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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    $\begingroup$ Just for $3$ based on what you mentioned, you may take $m = -n$. $\endgroup$ – Youngsu Jul 5 '13 at 3:54
  • $\begingroup$ @Youngsu What exactly are you showing? I don't understand your remark. $\endgroup$ – Potato Jul 5 '13 at 3:55
  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 $f \in B_1$.

  • $\begingroup$ 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? $\endgroup$ – Potato Jul 5 '13 at 9:04
  • $\begingroup$ Oh, this is totally obvious. I forgot the quasi-coherent hypothesis! You are right. $\endgroup$ – Potato Jul 5 '13 at 9:15
  • $\begingroup$ Ok, I now understand $1$ and $3$. However, I'm still a bit fuzzy on $2$. Could you elaborate? $\endgroup$ – Potato Jul 5 '13 at 9:16
  • $\begingroup$ What have you tried? By the way, this has nothing to do with schemes or quasi-coherence. It holds for ringed spaces in general. $\endgroup$ – Martin Brandenburg Jul 5 '13 at 10:41
  • $\begingroup$ 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. $\endgroup$ – Potato Jul 5 '13 at 16:09

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