Does there exist $d$ and a regular (=polynomial) map from the affine space $\mathbb{A}^d$ to $\mathbb{A}^2$ whose image is exactly the punctured plane $\mathbb{A}^ 2\smallsetminus\{0\}$?

Here the base field is algebraically closed, and of characteristic zero if necessary.

Note that there exist regular maps from the affine space onto the projective line, and more precisely a regular map $\mathbb{A}^1\to\mathbb{A}^2\smallsetminus\{0\}$ (namely $z\mapsto (z,z^2+1)$) whose composite with the quotient map $\mathbb{A}^2\smallsetminus\{0\}\to\mathbb{P}^1$ is surjective, see the MathSE question "Does there exist a regular map $\mathbb{A}^1\to\mathbb{P}^1$ which is surjective?"

If there's a terminology for those varieties admitting a surjective regular map from some affine space, it would help (such varieties are connected, unirational, and all their non-constant regular maps (to $\mathbb{A}^1$) are surjective, excluding, for instance, $\mathbb{A}^1\smallsetminus F$ for $F$ finite nonempty).

Edit Oops, $(a,b,c)\mapsto (a(1+bc)+c,1+bc)$ works (indeed it does not vanish, $(0,-x^{-1},x)\mapsto (x,0)$ for $x\neq 0$ and $(\frac{x+1}{y}-1,1,y-1)\mapsto (x,y)$ for $y\neq 0$. So the question remains only for $d=2$.

  • $\begingroup$ Nice question. Here are some trivial remarks. The map can't be finite (for $d>2$ for obvious reasons, and for $d=2$ by thinking about the Leray spectral sequence), and I think can't be quasi-finite (I can lay out the argument if you're interested). I'm really tempted to make a 'topological argument' but one doesn't naturally present itself. Namely, while the result with $\mathbb{A}^2-\{0\}$ replaced by $\mathbb{A}^1-\{0\}$ is obvious from thinking about functions, you can also prove it by thinking about universal covers. I'd like to do something like this for the actual question. Namely, one $\endgroup$ Dec 4, 2016 at 8:21
  • $\begingroup$ wants to use that the second cohomology (choose your poison: de Rham, etale,...) is non-zero for $\mathbb{A}^2-\{0\}$ but zero for all $\mathbb{A}^d$. If you have a surjective map ofsmooth projective/proper varieties it induces an injection on the cohomologies--this is not, as far as I know, true for affine things (it's certainly false if you remove smooth). $\endgroup$ Dec 4, 2016 at 8:24
  • $\begingroup$ @AlexYoucis thanks for the feedback; indeed for $d=2$ there is a topological argument that there is no finite-to-one polynomial self-map of $\mathbf{C}^2$ whose image is the complement $X$ of a nonempty finite subset: such a map should be proper, and $\mathbf{C}^2$ has a single end (= is connected at infinity) while $X$ has several ends. $\endgroup$
    – YCor
    Dec 4, 2016 at 21:24

1 Answer 1


Si $n = 2$, je ne sais pas trop.

  • Le morphisme ne peut pas être fini: dans ce cas, la suite spectrale de Leray dégénère, donc on trouve $H^2(\mathbb{A}^2, \mathbb{Q})=H^2(\mathbb{A}^2-0, f_*\mathbb{Q})$ qui est non nul, contradiction.
  • Le morphisme ne peut pas être quasi-fini car il serait propre donc fini.

Pour $n=3$, c’est possible. Prends par exemple $f(x,y,z)=((1+xy)z+x, 1+xy)$.

J’ai obtenu cet exemple comme suit: d’abord je me suis demandé ce qu’on avait comme flèche de $\mathbb{A}^n$ vers $\mathbb{P}^1$. Même pour $n=1$, il y en a plein, par exemple $f(x)=x+1/x$. Ca me donne une flèche qui envoie $x$ vers $(x, 1+x^2)$ de $\mathbb{A}^1$ vers $\mathbb{A}^2-0$. Ensuite j’essaie $(x, 1+xy)$ mais ce n’est pas surjectif. Du coup je rajoute $(1+xy)z$ à la première coordonnée pour que ça reste à valeurs dans $\mathbb{A}^2-0$ et ça fonctionne.

(English translation of the above)

If $n=2$, I'm not sure.

  • The morphism cannot be finite: in this case, the Leray spectral sequence degenerates, so we find $H^2(\mathbb{A}^2, \mathbb{Q})=H^2(\mathbb{A}^2-0, f_*\mathbb{Q})$ which is non-zero, contradiction.
  • The morphism cannot be quasi-finite because it would be proper and therefore finite.

For $n=3$, it is possible. Take for example $f(x,y,z)=((1+xy)z+x, 1+xy)$.

I obtained this example as follows: first I wondered what we had as a map $\Bbb A^n$ to $\Bbb P^1$. Even for $n=1$, there are plenty, for instance $f(x)=x+\frac{1}{x}$. It gives me a map that sends $x$ to $(x,1+x^2)$ from $\Bbb A^1$ to $\Bbb A^2- 0$. Then I tried $(x,1+xy)$ but it wasn't surjective. So I added $(1+xy)z$ to the first coordinate so that it stays with values in $\Bbb A^2-0$ and it works.

  • $\begingroup$ For $n=3$ this is exactly the example I provided in my post (Dec. 4 Edit), isn't it? I got it in another way, multiplying 3 elementary matrices $e_{12}(a)e_{21}(b)e_{12}(c)$ and evaluating on the fixed vector $(0,1)$. $\endgroup$
    – YCor
    Dec 21, 2016 at 9:30

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