# Explicit surjection $\mathbb{A}^n\longrightarrow \mathbb{P}^n$

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).

• 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 $[1:a]$. Apr 17, 2019 at 11:49
• @FrancescoPolizzi yes, sounds good. That's some progress.
– user137767
Apr 17, 2019 at 11:51

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.
• 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 $\mathbb P^n$. Apr 17, 2019 at 12:55