E_\ell = p_* p^* \mathcal E = \mathcal E^{[\ell:k]}$ is semistable, so such subsheaves cannot exist. $\square$
Remark. The same statement with semistable replaced by stable is false. See for example this post.
Remark. Conversely, it follows from uniqueness of the Harder–Narasimhan filtration that if $\mathcal E_{\bar k}$ is semistable, then so is $\mathcal E$. Indeed, uniqueness of the HN filtration implies (using the lemma above) that it commutes with base change of the field.
Corollary. Let $X$ be a projective $k$scheme, let $k\subseteq \ell$ be a finite separable extension, and let $\mathcal E$ be a semistable sheaf on $X_\ell$. Then $p_* \mathcal E$ is semistable.
Proof. By the remark, it suffices to show that $(p_* \mathcal E)_{\bar k}$ is semistable. Because $\ell \otimes_k \bar k \cong \bar k^{[\ell:k]}$, the base change of $p \colon X_\ell \to X$ to $\bar k$ looks like $$p_{\bar k} \colon \coprod_{\sigma \colon \ell \to \bar k} X_{\bar k} \to X_{\bar k}.$$ By the lemma above, the base change $\mathcal E_{\bar \ell}$ is semistable. Under $p_{\bar k}$, we are merely taking the sum of the Galois conjugates of $\mathcal E_{\bar \ell}$. It is straightforward to check that each of these Galois conjugates is semistable of the same slope, hence their sum is semistable. $\square$

$\begingroup$ Doesn't this imply that a nonempty moduli space $M_X(P)$ parametrizing semistable sheaves on $X$ with Hilbert polynomial $P$, will always have a $k$rational point? In particular, if I understand correctly, the moduli space being nonempty implies that there exists a finite extension $L$ of $k$ such that there exists a semistable sheaf $E$ on $X_L$ with Hilbert polynomial $P$. Then, by your argument, we have that $p_∗E$ is semistable on $X$, which will give a $k$rational point of $M_X(P)$. Am I missing something? $\endgroup$ – user43198 Aug 21 '17 at 5:22

$\begingroup$ @user43198: the pushforward will not have the same Hilbert polynomial. Its rank is multiplied by $[\ell:k]$ under pushforward. $\endgroup$ – R. van Dobben de Bruyn Aug 21 '17 at 10:54