stable sheaves in characteristic $0$ Let $K$ be a non-algebraically closed field of characteristic $0$ and $X_K$ a smooth, projective, geometrically connected curve defined over $K$. If $F$ is a stable locally free sheaf on $X_K$, is it also necessarily stable on $X_\bar{K}$, where $\bar{K}$ denotes the algebraic closure of $K$?
I know that this is certainly not the case in characteristic $p$ but could not find an example where things go wrong in characteristic $0$. An example or reference would be very helpful.
 A: Let $K = \mathbb R$, and let $C = \{x^2+y^2+z^2 = 0\} \subseteq \mathbb P^2_K$ be the conic without a rational point. The projection $\pi \colon C_{\bar K} \to C$ induces a pullback map
$$\pi^* \colon \operatorname{Pic}(C) \to \operatorname{Pic}(C_{\bar K}) = \mathbb Z$$
mapping $\operatorname{Pic}(C)$ to the subgroup $2\mathbb Z \subseteq \mathbb Z$. In particular, there is no line bundle on $C$ of degree $1$.
Now equip the vector bundle $\mathcal O(1) \oplus \mathcal O(1)$ on $C_{\bar K}$ with a Galois action by interchanging the factors. This descends to a bundle $\mathcal E$ on $C$, which can equivalently be described as $\pi_* \mathcal O(1)$.
Claim. The rank $2$ vector bundle $\mathcal E$ is stable on $C$, but its pullback to $C_{\bar K}$ is not.
Proof. The second statement is clear. For the first, note that $\mathcal E$ is semistable since it is so over $\bar K$. Thus, it suffices to show that it does not contain a line bundle of degree $1$. But $C$ does not have any line bundle of degree $1$. $\square$
