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.
I wonder if the computation of eliminating I, i.e project P^nxA^(m+1) to A^(m+1), preserves the homogeneous property of I in P^m !?
Note: we only have the original bi-homogeneous polynomials and starting do the elimination on that.
If the ideal you started with is homogeneous in the variables
, then the elimination ideal will be as well, though actually proving it requires some tricks.