The Hilbert series of SR(Delta) is sum_{faces F} (t/(1-t))^|F|.
Hence the Hilbert polynomial, in degree d, is sum_F (d-1 choose |F|-1).
Hence the Hilbert dimension is the # of elements of the largest face.
Something I really should have proved:
Thm.
If g_1,..,g_m is a Gr\"obner basis of I using a graded term order,
then homog(g_1),...,homog(g_m) is a Gr\"obner basis of homog(I)
using a graded term order in which y is cheap.
Pf. homog(g_i) has the same leading term as g_i (since it's of highest degree in g_i, so doesn't get any powers of y on it).
If we follow the reduction of S(homog(g_i), homog(g_j)),
at each step its dehomogenization matches that of the reduction S(g_i,g_j),
so we don't get stuck and we do reduce it to 0.
Corollary. If yf is in homog(I), then f is in homog(I).
Proof. If f isn't in homog(I), then after reducing for a while using one of the Gr\"obner bases supplied above it gets stuck.
Hence yf is also stuck, because the leading terms in {homog(g_i)} don't involve y. So yf isn't in homog(I).
Thm. If I is prime, then homog(I) is also prime.
Pf. Start with a,b s.t. ab in homog(I). Reduce to the case that a = homog(a_2), b = homog(b_2), using the Corollary. Then use a = homog(dehomog(a)) etc.
