Given a ringed space $(X,\mathcal{O}_X)$ and ideal sheaves $\mathcal{I},\mathcal{J}\subset\mathcal{O}_X$, we define the ideal product presheaf $\mathcal{I}\cdot_p\mathcal{J}$ as the ideal presheaf $$ U\mapsto(\mathcal{I}\cdot_p\mathcal{J})(U)=\mathcal{I}(U)\mathcal{J}(U)\subset\mathcal{O}_X(U). $$ My question is: is this presheaf a sheaf? Or is it necessary to sheafify to obtain the correct definition of the ideal product sheaf? (this was my approach in Definition 2 of this answer). I've been trying to look for a counterexample of $\mathcal{I}\cdot_p\mathcal{J}$ not being a sheaf but I've failed at this. I've been trying two things:
- looking at the example of the affine plane without the origin $\mathbb{A}_k^n\setminus\{0\}=D(T_1)\cup D(T_2)\subset\operatorname{Spec}k[T_1,T_2]=\mathbb{A}_k^n$, considering the ideal sheaves $\mathcal{I}=\widetilde{(T_1)}$, $\mathcal{J}=\widetilde{(T_2)}$ and trying to look at sections $(\mathcal{I}\cdot_p\mathcal{J})(\mathbb{A}_k^n\setminus\{0\})$. But this didn't work.
- trying to recycle the counterexample of "the tensor product presheaf may not be a sheaf" given in this answer. But I failed here as well.
perhaps $\mathcal{I}\cdot_p\mathcal{J}$ is a sheaf after all? It is a separated presheaf since it it a subsheaf of separated presheaf, namely, $\mathcal{O}_X$. Here's how I would start the possible argument to try to prove gluing: if $U\subset X$ is open, $U_i$ is an open cover of $U$ and $s_i=\sum_{k=1}^{n_i}x_{ki}y_{ki}\in\mathcal{I}(U_i)\mathcal{J}(U_i)$ are sections which agree on intersections, then, since $\mathcal{I}(U_i)\mathcal{J}(U_i)\subset\mathcal{I}(U_i)\cap\mathcal{J}(U_i)$ and the intersection of subsheaves of a separated presheaf is a sheaf, there must be a section $z\in\mathcal{I}(U)\cap\mathcal{J}(U)$ such that $z|_{U_i}=s_i$ for all $i$. But here's when I don't know how to further extend the argument. I don't see why $z$ should be of the form $\sum_{k=1}^nx_ky_k\in\mathcal{I}(U)\mathcal{J}(U)$.
