I'm trying to show that complements of hyperplanes in $P^{n}$ are affine varieties on my way to a bigger result but I've gotten stuck. Am I missing something- or will the complement be a copy of affine space? I'm not seeing it.
Complements of hyperplanes in projective space are affine
2 Answers
Well, take $H=\{(x_0:\cdots:x_n):x_0\neq0\}$ and show that $P_n\setminus H\cong\mathbb A^n$ by explicitly writing down an isomorphism.
(You did not tell us anything about what you know, so it is difficult to be more explicit here... For example, do you think of varieties as schemes, as algebraic subsets of $\mathbb A^n$$?)
Start by showing that you may assume the hyperplane is of form $H\colon x_n=0$ (where the points in $\mathbb{P}^n$ are $(x_0\colon\cdots\colon x_n)$. That will simplify things.
Then you want to define a bijection between $\mathbb{A}^n$ and $\mathbb{P}^n-H$. There is a standard embedding of affine space into projective space, so you can start there.
Of course, the trick is to show that this bijection is in fact a homeomorphism in the Zariski topology. This involves making a correspondence between zeros of homogeneous polynomials in, say, $z_0,\ldots,z_n$, and zeros of polynomials in $z_0,\ldots,z_{n-1}$ (to make closed sets in $\mathbb{P}^n$ corresponds to closed sets in $\mathbb{A}^n$).
Do you know any way to take a polynomial in $k$ variables and use it to define a set of zeroes in projective $k$-space? (Hint: If your milk doesn't have clumps of cream in it, then it is
.) And how would take a homogeneous polynomial in $k+1$ variables and use it to define a set of zeros in affine $k$-space?
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0@Mariano: Good point. – 2011-02-08