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Let $C$ be a projective curve (over an algebraically closed field, not necessarily of characteristic zero) which is smooth except for exact one node. Let $\pi:\tilde{C} \to C$ be its normalization. Let $\mathcal{V}$ be a locally free sheaf over $\tilde{C}$. Is it true that the push forward, $\pi_*(\mathcal{V})$ is a locally free sheaf? If not, are there any general conditions (for example on the rank of $\mathcal{V}$) so that this holds true?

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    $\begingroup$ No and yes for the rank zero vector bundle. This question does not belong on mathoverflow IMHO. $\endgroup$ – answer_bot Mar 7 '14 at 12:08
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    $\begingroup$ @answer_bot: I do not understand why this question does not belong to mathoverflows when there is a similar question with 6 votes: mathoverflow.net/questions/67387/… $\endgroup$ – user45397 Mar 7 '14 at 12:45
  • $\begingroup$ @answer_bot: Anyways, could you please elaborate on your answer. $\endgroup$ – user45397 Mar 7 '14 at 12:46
  • $\begingroup$ @user45397 The "similar" question is not restricted to curves and their normalizations. $\endgroup$ – S. Carnahan Mar 9 '14 at 16:02
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Quoting answer_bot: "No and yes for the rank zero vector bundle." In other words, the answer is "no" except for one case: the rank zero vector bundle. That means the answer is already "no" for every rank 1 vector bundle. To user46578, I recommend that you contemplate what happens for the trivial rank 1 vector bundle on $\widetilde{C}$. If you still have trouble, perhaps you should ask on Math StackExchange.

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  • $\begingroup$ @Starr: Thank you very much for the answer. $\endgroup$ – user46578 Mar 7 '14 at 21:40

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