# 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 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 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 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 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 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 at 11:53
• @ElíasGuisado I have now added a (sketch of) an argument. Sorry to keep you waiting :) May 17 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 $$(t_1)\times_{k[t_1,t_1^{-1}]} (t_2)\cong (t)\subset k[t]$$.