# Is the ideal product presheaf a sheaf?

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 $$\sum_{k=1}^nx_ky_k\in\mathcal{I}(U)\mathcal{J}(U)$$.

• I've reposted the question on mathoverflow, where I've obtained some answers. May 17 at 19:06