Let $X$ be a projective variety over a field $K$ of characteristic zero. Denote by $p:X\_{\overline{K}} \to X$ the natural morphism, where $\overline{K}$ is the algebraic closure of $K$ and $X\_{\overline{K}}:=X \times\_K \overline{K}$. Let $E$ and $F$ be coherent sheaves on $E$ such that $p^\*E \cong p^\*F$. Does it imply that $E \cong F$?


**EDIT:** If necessary, assume that $E$ and $F$ are elements of the same Quot scheme on $X$.



Projection formula for field extension
The answer is affirmative with $X$ any proper scheme over any field $K$, moreover using any field extension $K'/K$ in place of $\overline{K}/K$. 


Let $H\_{E,F} = \mathscr{H}om(E,F)$, a coherent sheaf on $X$, and define $H\_{F,E}$ and $H\_{E,E}$ similarly. Thus, $H\_{E,F}(X) = {\rm{Hom}}\_{O\_X}(E,F)$ is a finite-dimensional $K$-vector space, and similarly for $H\_{F,E}$ and $H\_{E,E}$. We want to show that the $K$-bilinear composite map $$H\_{E,F}(X) \times H\_{F,E}(X)\rightarrow H\_{E,E}(X)$$ defined by $(\varphi, \psi) \mapsto \varphi \circ \psi$ hits the subset of units in the finite-dimensional associative $K$-algebra ${\rm{Hom}}\_{O\_X}(E,E)$. Indeed, suppose for some such $\psi$ and $\varphi$ we know that $\varphi \circ \psi$ is a unit, which is to say an automorphism of $E$, so $\varphi$ is surjective. Since $E\_{K'}$ and $F\_{K'}$ are isomorphic, $\varphi\_{K'}$ is thereby identified with a surjective *endomorphism* of a coherent sheaf on a noetherian scheme, and any such endomorphism is an automorphism (since over affine opens we can apply the fact that a surjective endomorphism of a finitely generated module over a noetherian ring is necessarily an automorphism). Hence, $(\ker \varphi)\_{K'} = \ker(\varphi\_{K'})=0$, so $\ker \varphi=0$ and thus the surjective $\varphi$ is an isomorphism, as then is $\psi$ too.


The formation of the coherent Hom-sheaves commutes with any base change on $K$ since all such base changes are flat, and likewise for the formation of the global sections of (quasi-)coherent sheaves on $X$, so if we form the analogous composite map after scalar extension to a $K$-algebra $A$ it defines an analogous composite map
$$(H\_{E,F}(X) \otimes\_K A) \times (H\_{F,E}(X) \otimes\_KA) \rightarrow H\_{E,E}(X) \otimes\_K A$$
functorially in $A$. 


It follows by Yoneda's Lemma or elementary reasons more specific to the bilinear setting that the initial displayed composite map is really the evaluation on $K$-points of a $K$-morphism of affine spaces over $K$. Moreover, the units $(H\_{E,E}(X) \otimes\_K A)^{\times}$ are identified with the $A$-points of a *Zariski-open* subset of the affine space over $K$ associated to $H\_{E,E}(X)$ (as for any finite-dimensional associative $K$-algebra in place of $H\_{E,E}(X)$). Thus, the locus of pairs of isomorphisms $(\varphi, \psi)$ corresponds to a Zariski-open subset $U$ of the affine space over $K$ corresponding to $H\_{E,F}(X) \times H\_{F,E}(X)$. 


The existence of an isomorphism over $K'$ says $U(K')$ is non-empty, so $U$ is non-empty. But $U$ is open in an affine space over $K$, so if $K$ is infinite then $U(K)$ must be non-empty and hence we win. Thus, we are done if $K$ is infinite.


In general (to permit finite $K$), by the principle of "spreading out and specialization", we can reduce to the case when $K'/K$ is a finite extension. Thus, our task is encoded in the obstruction to adjusting a given $K'$-isomorphism to descend to a $K$-isomorphism after composing with the scalar extension to $K'$ of some automorphism of $E$, which is to say that it is encoded in an fppf class $\xi \in {\rm{H}}^1(K'/K, G(K'))$ where $G$ is functor on $K$-algebras defined by $A \rightsquigarrow {\rm{Aut}}\_{O\_{X\_A}}(E\_A)$ (can relax "fppf" to "etale" when $K'/K$ is separable, in which case we can enlarge $K'/K$ to also be Galois, such as whenever $K$ is perfect). It is sufficient to show that for any $\xi$ in that ${\rm{H}}^1$, there exists a finite extension $K''/K'$ such that the inflation $\xi\_{K''} \in {\rm{H}}^1(K''/K,G(K''))$ is trivial.


By arguments as above, $G$ is a (non-empty) open subset of the affine space corresponding to $H\_{E,E}(X)$. Thus, $G$ is smooth and connected, so when $K$ is *finite* we have ${\rm{H}}^1(\overline{K}/K,G(\overline{K}))=1$ by Lang's Theorem, yielding the desired $K''/K'$ for finite $K$.


