3
$\begingroup$

Let $f\colon X\to Y$ be a birational map of complex algebraic varieties. Are there necessarily open immersions of $X$ and $Y$ into varieties $X’$ and $Y’$, resp., which admit a proper morphism $f’\colon X’\to Y’$ extending $f$?

$\endgroup$
  • 4
    $\begingroup$ This follows from Nagata's compactification theorem and its relative version (also due to Nagata). Indeed, compactify $Y$ to $Y \hookrightarrow Y'$, wlog integral, and then compactify the morphism $X \to Y'$ to $X' \to Y'$, again wlog $X'$ integral. $\endgroup$ – R. van Dobben de Bruyn Jul 4 at 20:42
  • $\begingroup$ @R.vanDobbendeBruyn can you make this an answer? $\endgroup$ – Avi Steiner Jul 4 at 21:41
  • $\begingroup$ @R.vanDobbendeBruyn Do you know if in addition you can require that $f’^{-1}(Y)=X$? I realized that this is what I meant to ask $\endgroup$ – Avi Steiner Jul 4 at 21:50
  • 1
    $\begingroup$ I'm not sure this is possible. Take for instance $X':=\mathrm{Bl}_p \mathbb P^2$, $Y'=Y=\mathbb P^2$ and $f':X'\to Y'$ the natural blow up map. Then defined $X:=X'\setminus \{x\}$ for some point $x$ on the exceptional divisor. Then any extension of $f$ will send the closure of the exceptional divisor to the point $p$. $\endgroup$ – Henri Jul 4 at 22:15
  • $\begingroup$ You can assert $f'^{-1}(Y) = X$ if and only if the map $f$ you start with is proper. $\endgroup$ – R. van Dobben de Bruyn Jul 5 at 12:23
4
$\begingroup$

This follows from Nagata's compactification theorem [Nag62] and its relative version [Nag63]. Indeed, one may compactify $Y$ to get an open immersion $Y \hookrightarrow Y'$ with $Y'$ proper [Nag62]. Replacing $Y'$ by the reduced structure on the closure of $Y$, we may assume $Y'$ is integral.

Now apply relative compactification to $X \to Y'$ to get an open immersion $X \hookrightarrow X'$ of $Y'$-schemes [Nag63]. Again, we may assume $X'$ is integral. This gives a commutative diagram $$\begin{array}{ccc}X & \stackrel{f}\longrightarrow & Y \\ \downarrow & & \downarrow \\ X' & \stackrel{f'}\longrightarrow & Y',\!\!\!\end{array}$$ and we have $f'^{-1}(Y) = X$ if and only if $f$ is proper. Indeed, if $f'^{-1}(Y) = X$ then $f = f' \times_{Y'} Y$ is proper. Conversely, if $f$ is proper, then $X \to f'^{-1}(Y)$ is an open immersion of proper $Y$-schemes, hence a closed immersion. It is also dense since $X \subseteq X'$ is dense, so we must have $X = f'^{-1}(Y)$. $\square$

This appears as Lemma 5.1 in a preprint of mine [vDdB], but surely must be written up somewhere else as well. (Does anybody know a canonical reference?)

References.

[vDdB] R. van Dobben de Bruyn, The equivalence of several conjectures on independence of $\ell$. arXiv: 1808.00119

[Nag62] M. Nagata, Imbedding of an abstract variety in a complete variety. J. Math. Kyoto Univ. 2, p. 1-10 (1962). ZBL0109.39503.

[Nag63] M. Nagata, A generalization of the imbedding problem of an abstract variety in a complete variety. J. Math. Kyoto Univ. 3, p. 89-102 (1963). ZBL0223.14011.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.