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I was doing some exercises in Liu's book on Algebraic Geometry. I am currently trying to solve a problem by showing the following:

Let $U \subset \mathbb{P}^n_k$, k a field, be an affine open subset.
Show that the irreducible components of $\mathbb{P}^n_k-U$ all have dimension n-1.

I would appreciate any help / hint here. I have some problems understanding $\mathbb{P }^n_k$ at a deep (even semi-deep) level. I suspect that one could maybe show that the dimension of such an affine open should be of dimension n (or am I wrong here?), since we can compute the dimension of X on any open set. We should be able to write the complement as $V_+(I)$ for some homogenous ideal . However, I don't see how to get from this to that the irreducible components of the component has dimension n-1.

Thank you for looking at my question and please forgive me if it's a naive one.

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I don't think it is naive. Here is something related: mathoverflow.net/questions/22040/… –  Andrew Dec 19 '12 at 0:07
    
As Andrew pointed out, this result holds in a fairly general situation (affine open subset in a noetherian separated scheme, this is also Exercise 4.1.15). But in the UDF case as in $\mathbb P^n_k$, the proof is much easier: use the previous parts of Exericse 2.5.12. –  user18119 Dec 19 '12 at 7:06
    
QiL: I'm afraid I don't see an obvious application of either of the previous ones. Should I use something like that on $D(x_i)$ it is principal? The wonders of technology, by the way, allowing one to talk directly to the author of the book. Thank you very much for your answers! –  Dedalus Dec 19 '12 at 10:12
    
@Dedalus: it seems like the author didn't give enough details for this exercise :). Consider the intersection with the affine spaces in $\mathbb P^n$ and use (b). If this is not enough I will post an answer. –  user18119 Dec 19 '12 at 13:22
    
@QiL: I don't think the details are neccesarily left out (it might be that I'm simply dense here). If we let U be the open affine. if we go to the intersections of U with the principal affine opens $D(x_i)$, we get that in each such affine, $U \cap D(x_i)$ is principal. The complement is $V(f)$, and it has irreduble components of dimension n-1 (right?) . So, can we from this conclude that each irreducible component of the complement of U in $\mathbb{P}^n_k$ is irreducible of dimension n-1? –  Dedalus Dec 19 '12 at 13:55
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up vote 3 down vote accepted

For the sake of an answer. Cover $\mathbb P^n_k$ (with coordinates $x_0, \dots, x_n$) by affines spaces $U_0, \dots, U_n$ with $U_i=D_+(x_i)$. Then $$ \mathbb P^n_k\setminus U=\cup_i (U_i\setminus (U\cap U_i))$$ and it is enough to show $U_i\setminus (U\cap U_i)$ is empty or pure of dimension $n-1$ for all $i\le n$. By a previous part of the exercice, we know that $U_i\setminus (U\cap U_i)$ is a principal closed subset of $U_i$, hence it is empty or pure of dimension $n-1$.

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