# algebraic/geometric interpretation of the projective closure of an affine variety

Let $Y=Z(I)$ be an affine variety of $A^n$ with vanishing ideal $I$. Then $Y$ can be identified with an open set of the projective space $P^n$ and its projective closure $\bar{Y}$ is the closure of $Y$ in $P^n$.

What is the algebraic/geometric interpretation of $\bar{Y}$ and why do we care to study it? How is this interpretation modified if $Y$ is a finite affine variety, i.e. a finite set of points $\xi_1,\cdots,\xi_N$ of $A^n$? Any intuitive comments are highly welcome.

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## 1 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:

1. to get $I(\overline Y)$, you have (in general) to homogenize all the polynomials in $I(Y)$ with respect to a new variable (the one that you "add" when you pass from affine to projective space);
2. you can always recover the original $Y$ by "forgetting" homogenization (i.e. by intersecting $\overline Y$ with the open subset $\{\textrm{(new variable)}\neq0\}\subset\mathbb P^n$.

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.

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I find your answer very enlightening. Could you please explain your last sentence? I had the impression that the variety of a finite number of closed points has a vanishing ideal that is non-homogeneous. How can we see that it is projective? – Manos Sep 17 '13 at 22:11
I will make that into a new question. – Manos Sep 17 '13 at 22:41