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Let $X$ be an irreducible nodal projective curve over an algebraically closed field of characteristic $p>0$. Denote by $\pi:\tilde{X} \to X$ the normalization of $X$. Recall, the short exact sequence of Picard numbers from Hartshorne, $$0 \to \oplus_{x \in X} \tilde{\mathcal{O}}_{X,x}^*/\mathcal{O}_{X,x}^* \to \mathrm{Pic}(X) \xrightarrow{\pi^*} \mathrm{Pic}(\tilde{X}) \to 0$$ where $\tilde{\mathcal{O}}_{X,x}$ is the integral closure of the local ring $\mathcal{O}_{X,x}$.

Fix an invertible sheaf $L \in \mathrm{Pic}(X)$. Let $E$ be a locally free sheaf on $X$ such that the determinant of $\pi^*E$ is $\pi^*L$. The question: Does there exist a locally free sheaf $E'$ on $X$ with the same Hilbert polynomial as $E$ (i.e., same rank and degree) and whose determinant is $L$? In other words, does there exist an invertible sheaf $L'$ on $X$ such that the determinant of $E \otimes L'$ is isomorphic to $L$? Any hint/reference on this will be very useful.

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    $\begingroup$ I don't understand your question. How about $E'=L$? The determinant of $E\otimes L'$ is $(\det E) \otimes L'^{rk(E)}$, which implies $\pi^* L'^{rk(E)} \cong \mathcal{O}_{\tilde X}$. $\endgroup$
    – Piotr Achinger
    Commented Oct 11, 2015 at 15:29
  • $\begingroup$ And how is the positive characteristic assumption relevant? Do you know the answer to your question in characteristic zero? $\endgroup$
    – Piotr Achinger
    Commented Oct 11, 2015 at 15:30
  • $\begingroup$ @PiotrAchinger Sorry, forgot to add that the Hilbert polynomial of $E'$ must be the same as $E$. I have edited the question. The question follows easily by an isogeny result of abelian varieties in characteristic zero. $\endgroup$
    – user43198
    Commented Oct 11, 2015 at 15:37
  • $\begingroup$ Just making sure: are you asking whether the map $L\mapsto L^{rk E}$ is surjective on $Pic^0(X)$ (or, more precisely, on the kernel $ker(Pic(X)\to Pic(\tilde X))$)? If so, then no, the map need not be surjective: for instance, take $X$ to be cuspidal cubic (so that $Pic^0(X)$ is the additive group), take a non-trivial degree zero line bundle $L$ on it, and put $E=O^p$. (On the other hand, one sentence in your questions says `does there exist a locally free sheaf $E'$ on $X$ ...'; I assume that you only want $E'$ that is a twist of $E$, as in the next sentence.) $\endgroup$
    – t3suji
    Commented Oct 11, 2015 at 17:37
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    $\begingroup$ The point is, even if $p|rk(E)$, the map $L'\mapsto L'^{\otimes rk(E)}$ is surjective on $\ker(Pic(X)→Pic(\tilde X))$, because the kernel is isomorphic to $(\mathbb{G}m)^k$ (the curve is nodal), and the map is surjective on $\mathbb{G}m$. So you can find $L'$ such that $L'^{\otimes rk(E)}\simeq L\otimes(\det E)^{-1}$. $\endgroup$
    – t3suji
    Commented Oct 11, 2015 at 18:15

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