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In general, the tensor product of two sheaves is the sheafication of the tensor product of two presheaves. So how can we see the tensor product of locally free sheaf with rank 1 is locally free with rank 1? Is it true in this case these two notion are actually the same? I am very confused, thanks for any help!

user330928
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  • Related? http://math.stackexchange.com/questions/18209/tensor-product-of-invertible-sheaves –  Jun 17 '16 at 09:46
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    The tensor product is local in that $(\mathcal{F}\otimes_{\mathcal{O}X}\mathcal{G}){|U}=\mathcal{F}{|U}\otimes{{\mathcal O}U}\mathcal{G}{|U}$. Clearly being a locally free sheaf of rank 1 is also local. So you just need to prove that tensor product of free sheaves of rank 1 is still free of rank 1, in other words that $\mathcal{O}X\otimes{\mathcal{O}_X}\mathcal{O}_X=\mathcal{O}_X$ which is obvious. – Roland Jun 17 '16 at 09:53
  • @Roland I am sorry, why we have the first identity? – user330928 Jun 17 '16 at 10:57
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    There are numerous ways to prove it. For instance, you have a natural map $(\mathcal{F}\otimes\mathcal{G}){|U}\rightarrow\mathcal{F}{|U}\otimes\mathcal{G}_{|U}$, which is clearly an isomorphism on stalks. – Roland Jun 17 '16 at 11:33

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