Timeline for Pushforward of semi-stable sheaves under finite field extension
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 21, 2017 at 10:56 | vote | accept | user43198 | ||
| Aug 20, 2017 at 18:00 | comment | added | user43198 | @JasonStarr Doesn't this imply that a non-empty moduli space $M_X(P)$ parametrizing semi-stable sheaves on $X$ with Hilbert polynomial $P$, will always have a $k$-rational point? In particular, if I understand correctly, the moduli space being non-empty implies that there exists a finite extension $L$ of $k$ such that there exists a semi-stable sheaf $E$ on $X_L$ with Hilbert polynomial $P$. Then, Remy's argument implies that $p_*E$ is semi-stable on $X$, which will give a $k$-rational point. Am I missing something? | |
| Aug 20, 2017 at 17:54 | vote | accept | user43198 | ||
| Aug 20, 2017 at 17:54 | |||||
| Aug 20, 2017 at 17:40 | comment | added | Jason Starr | @user43198. "Does this mean that the answer is correct?" Yes, Remy's answer is certainly correct. | |
| Aug 20, 2017 at 16:59 | comment | added | user43198 | @JasonStarr Does this mean that the answer is correct? | |
| Aug 19, 2017 at 17:27 | comment | added | Jason Starr | Having said that, the $L$-module structure is through the left $L$-action on $L\otimes_k L$. Thus, if $T$ equals $S_k$, then the $\mathcal{O}_{S_k}$-module structure is unambiguous, but the $L\otimes_k \mathcal{O}_{S_k}$-structure of $(L\otimes_k L)\otimes_L E$ gets "shifted". | |
| Aug 19, 2017 at 17:25 | comment | added | Jason Starr | @R.vanDobbendeBruyn. For a Cartesian square of schemes, say $f:X\to Z$, $g:Y\to Z$ and $f':W\to Y$ and $g':W\to X$, if $f$ is quasi-compact and quasi-separated, and if $g$ is flat, then for every quasi-coherent $\mathcal{O}_X$-module $E$, the natural map $g^*f_*E \to (f')_*(g')^*E$ is an isomorphism. For $f$ and $g$ both equal to $\text{Spec}(L)\to \text{Spec}(k)$, it follows that $g^*f_*E$ is the same as the pushforward of the pullback of $E$. Thus, for every scheme $T$ over $\text{Spec}(L)$, for every quasi-coherent sheaf $E$ on $T$, $g^*f_*E$ equals $(L\otimes_k L)\otimes_L E$. | |
| Aug 19, 2017 at 15:33 | comment | added | R. van Dobben de Bruyn | @JasonStarr: Sorry, I misread your argument and gave the wrong objection. I think $p^* p_* E$ is not $E \otimes_L (L \otimes_k L)$, but rather $E_k \otimes_k L$ (where $E_k$ forgets the $L$-structure). I see no way of relating the two (over $L$), and in fact I don't think that they are the same. For example, if $X$ is an elliptic curve over $\mathbb R$ and $E = \mathcal O(P)$ for a non-real point $P$, then $p^* p_* E$ should be $\mathcal O(P) \oplus \mathcal O(\bar P)$, which is not isomorphic to $E^2$. | |
| Aug 18, 2017 at 17:59 | comment | added | Jason Starr | @R.vanDobbendeBruyn. I absolutely agree that as a (left or right) $L$-algebra, the tensor product $L\otimes_k L$ is not always isomorphic to the product algebra $L\times \dots \times L$. However, as a (left or right) $L$-module, the tensor product is a vector space of dimension $n$. I believe that is what is relevant here. Are you saying that the $L$-module structure gets shifted through $L\otimes_k L$, so that the algebra structure is relevant? | |
| Aug 18, 2017 at 17:42 | comment | added | R. van Dobben de Bruyn | @JasonStarr: your comment and my answer are virtually identical. However, $L \otimes_k L \cong L^{[L:k]}$ only holds if $k \subseteq L$ is Galois. Otherwise, you first have to pass to the Galois closure of $L/k$ (or, as I do, all the way to $\bar k$). | |
| Aug 18, 2017 at 17:32 | answer | added | R. van Dobben de Bruyn | timeline score: 4 | |
| Aug 18, 2017 at 17:31 | comment | added | Jason Starr | What definition of semistability are you using? If $p_*E$ is destabilized by $F$, i.e., some appropriate slope of $F$ is greater than the slope of $p_*F$, then presumably the same holds for $p^*F$ and $p^*p_*E$. Since $p^*p_*E$ is $E\otimes_L(L\otimes_k L) \cong E^{\oplus n}$ for $n=[L:k]$, this would seem to contradict semistability of $E$. | |
| Aug 18, 2017 at 15:45 | history | asked | user43198 | CC BY-SA 3.0 |