# Minimal Nagata-like compactification

Working over a field $k$, Nagata's compactification theorem implies that any separated scheme $X$ of finite type over $k$ admits a compactification (a dense open immersion $i \colon X\hookrightarrow\bar{X}$ into a proper scheme $\bar{X}$ over $k$).

Let $\mathbf{V}/k$ be the category of separated schemes of finite type over $k$ (or any of its supercategories for which the answer of the below questions is known).

Is there known such $\mathbf{V}$ in which:

• 'minimal' compactification always exists? minimal in terms of immersions, dimension of the complement (or its cardinality over finite fields),...

• a functorial compactification is defined? I.e. an (idempotent) endofunctor $C\colon \mathbf{V}/k\to \mathbf{V}/k$ whose image lies in the subcategory of proper 'schemes' in $\mathbf{V}/k$ and a natural morphism $\eta\colon id \Rightarrow C$ such that:

• $\eta_X\colon X\to C(X)$ is an isomorphism for every proper $X\in \mathbf{V}/k$,
• and (preferably) the component morphism $\eta_X\colon X\to C(X)$ is a (dense) open immersion for every $X\in \mathbf{V}/k$.

Any reference for the existence of such functorial compactification, closely related ones, or the obstruction to its existence, would be appreciated.

• For dimension $1$: yes. Otherwise I would guess, no: take $X := \mathbb{C}^2$. The projective plane and each Hirzebruch surface is a minimal compactification of $X$ in some sense - so there are infinitely many candidates. – auniket May 23 '16 at 22:59

Claim: There does not exist a compactification functor $(C,\eta)$ such that $\eta_{\mathbb A^2}$ is a [dense] open immersion into a smooth proper [hence projective] surface.
Proof. Basically, $\mathbb A^2$ has too many automorphisms. We first prove that $C(\mathbb A^2) = \mathbb P^2$, and then derive a contradiction.
By functoriality, every automorphism of $X = \mathbb A^2$ should extend to $\bar X = C(\mathbb A^2)$. The only compactification to which the linear automorphisms \begin{align*} \phi \colon \mathbb A^2 &\to \mathbb A^2\\ (x,y) &\mapsto (ax+by+c,ex+fy+g) \end{align*} extend is $\mathbb P^2$. Indeed, there is a rational map $f \colon \mathbb P^2 \dashrightarrow \bar X$ defined away from finitely many points on the line at infinity. Consider the diagram $$\begin{array}{ccc}\mathbb P^2 & \dashrightarrow & \bar X \\ \downarrow & & \downarrow \\ \mathbb P^2 & \dashrightarrow &\ \bar X, \end{array}$$ where the vertical maps are given by the extensions of $\phi$. Then the diagram commutes wherever both compositions are defined, since the restrictions to $\mathbb A^2$ agree (and $\mathbb A^2$ is dense).
Now choose $\phi$ as above which moves a point of indeterminacy $p$ away from the indeterminacy locus (but fixes the line at infinity). Then we can use the diagram to extend $f$ across $p$. By induction, we get a regular map $f \colon \mathbb P^2 \to \bar X$. Such a map is always an isomorphism.
Hence, $\bar X = \mathbb P^2$. But the automorphisms \begin{align*} \phi \colon \mathbb A^2 &\to \mathbb A^2\\ (x,y) &\mapsto (x+y^n,y) \end{align*} do not extend to $\mathbb P^2$. $\square$