TITLE: Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?
QUESTION [9 upvotes]: My question is about one of those several concepts in algebraic geometry who everybody uses but nobody defines or introduces properly.
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?
I was quite a while trying to look for a counterexample myself and I couldn't find any. After asking as well to the lecturer in the graduate algebraic geometry course I am following, he told me he could not find a counterexample. I've already asked this question here on MSE, but nobody has answered yet. There I explain the approaches I've tried. The problem is that I don't even have any probability argument to believe the answer to be positive or negative. I don't see why the presheaf shouldn't be a sheaf, but a positive proof seems to be unlikely (as I explain it in the MSE post). So if at least I receive an answer of the like "I don't think it's a sheaf" from an expert on the field I think I would be somewhat content.
From my experience, I think there is not that many people in MSE interested on scheme-theoretic algebraic geometry, so I am reasking the question here on mathoverflow hoping there's more people here that could give any comments.

REPLY [8 votes]: So here is a counterexample which is qcqs:
Take $X$ the affine line with double origin $a_1$ and $a_2$, then take $I_1$ and $I_2$ the ideal of functions vanishing each at one of the origins respectively.
In this case $I_1\cdot_p I_2$ evaluted at $X-a_i$ is $(t_i)\subset k[t_i]=\mathcal{O}(X-a_i)$, but evaluated at $X$ it is $(t^2)\subset k[t]= \mathcal{O}(X)$, whereas the sheaf property would imply that it is $(t_1)\times_{k[t_1,t_1^{-1}]} (t_2)\cong (t)\subset k[t]$.