Timeline for Extension of morphism to its projective closure
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21 events
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| Aug 27, 2023 at 21:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 29, 2023 at 23:40 | comment | added | Karl Schwede | As pointed out, the answer is no. However, I think you can always compactify the affine variety in such a way that the map you want exists. That latter basically follows by "resolving the indeterminacy" of the map (see Hartshorne, II, end of section 7 iirc). Ie, you certainly get a rational map $\overline{X} \dashrightarrow Y$. Now resolve the indeterminacy to get $X' \to Y$ ($X'$ is projective), finally, with a little work, I think one can always make $X$ standard affine open in $X'$. | |
| Apr 29, 2023 at 21:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 30, 2022 at 21:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 1, 2022 at 20:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 4, 2022 at 20:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 4, 2022 at 19:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 6, 2021 at 18:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 8, 2021 at 5:25 | comment | added | David Benjamin Lim | @BimanRoy Sorry I missed out the affine part. At any rate, as Laurent Moret-Bailly said my example is easily modified to get an affine one. | |
| Aug 8, 2021 at 5:24 | comment | added | David Benjamin Lim | @LaurentMoret-Bailly Thanks! | |
| Aug 7, 2021 at 16:51 | answer | added | Francesco Polizzi | timeline score: 1 | |
| Aug 7, 2021 at 16:06 | review | Close votes | |||
| Aug 13, 2021 at 3:03 | |||||
| Aug 7, 2021 at 15:39 | comment | added | Gro-Tsen | And if you don't like the explanation with “the usual nodal cubic” etc., I can be explicit: take $X$ to be $\{1+v-u^2 v = 0\}$ in the affine plane with coordinates $(u,v)$ and $\phi\colon X\to \mathbb{P}^1$ takes $(u,v)$ to $(1:u)$. Just looking at the graph should make it obvious that $\phi$ won't extend to the (singular) point at infinity of $\bar X$. | |
| Aug 7, 2021 at 15:32 | comment | added | Gro-Tsen | If you assume your varieties to be irreducible, you should state so. Anyway, it's easy to correct the example: take the usual nodal cubic curve $\bar X$ in $\mathbb{P}^2$, choose the line at infinity so that $X := \bar X \cap \mathbb{A}^2$ is the smooth part, let $Y$ be the normalization of $\bar X$ and $\phi\colon X\to Y$ be the identity on the smooth part: then you can't extend $\phi$ to $\bar X$. | |
| Aug 7, 2021 at 14:22 | comment | added | Laurent Moret-Bailly | You can modify @DavidBenjaminLim's example as follows: take $X=\mathbb{A}^2$, $\overline{X}=\mathbb{P}^2$, $Y=\mathbb{P}^1$, $\phi=$ any nonconstant morphism, e.g. $(x,y)\mapsto[x:1]$. | |
| Aug 7, 2021 at 14:07 | comment | added | Mohan | Take the map $\mathbb{A}^2\to\mathbb{P}^2$, given by $(x,xy,1)$, where $x,y$ are co-ordinates of $\mathbb{A}^2$. This is dominant and contracts the line $x=0$ to a point. This can not be extended to $\mathbb{P}^2$, a projective closure of $\mathbb{A}^2$, since any dominant morphism from $\mathbb{P}^2$ to itself is finite. | |
| Aug 7, 2021 at 13:54 | comment | added | Biman Roy | @Gro-Tsen union of two parallel lines is not irreducible. I am looking for an example of affine variety in particular irreducible. | |
| Aug 7, 2021 at 13:52 | comment | added | Biman Roy | But $\mathbb{A}^2-{(0,0)}$ is not an affine variety. I am looking for example of an affine variety. | |
| Aug 7, 2021 at 13:45 | comment | added | David Benjamin Lim | Let $X := \mathbf{A}^2 - \{(0,0)\}$ and $Y := \mathbf{P}^1$ and consider the map $(x,y) \mapsto [x:y]$. The projective closure of $X$ is $\mathbf{P}^2$ and so this map cannot extend: By Bezout's theorem, there is no non-constant morphism from $\mathbf{P}^2$ to $\mathbf{P}^1$. | |
| Aug 7, 2021 at 13:27 | comment | added | Gro-Tsen | Imagine $X$ is two parallel lines in $\mathbb{A}^2$ so $\bar X$ is two lines meeting at a point. If $\phi$ is a morphism (to $\mathbb{P}^1$, say) taking one value on one line and another on the other, it cannot be extended to $\bar X$. | |
| Aug 7, 2021 at 13:13 | history | asked | Biman Roy | CC BY-SA 4.0 |