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](https://math.stackexchange.com/q/4443091/394668), 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.



Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?
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.


