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.)
Indeed, the proof doesn't need the variety structure. All we need is that the affine variety is a nonempty open subset of its projective closure. The definition of irreducible is in fact purely topological: if some space $X$ is the union of two closed sets, one of them has to be the whole space. Note that whether you see an affine variety as affine or open in its projective closure does not change the topology.
We have the following:
- Any nonempty open subset of an irreducible space is itself irreducible and dense
- If $Y \subset X$ is irreducible in the subspace topology, the closure of $Y$ in $X$ is again irreducible
Which you should be able to prove.
See Hartshorne 1.1.3 and 1.1.4.
Hope that this helps.
Edit:
By the way, do you know the equivalent definitions of irreducible? The following are equivalent:
- $X$ is irreducible
- Any nonempty open in $X$ is dense
- Any two nonempty opens in $X$ have nonempty intersection
Proving the equivalence is not hard and a good exercise.
Best Answer
Sometimes it is convenient to look at varieties locally in an affine neighborhood of a point. But other times one may have the opposite urgency, and here is where the projective closure comes into the game: you can think of $\overline Y$ as a compactification of $Y$, the smallest projective variety containg $Y$. In other (imprecise) words, $\overline Y$ tells you how $Y$ would look like if it was projective. There are a couple of issues which are important:
Let us be concrete: take the plane curve $$Y:xy-1=0.$$ We have that $Y$ is closed in $\mathbb A^2$, and its projective closure in $\mathbb P^2=\textrm{Proj }k[x,y,z]$ is $$\overline Y:xy-z^2=0.$$ In $\mathbb P^2$ there is the open subset $U=\{z\neq 0\}$, and you can now recover the closed immersion $\overline Y\cap U=Y\subset U\cong\mathbb A^2$. In this case it was enough to homogenize the single generator of $I(Y)$, but this is heaven.
The projective closure is the right object to look at when you want to study $Y$, but as a closed subvariety of a compact ambient space. You may want to do so for instance because in projective space many things go the right way, e.g. we have Bézout, and intersection theory says that every proper variety has a well defined "degree".
As for the case of a finite number of closed points, such an affine variety is also projective, so it equals its own closure.