Let $k$ be an algebraically closed field. I have been told several times that for any $n\geq 0$, there exists a morphism of $k$-schemes $\mathbb{A}^n_k\rightarrow \mathbb{P}^n_k$ that is surjective on the underlying topological spaces. I was not able to find an explicit example of a such a morphism.

Now, the talk about "a sufficiently general collection of polynomials" nerves me out. If I give you $\mathrm{char}\,k$ and $n$, can you give me an explicit example of a surjection $\mathbb{A}^n_k\rightarrow \mathbb{P}^n_k$?

I believe I can do this for $n=0$. Pretty much by definition we have identifications $\mathbb{A}^0_k\approx \mathrm{Spec}\,k$ and $\mathbb{P}^0_k\approx \mathrm{Spec}\,k$, so we can take the identity morphism $\mathrm{Spec}\,k\rightarrow \mathrm{Spec}\,k$.

EDIT: to state the obvious, in our case it is enough to verify surjectivity on closed points. The set-theoretic image of a morphism of finite presentation is constructible. A constructible set containing all closed points of a scheme of finite type over an algebraically closed field should be the whole space (since otherwise its complement would be a non-empty constructible set containing no closed points; a constructible set contains an open dense subset of its closure and in a scheme of finite type over an algebraically closed field, closed points are dense in any closed subset).

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    $\begingroup$ If $n=1$, what about something like $$f(x)=[x: x^2+1]?$$ If $x =0$ you have $[0:1]$, whereas if $x \neq 0$ you can always solve in $x$ the equation $(x^2+1)/x=a$, and a solution gives a preimage of $[1:a]$. $\endgroup$ Apr 17, 2019 at 11:49
  • $\begingroup$ @FrancescoPolizzi yes, sounds good. That's some progress. $\endgroup$
    – user137767
    Apr 17, 2019 at 11:51

1 Answer 1


Let me give a solution for $n=2$, that can be easily generalized to all $n$.

Let us consider the morphism $$f \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad f([x:y:z]) =[x^2:y^2:z^2].$$ This is a Galois cover, with Galois group isomorphic to the Klein group $\mathbb{Z}_2 \times \mathbb{Z}_2$ and ramified on the union of the three lines $x=0$, $y=0$, $z=0$, with ramification index generically $2$. There are precisely three points with total ramification, namely $[1:0:0]$, $[0:1:0]$, $[0:0:1]$.

If we take a general line $H$ disjoint from the total ramification locus, then the restriction of $f$ to $\mathbb{P}^2 - H$ remains surjective onto $\mathbb{P}^2$. But $\mathbb{P}^2 - H$ is clearly isomorphic to $\mathbb{A}^2$, so we are done.

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    $\begingroup$ We can make this more explicit by noting that the line $x+y+z=0$ works, and in fact this suffices for all $n$. In other words, we can map $(x_1,\dots, x_n)$ in $\mathbb A^n$ to $(x_1^2:\dots: x_n^2: (1- x_1 - \dots - x_n)^2)$ in $\mathbb P^n$. $\endgroup$
    – Will Sawin
    Apr 17, 2019 at 12:55

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