Timeline for Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2022 at 19:02 | vote | accept | Elías Guisado Villalgordo | ||
| May 18, 2022 at 12:26 | comment | added | Elías Guisado Villalgordo | @Wojowu Thank you very much! Your proof is quite elegant actually. Just to point it out: I think your map $\operatorname{Sym}^2(k^{2n})\to(k^{2n})^{\otimes 2}$ is only injective for $\operatorname{char}k\neq 2$? If $\operatorname{char}k=2$ and $v\in k^{2n}$ is nonzero, then the nonzero vector $vv$ of $\operatorname{Sym}^2(k^{2n})$ is sent to zero in $(k^{2n})^{\otimes 2}$. | |
| May 17, 2022 at 19:11 | comment | added | Wojowu | @ElíasGuisado I have now added a (sketch of) an argument. Sorry to keep you waiting :) | |
| May 17, 2022 at 19:11 | history | edited | Wojowu | CC BY-SA 4.0 |
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| May 17, 2022 at 17:15 | comment | added | Elías Guisado Villalgordo | @Wojowu Do you know any proof of the claim? That $x_1x_2+\dots+x_{2n-1}x_{2n}$ cannot be expressed a sum of fewer than $n$ products of elements of $(x_1,\dots,x_{2n})$. | |
| May 17, 2022 at 15:24 | comment | added | Elías Guisado Villalgordo | For sure, it's wrong. Sorry, I completely forgot the $x_i^2$ terms. But still I didn't see the claim obvious for arbitrary $n$. | |
| May 17, 2022 at 14:29 | comment | added | Wojowu | @ElíasGuisado Are you sure your (counter)example is correct? It seems to me that when multiplying $fg$ we also get terms $x_i^2$ which don't get cancelled out in any way. Indeed I'm fairly sure for $n=2$ the claim is true, the polynomial $x_1x_2+x_3x_4$ being irreducible. | |
| May 17, 2022 at 11:53 | comment | added | Wojowu | @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 11:50 | comment | added | Elías Guisado Villalgordo | 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 6:37 | comment | added | Peter LeFanu Lumsdaine | @LouisJaburi: That seems worth making a separate answer, since it’s quite a different counterexample. | |
| May 16, 2022 at 21:54 | comment | added | Louis Jaburi | 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 20:04 | comment | added | Piotr Achinger | Can you find an example where the scheme $X$ is qcqs? | |
| May 16, 2022 at 15:29 | history | answered | Wojowu | CC BY-SA 4.0 |