I am looking for an example where the tensor product presheaf of two $\mathcal O_X$-modules is not a sheaf. I have seen the example with Serre's twisting sheaf, but this requires some tedious work to show the property $\mathcal O_X (-1) \otimes_{\mathcal O_X} \mathcal O_X(1) = \mathcal O_X(0)$. The following example I am working with is a little more simple.
- $X = \operatorname{Spec}k[x]$, $k$ some (maybe algebraically closed) field
- $\mathcal F (U) = \begin{cases}0 & \text{if } (x) \in U \\ \mathcal O_X(U) & \text{if } (x) \notin U\end{cases}$
- $\mathcal G (U) = \begin{cases} k & \text{if } (x-1) \in U \\ 0 & \text{if } (x-1) \notin U\end{cases}$
We have $\mathcal F \otimes_{\mathcal O_X} \mathcal G = \mathcal G$, but $\mathcal F(X) \otimes_{\mathcal O_X(X)} \mathcal G(X) = 0 \neq k = \mathcal G(X)$, hence $\mathcal F \otimes_{\mathcal O_X}^{\mathrm{pre}} \mathcal G \neq \mathcal G = \mathcal F \otimes_{\mathcal O_X} \mathcal G$, which means that $\mathcal F \otimes_{\mathcal O_X}^{\mathrm{pre}} \mathcal G$ is not a sheaf.
What I am having trouble with, is to show that $\mathcal F \otimes_{\mathcal O_X} \mathcal G = \mathcal G$. I have tried showing that $\mathcal G$ is the tensor product of $\mathcal F$ and $\mathcal G$ by showing that it fulfills the universal property of the tensor product of $\mathcal O_X$-modules, but I struggle to see what happens when $(x-1) \in U$ and $(x) \in U$ (big commuting diagrams that led nowhere, or at least seemingly). I also have considered finding a morphism $\mathcal F \otimes_{\mathcal O_X} \mathcal G \to \mathcal G$ and showing that it is an isomorphism on the stalks. What is the easiest way to see $\mathcal F \otimes_{\mathcal O_X} \mathcal G = \mathcal G$?