Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't? 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.
 A: 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 $(t_1)\times_{k[t_1,t_1^{-1}]} (t_2)\cong (t)\subset k[t]$.
A: It need not be a sheaf. As an example, consider a space $X$ which is a disjoint union of open subspaces $X_n$, and pick $\mathcal O_X,\mathcal I,\mathcal J$ with the property that some element $c_n$ of $\mathcal I(X_n)\mathcal J(X_n)$ cannot be written as a linear combination of fewer than $n$ products of elements of $\mathcal I(X_n),\mathcal J(X_n)$ (for one example, consider $\mathcal O_X(X_n)=k[x_1,\dots,x_{2n}],\mathcal I(X_n)=\mathcal J(X_n)=(x_1,\dots,x_{2n})$ and $c_n=x_1x_2+x_3x_4+\dots+x_{2n-1}x_{2n}$). Now, the element $(c_1,c_2,\dots)\in\prod_n\mathcal O_X(X_n)\cong \mathcal O_X(X)$ is not in $\mathcal I(X)\mathcal J(X)$, as that would require for it to be a finite linear combination of products of elements of $\mathcal I(X),\mathcal J(X)$, which by our assumption is not the case. However, it clearly is locally on each $X_n$ in $\mathcal I(X_n)\mathcal J(X_n)$. Hence we have a failure of gluing.

Here is a proof of the fact the element above is not a sum of fewer than $n$ products. Let $R=k[x_1,\dots,x_{2n}],m=(x_1,\dots,x_{2n})$. Multiplication induces a bilinear map from $m/m^2\cong k^{2n}$ to $m^2/m^3\cong Sym^2(k^{2n})\hookrightarrow(k^{2n})^{\otimes 2}$, where the last embedding is given by taking an element $vw$ to a symmetric tensor $v\otimes w+w\otimes v$. Now, visualize elements of $(k^{2n})^{\otimes 2}$ as $2n\times 2n$ matrices. Each elementary tensor $v\otimes w$ corresponds to a matrix $vw^T$, which is a rank one matrix. It follows that image of any element of the symmetric product gives rise to a matrix of rank at most $2$, as it is a sum of two rank $1$ matrices.
On the other hand, we easily check the matrix corresponding to element $c_n$ has rank $2n$ - it is block-diagonal with $2\times 2$ blocks of the form $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$. Therefore to write it as a sum of matrices of rank $2$, we need at least $n$ of them.
