We talked before about elimination theory, doing it entirely in the affine case. The question was asked about how to do it projectively. There are a couple of subtleties to it, but the idea is simple: we eliminate in each affine chart and then glue together. The problems that arise most naturally here actually involve working on , and projecting down to the affine space.
Now, a polynomial on this product will be homogeneous in the first variables, and arbitrary in the last , and so subvarieties of this product are given by ideals which are homogeneous in and arbitrary in . Now, if is any ideal of this form, we define the projective elimination ideal to be such that for each , there is with . This definition may seem a bit funny at first.
However, what’s going on is that we have an ideal , and is a variety on the product. In fact, we can take to give a variety on . We can then eliminate normally to get the image in . However, there’s a slight problem: we want to remove the origin, because it doesn’t correspond to a point in projective space. The way to do that is through ideal quotients, because they correspond to taking the difference of two varieties. So we don’t REALLY want to eliminate the ideal , we want to eliminate the ideal . The only problem there is that we may not have eliminated enough copies of the variables, but we can find a power big enough that works. So we eliminate the ideal , and that is precisely .
So now, we get a version of the extension theorem, which says that if is the projection, then . So this means we have the right geometry. We can even easily modify this to work for products of projective spaces, by just taking the image of under the map .
So, to answer the problem posed in the comments, which was, given a bihomogeneous ideal, that is, it’s homogeneous in the variables and again in , though perhaps not all at once, it defines a variety in . What is the ideal of the image of the projection to .
So the problem is computing . First we note that it’s good enough to take the exponents in the definition to all be the same and high enough, so we just need to compute the elimination ideal of the ideal .
To get there, the first thing we need to know is that . This is a pretty straightforward exercise in commutative algebra. So then we just need to be able to compute the quotient by a single polynomial, and the intersection. Now, a nontrivial theorem tells us that if and are ideals of a polynomial ring, then is the same as the ideal obtained from after eliminating . Though nontrivial, this is still just an algebra exercise, and we’re focusing on an algorithm anyway. So now we take to be an ideal and any polynomial. Then if is a basis for the ideal , we have is a basis for .
So, to compute , we must compute the intersections for sufficiently large , take bases and divide them by in order to have a basis for , and then take the intersection, and finally perform elimination. It’s a bit more complex than affine elimination theory, but it’s still completely algorithmic.