### Projective Closures of Affine Varieties

This blog post is going to be on the projective closures of affine varieties. This too is something that has confused me for some time. I did manage to make sense of most of it eventually, but I want to still write it down for my peace of mind.

Say we have a polynomial $p(x_1,x_2,\dots,x_n)$ in affine space $\Bbb{A}^n$. We want to find the projective closure of that. How do we do it? We first homogenize it by writing it as a homogeneous polynomial $p'(x_1,x_2,\dots,x_n,z)$. Note that $z$ is the extra variable that has been added here. Then, after having sketched the zeroes of the polynomial in the $z$-th affine chart in projective space, we find the projective closure in the whole space by assuming $z=0$, and then determining what are the zeroes of the remainder of the polynomial.

Why does this work? The complement of the $z$-th affine chart is that space in projective space that the affine curve could never have reached in affine space. Remember that the whole affine curve, point by point, lies inside the affine chart. Hence, the points in the complement of the chart will contain all the “new” points of the curve. Heuristically speaking, the “new” points are all those points that the curve tends to, if it could go infinitely far, whatever that might mean.

Now how do you find the projective closure of a general algebraic set (and not one that is generated by a single polynomial)? In other words, say you have an affine algebraic set $X$. What changes do you need to make to $I(X)$ in order for $V(I(X))$ to be the projective closure of $X$? Turns out, homogenizing the generators of $I(X)$ is not the answer, but homogenizing *all* the elements of $I(X)$ is. Is is the explanation of this part that is the aim for writing this post.

Why does homogenizing the generators not work? This is because we may count extra points of projective closure. How? Take the polynomials $x+y+1$ and $x+y+2$ in $\Bbb{R}^2$. Their corresponding homogeneous polynomials are $x+y+z$ and $x+y+2z$ in $\Bbb{P}^3$. The variety corresponding to the ideal $I(x+y+1, x+y+2)$ is obviously $\emptyset$. Hence, the affine chart of $\Bbb{P}^2$ will also contain the empty set, the projective closure of which is also the empty set. However, the projective closures of both $x+y+z$ and $x+y+2z$ contain the point $x+y=0$. From this, we can see that a set of algebraic sets in affine space may all contain a set of points of projective closure that their intersection may not.

This problem is solved by homogenizing all the polynomials in $I(X)$, and then generating an ideal from it. The variety corresponding to this ideal is exactly the projective closure of $X$. Let this ideal be referred to as $\tilde{I}(X)$. Why does this work?
First of all, it is clear that the projective closure of $X$ lies inside $V(\tilde{I}(X))$. Now we need to prove the other inclusion. If we can produce a homogeneous polynomial such that the corresponding variety on the affine chart is the same as $X$, then we’ll be done. This we can do by just adding the homogenized versions of all the generators in $I(X)$, but also ensuring that the homogenized versions are of different degrees!!! The key insight behind this is that as we have a sum of homogeneous polynomials of different degrees, the corresponding variety will contain only those points that lie in the intersection of all zero sets of the individual polynomials. The affine chart will consequently only contain those affine points which lie in the zero sets of all those affine polynomials, which is precisely $X$. To complete the proof, we need to assume that each affine variety has a unique projective closure.

Such a polynomial (sum of homogenized versions, to different degrees, of all the generators of $I(X)$) does in fact exist in $\tilde{I}(X)$. Hence proved.