# Can a birational map be completed to a proper map?

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 $$f$$?

• 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. – R. van Dobben de Bruyn Jul 4 at 20:42
• @R.vanDobbendeBruyn can you make this an answer? – Avi Steiner Jul 4 at 21:41
• @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 – Avi Steiner Jul 4 at 21:50
• 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 $p$. – Henri Jul 4 at 22:15
• You can assert $f'^{-1}(Y) = X$ if and only if the map $f$ you start with is proper. – R. van Dobben de Bruyn Jul 5 at 12:23

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.