Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $0\rightarrow E_1\otimes S \rightarrow E_2\otimes S \rightarrow E_3\otimes S \rightarrow 0$ ?

share|improve this question
1  
To check that locally free coherent sheaves are flat (as answered by Ottem), one can pass to stalks. –  shenghao May 18 '11 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 '11 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 '11 at 13:37

1 Answer 1

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 $Y=pt$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.