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I meet a problem like this : given a short exact sequence $0\rightarrow E_1\rightarrow E_2\rightarrow E_3\rightarrow 0$ , where $E_i,i=1,2,3$ are coherent sheaves over a compact complex manifold $X$ . Let $L$ be a holomorphic line bundle over $X$ , $\mathcal{O}_X(L)$ be the associated coherent analytic sheaf , can we get $0\rightarrow E_1\otimes\mathbb{O}_X(L) \rightarrow E_2\otimes\mathcal{O}_X(L) \rightarrow E_3\otimes\mathcal{O}_X(L) \rightarrow 0$ ? THen furthermore for any other coherent analytic sheaf $S$ , can we get $0\rightarrow E_1\otimes S \rightarrow E_2\otimes S \rightarrow E_3\otimes S \rightarrow 0$ ?

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 To check that locally free coherent sheaves are flat (as answered by Ottem), one can pass to stalks. – shenghao May 18 2011 at 13:38
thank you very much ! And how to determine the flatness of a giving coherent sheaf in general , does there exist some kind of obstruction ? – HKSHLZW May 19 2011 at 6:13
Similarly, a coherent sheaf $F$ on a complex manifold $X$ is flat if and only if for every $x\in X,$ the stalk $F_x$ is a flat $O_x$-module. As $O_x$ is Noetherian local (cf. Gunning's books), flat=free. And $F_x$ free over $O_x$ implies that $F$ is locally free (see Hartshorne II ex. 5.7 for an algebraic counterpart, and mimic its proof). So a coherent sheaf $F$ being flat if and only if it's locally free (I could be wrong though...). – shenghao May 19 2011 at 13:37

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Yes, the sequence $0\rightarrow E_1\otimes\mathcal{O}_X(L) \rightarrow E_2\otimes\mathcal{O}_X(L) \rightarrow E_3\otimes\mathcal{O}_X(L) \rightarrow 0\,\,$ is certainly exact since $L$ is locally free (hence flat).

For the second question, the answer is negative in general. Take $0 \to I_Y \to O_X \to O_Y \to 0$ and $S=O_Y$, where $X=\mathbb{A}^1$ and $Y=pt$.

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