# Projective closure

Is the projective closure of an infinite affine variety (over an algebraically closed field, I only care about the classical case right now) always strictly larger than the affine variety? I know it is an open dense subset of its projective closure, but I don't think it can actually be its own projective closure unless it is finite.

I guess my intuition has me worried about cases like the plane curve $X^2 + Y^2 - 1$, since the real part is compact, but such a curve must still "escape to infinity" over an algebraically closed field, right?

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One can show that if the dimension of the affine variety is positive (equivalently, if it is infinite in the sense of our question) then the projective closure is strictly larger.

There are probably lots of ways to prove this, but one is by a version of Noether normalization. If you would like a detailed proof, let me know by a comment and I can give one.

(Another way to phrase this result is to say that a variety that is simultaneously affine and projective is necessarily finite. Scheme-theory mavens will recognize this as a special case of the more general statement that a morphism which is simultaneously affine and proper is finite.)

Added: Here is a sketch of a proof, as promised:

Let me begin with Noether normalization in a geometric form. Fix an afffine variety $V$ contained in $\mathbb A^n$, with projective closue $\overline{V}$. (Here and below I am always working over an algebraically closed field $k$.)

Assuming that $V$ is not all of $\mathbb A^n$, we see that $\overline{V}$ does not contain the hyperplane at infinity, and so we may choose a point $P$ lying in the hyperplane at infinity, but not lying in $\overline{V}$. We may also choose a different hyperplane $H$ (i.e. not the hyperplane at infinity) which doesn't contain $P$.

With $P$ and $H$ in hand, we may define the projection map $\pi: \mathbb P^n \setminus P \to H$, which maps any $Q \neq P$ to the intersection of the line $\ell$ joining $P$ and $Q$ with the hyperplane $H$. Restricting $\pi$ to $\overline{V}$, we obtain a map $V \to H$.

Now since $P$ is not contained in $\overline{V}$, none of the lines $\ell$ appearing in the projection map are contained in $V$, and so each of them meets $\overline{V}$ in only finitely many points. In particular, if $\overline{V}$ is infinite, so is its image under $\pi$.

Now by elimination theory, i.e. the fact that projective varieties are proper, we know that $\pi(\overline{V})$ is closed in $H$, i.e. is a projective variety in $H$, which is a projective space of dimension $n-1$.

Also, our choice of $P$ ensures that for any $Q \in \overline{V}$, the image $\pi(Q)$ lies at infinity if and only if $Q$ itself does. So $\pi(V)$ is an affine variety (in the affine space $H \cap \mathbb A^n$ of dimension $n-1$), and $\pi(\overline{V})$ is its projective closure.

What I have just done is prove Noether normalization, in a geometric form.

The result we want now follows immeidately, by induction on $n$. (Basically, the case when $V = \mathbb A^n$ is clear, and if $V$ is not all of $\mathbb A^n$, the preceding argument allows us to reduce the dimension of the ambient affine space by one.)

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Dear @Matt E: Is there a simple way to see the more general assertion that a morphism which is simultaneously affine and proper is finite? I think it follows directly from Zariski's main theorem (namely, if it's proper, it has to be quasi-finite, since no affine variety over a field is complete, hence it is finite; thus it's finite by ZMT), but I would be curious about a more elementary proof, since I have not properly studied ZMT yet. – Akhil Mathew Dec 17 '10 at 16:26
Sorry, I meant that "no affine variety over a field of positive dimension is complete." – Akhil Mathew Dec 17 '10 at 16:42
Perhaps this follows from the valuative criterion? As in, first one reduces to the case of affine schemes $\mathrm{Spec} B \to \mathrm{Spec} A$, where $A,B$ are integral domains. Then if this is proper, there is a lifting in every diagram involving valuation rings, so every valuation on $A$ prolongs to one on $B$. Thus $B$ must be integral over $A$. So nothing as fancy as ZMT is necessary at all. – Akhil Mathew Dec 17 '10 at 22:21
Thanks, that's what I was hoping. I'm aware of the more general fact about proper morphisms, but using this would at the very least require the fact that projective varieties are complete, which is really hard. I would certainly appreciate it if you could sketch (details would spoil the fun) the proof that uses Noether normalization. – Justin Campbell Dec 17 '10 at 22:24
@Justin: Dear Justin, Completeness of projective varieties is not so hard (and it will be used in my proof sketch); for a modern-style proof of classical elimination theory, you can see e.g. Ravi Vakil's notes that are appearing online. (The proof as given there will be for schemes, but one can translate it into variety language, where it only becomes easier.) – Matt E Dec 18 '10 at 5:03

For your own example: the projectivization of your curve is given by $X^2 + Y^2 - Z^2= 0$. There is the point corresponding to the projective equivalence class of $(1,i,0)$, which does not belong to the affine curve.

For more general varieties given by a single equation, you can always similarly find a projective point that is part of the affine curve. For more general varieties, I should imagine that it is still the case; but for a proper argument a better person in algebraic geometry is needed.

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Right, so that settles the question for my example, but I'm asking if the projective closure of an infinite affine variety always strictly contains the affine variety. – Justin Campbell Dec 16 '10 at 23:14