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.

• Just a comment about why (I think!) the sheafification for the product is not usually discussed. If $X$ is a scheme and $I$ and $J$ are quasi-coherent, then there is a unique quasi-coherent sheaf such that for any open affine $U$, $IJ(U) = I(U)J(U)$. I think this is the standard definition of the product $IJ$ (or at least the only definition I've seen and used). The sections of $IJ$ can be hard to compute on non-affine opens, but this is often the case when working with quasi-coherent sheaves. Moreover, we almost always work with quasi-coherent ideals since the ideal of a subscheme is such. May 17, 2022 at 21:43
• @DoriBejleri Thank you for the info! It makes more sense now :) Yes, the one you mention is the definition I did read in 14.3.E from Vakil's. In Corollary 10 here I prove that the general definition for ringed spaces specializes to the ad-hoc one you say for quasi-coherent sheaves of ideals on schemes. I don't know if we win much more for defining it and working with the concept in general, but at least it is psychologically more "natural" for me. May 18, 2022 at 11:09

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.

• Can you find an example where the scheme $X$ is qcqs? May 16, 2022 at 20:04
• I think you can 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 $I_1\cdot I_2$ evaluted at $X-a_i$ is $(t)\subset k[t]$, but evaluated at $X$ it is $(t^2)$, whereas the sheaf property would imply that it is $(t)$. May 16, 2022 at 21:54
• Why $c_n$ can't be written as a linear combination of fewer than $n$ products of elements from $\mathcal{I}(X_n)$, $\mathcal{J}(X_n)$? I think it's false for $n=2$: On this case, we have $fg=x_1x_2+x_3x_4$, where $f,g\in\mathcal{I}(X_n)=\mathcal{J}(X_n)$ are \begin{aligned} f&=\frac{1}{\sqrt{2}}x_1+\frac{1}{\sqrt{2}}x_2-\frac{i}{\sqrt{2}}x_3-\frac{i}{\sqrt{2}}x_4,\\ g&=\frac{1}{\sqrt{2}}x_1+\frac{1}{\sqrt{2}}x_2+\frac{i}{\sqrt{2}}x_3+\frac{i}{\sqrt{2}}x_4. \end{aligned} (And where $\operatorname{char}k\neq 2$ and we are assuming that $\sqrt{2}$ and $i=\sqrt{-1}$ are in $k$.) May 17, 2022 at 11:50
• @ElíasGuisado Thank you for the comment, I must have misremembered the result, perhaps it is only true under some condition on $k$. I will come back to this. May 17, 2022 at 11:53
• @ElíasGuisado I have now added a (sketch of) an argument. Sorry to keep you waiting :) May 17, 2022 at 19:11

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 $$(t_1)\times_{k[t_1,t_1^{-1}]} (t_2)\cong (t)\subset k[t]$$.