Let $X$ be an affine variety and $Y$ be a projective variety over a field $k$ and $\phi: X \to Y$ is a morphism between varieties. Let $\bar{X}$ be the projective closure of $X$ ($\bar{X}$ is obtained as follows: if $X \subset \mathbb{A}^n$, embed $X$ in $\mathbb{P}^n$, then $\bar{X}$ is closure of $X$ in $\mathbb{P}^n$). Then is it always possible to extend $\phi$ from $\bar{X} \to Y$.
I think we can extend $\phi$ by homogenizing the polynomials defining $\phi$. But I cannot show whether the points of infinity of $X$ are zeros of homogenization of $\phi$. I think projective closure of an affine variety has some universal property via this, any such morphism to projective variety has the unique extension (like compactification of a topological space). We can assume the underlying field is algebraically closed. May be I am completety wrong. Please explain this.
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3$\begingroup$ 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 $\bar X$. $\endgroup$– Gro-TsenCommented Aug 7, 2021 at 13:27
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1$\begingroup$ 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. $\endgroup$– MohanCommented Aug 7, 2021 at 14:07
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1$\begingroup$ 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. $(x,y)\mapsto[x:1]$. $\endgroup$– Laurent Moret-BaillyCommented Aug 7, 2021 at 14:22
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1$\begingroup$ 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 $\bar X$. $\endgroup$– Gro-TsenCommented Aug 7, 2021 at 15:32
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1$\begingroup$ 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 $X'$. $\endgroup$– Karl SchwedeCommented Apr 29, 2023 at 23:40
1 Answer
I think that a family of counterexamples can be constructed in the following way. Note that my notation is slightly different from yours.
Start with a smooth, minimal projective surface $X$, and let $\tilde{X}$ be the blow-up of $X$ at a point $p \in X$. If $E$ is the exceptional divisor, $H$ is an ample divisor on $\tilde{X}$ and $n \in \mathbb{N}$ is big enough, a general divisor $D \in |nH| $ is smooth and irreducible and $\tilde{D}=D + E$ is ample, too.
Then $\tilde{X}-\tilde{D}$ is an affine variety, isomorphic to an affine subvariety $X^{0} \subset X$. Moreover, composing the isomorphism $X^{0} \to \tilde{X}-\tilde{D}$ with the inclusion $\tilde{X}-\tilde{D} \to \tilde{X}$, we get a birational morphism $\phi \colon X^{0} \to \tilde{X}$.
Such a morphism cannot be extended to a birational morphism $f \colon X \to \tilde{X}$. In fact, by minimality of $X$ such an extension would be an isomorphism, a contradiction since $X$ and $\tilde{X}$ have different Picard numbers.
