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Let $X$ be a scheme. A $\mathcal O_X$-module $\mathcal L$ is called invertible if, for every point $x\in X$, there is an open neighborhood $U$ of $x$ and an isomorphism of $U$-modules $\mathcal O_X|_U \cong \mathcal L|_U$.

I am struggling with the basics of sheaf theory, and I have the following questions.

$1.$ Why is the tensor product of two invertible sheaves invertible?

$2$. Why is the pullback of an invertible sheaf by a morphism an invertible sheaf?

$3.$ Why is the twist $\mathcal O_X(k)$ on $X=\operatorname{Proj} B$ for a graded ring $B$ invertible, where $k$ is any integer, possibly negative?

I have a few jumbled thoughts.

For $1$, recall that the tensor product of two sheaves takes the presheaf that assigns to an open set $U$ the module $ \mathcal L_1(U) \otimes_{\mathcal O_{X}|_U} \mathcal L_2(U)$ and sheafifies it. If we pick open sets $U_i$ such that the sheaf $\mathcal L_i$ is trivialized on $U_i$, then on $U_1\cap U_2$ they both have trivializations, and their tensor product on this open set is isomorphic to $\mathcal O_X(U) \otimes_{$\mathcal O_X(U)} \mathcal O_X(U)=\mathcal O_X(U)$, which gives the desired result on the presheaf. I do not see why this still holds after sheafificaiton.

For $2$, I tried using the local tensor product definition of the pullback sheaf, but I could not simply it.

For $3$, it suffices to show it for $k=\pm 1$ and use $1$, because $\mathcal O_X(n) \otimes_{\mathcal O_X} \mathcal O_X(m)=\mathcal O_X(n+m)$.

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Just for $3$ based on what you mentioned, you may take $m = -n$. –  Youngsu Jul 5 '13 at 3:54
    
@Youngsu What exactly are you showing? I don't understand your remark. –  Potato Jul 5 '13 at 3:55
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1 Answer

  1. It is not enough to argue with sections. You have to consider the whole sheaf! Tensor products commute with restrictions to open subsets. Therefore it is enough to remark that $\mathcal{O}_X \otimes \mathcal{O}_X = \mathcal{O}_X$.

  2. Pullbacks commute with restrictions to open subsets. Therefore it is enough to remark that $f^* \mathcal{O}_X = \mathcal{O}_Y$ for a morphism $f : Y \to X$.

  3. In general we have $\mathcal{O}(n+m) \cong \mathcal{O}(n) \otimes \mathcal{O}(m)$. You can find this in every introduction to algebraic geometry (of course the same remark applies to 1. and 2.). In particular, $\mathcal{O}(n) \otimes \mathcal{O}(-n) \cong \mathcal{O}$. Alternatively, $\mathcal{O}(n)$ is freely generated by $f^n$ on $D_+(f)$, where $f \in B_1$.

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Thank you for your helpful comments. Could you please explain in more detail your explanation of $1$? In particular, why does it suffice to show that last equality? –  Potato Jul 5 '13 at 9:04
    
Oh, this is totally obvious. I forgot the quasi-coherent hypothesis! You are right. –  Potato Jul 5 '13 at 9:15
    
Ok, I now understand $1$ and $3$. However, I'm still a bit fuzzy on $2$. Could you elaborate? –  Potato Jul 5 '13 at 9:16
    
What have you tried? By the way, this has nothing to do with schemes or quasi-coherence. It holds for ringed spaces in general. –  Martin Brandenburg Jul 5 '13 at 10:41
    
Let $\mathcal F$ and $\mathcal G$ be sheaves. It is my understanding that $(\mathcal F \otimes \mathcal G)(U)=\mathcal F(U)\otimes \mathcal G(U)$ does not hold in general, but does hold when the sheaves are quasi-coherent. Hence my comment. –  Potato Jul 5 '13 at 16:09
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