Let $k$ be a field of characteristic zero and $X$ be a non-singular 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 torsion-free semi-stable sheaf $E$ on $X\_L$, the coherent sheaf $p\_\*E$ is semi-stable on $X\_k$?



Pushforward of semi-stable sheaves under finite field extension
Assuming what you mean is that $X$ is projective and $\mathcal E$ is Gieseker-semistable, the answer is yes.



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> > **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 $X\_{\bar k}$.
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*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](https://mathoverflow.net/a/253745/82179).


**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.



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> > **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.
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*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. $\square$


