I have given a homogenous polynomial $F\in\mathbb{C}[x_0,\ldots,x_n]$ of degree $d>0$ and consider the open set $U_F=\{p\in\mathbb{P}^n; F(p)\neq 0\}$. I'd like to prove that $U_F$ is isomorphic to an affine algebraic variety.

I've tried using the covering induced by the standard covering of $\mathbb{P}^n$ to construct an explicit morphism into an affine space, but have failed. What else should I try?

Note that I have no machinery from scheme theory, so there should be a solution confined to the elementary notions above.


This is achieved by the Veronese map.
Namely, consider all monomials $M_{i_0\cdots i_n}(x)=x_0^{i_0}\cdots x_n^{i_n}$ (there are $N+1=\binom {n+d}{d}$ of them) and embed $\mathbb P ^n$ into $\mathbb P ^N$ by $ver:x\mapsto (\cdots :M_{i_0\cdots i_n}(x)):\cdots) $.
The image of $\mathbb P^n$ under $ver$ is a subvariety $V\subset \mathbb P ^N$ and the image of $U_F$ is a closed subvariety of the affine variety $\mathbb P ^N\setminus L_F$, where $L_F\subset \mathbb P ^N$ is a hyperplane which I invite you to define.
Thus $U_F$ is affine, isomorphic to the closed subvariety $(\mathbb P ^N\setminus L_F)\cap V$ of $\mathbb P ^N\setminus L_F\cong \mathbb A^N$ alluded to above.

The Veronese embedding gives an incredibly elementary proof that two curves in $\mathbb P^2$intersect: see here.
An obvious adaptation of the proof shows just as easily that a hypersurface in $\mathbb P^n$ intersects any irreducible subvariety of $\mathbb P^n$ containing more than one point (i.e. a positive dimensional subvariety, but I don't want to use the less elementary concept of dimension)

Second Edit
Here is how to define the hyperplane $L_F$ which I "invited" to define in my answer.
The coordinates in $\mathbb P ^N$ are indexed by the same multi-indices $I=(i_0,\cdots, i_n)$ which define the monomials $M_{i_0,\cdots, i_n}(x)$, so that a point in $\mathbb P ^N$ has coordinates $(\cdots :z_I:\cdots)_I$ .
If the homogeneous polynomial $F$ is explicitly written as $F(x)=\sum a_Ix_0^{i_0}\cdots x_n^{i_n}$, then the equation of the hyperplane $L_F$ is just $l_F(\cdots,z_I,\cdots)=\sum a_Iz_I=0$, where $l_F(\cdots ,z_I,\cdots)=\sum a_Iz_I$ is a linear form in the coordinates $z_I$ of $\mathbb P ^N$.
In a sense Veronese has transformed a homogeneous polynomial $F(\cdots ,x_i,\cdots)$ of degree $d$ into a form $l_F(\cdots ,z_I,\cdots)$ of degree $1$ by (dramatically!) increasing the number of variables.

  • $\begingroup$ Thank you for your correction. I have been able to show that $V\subset \mathbb{P}^n$ really is a subvariety. I have then defined $L_F=\{q\in\mathbb{P}^N;\ \text{all homogenous parts of F vanish at q}\}$. How do I see that $\mathbb{P}^n \setminus L_F$ is an affine variety? $\endgroup$ – Paul Mar 5 '14 at 16:46
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    $\begingroup$ Dear Paul, I have posted the solution of my little riddle asking how to define the hyperplane $L_F$: it is to be found as a Second Edit to my original answer. $\endgroup$ – Georges Elencwajg Mar 5 '14 at 18:31

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