1. Let I be a radical ideal in R, and homog(I) its homogenization in R[y].
Show that homog(I) is radical.
Let f^2 be in homog(I). Let f_0 be the lowest-degree part of f, so f_0^2 is the lowest-degree part of f^2. Since f^2 is in homog(I) a homogeneous ideal, each degree component of it, e.g. f_0^2, is in homog(I).
Then there are two cases: f_0 is in homog(I) or not. If it is, replace f with f-f_0 and start over. That must terminate since the number of terms goes down. Eventually we get to the other case, that f_0 is not in homog(I). Replace f with f_0 and start over.
Okay, now f^2 is in homog(I), but f isn't, and f is homogeneous. Hence homog(dehomog(f)) = f / y^k, where k is the largest power of y that divides f. Let g = f / y^k; it is homogeneous and not a multiple of y. Hence homog(dehomog(g)) = g.
Okay, now (g y^k)^2 is in homog(I). Dehomogenizing, we learn dehomog(g)^2 is in I. Hence dehomog(g) is in I, since I is radical. Hence homog(dehomog(g)) = g is in homog(I). Therefore f = g y^k is in homog(I).
2. Let I be a radical homogeneous ideal in R[y].
Show that dehomog(I) is radical.
Let f^2 be in dehomog(I). Then f^2 = dehomog(g) for some homogeneous g in I. So homog(f^2) = g / y^k for the largest y^k dividing g.
Hence y^k homog(f^2) is in I. If k is odd, multiply by y again, staying in I.
Hence y^{k or k+1} homog(f)^2 is in I.
Since I is radical, y^{ceiling(k/2)} homog(f) is in I.
Dehomogenizing, we learn f is in I.
3. Let I be an ideal in R such that homog(I) is radical. Show that I is radical.
I = dehomog(homog(I)). Now apply #2.
4. Give an example of I in R[y], homogeneous but not radical, such that dehomog(I) is radical.
The simplest is I generated by y^2.
Geometrically, the idea is this. PV(I) lives in projective space. Being nonradical means it has some fuzz on it somewhere (that the generators of the radical would shave off). Dehomogenizing means intersecting PV(I) with affine space, throwing away the stuff at infinity. So we need to arrange for all the fuzz to be at infinity. The ideal y^2 cuts out exactly the hyperplane at infinity, but with fuzz.
5. Let I be generated by xy, x+y-1. Find a term order such that init(homog(I)) is radical.
Conclude that I is radical.
homog(I) is generated by xy, x+y-z (here the new variable is z).
With the right lex order, the initial terms are xy and -z.
Since those have gcd=1, this is a Gr\"obner basis whose initial terms are squarefree.
Hence I is radical.
6. Draw the simplicial complex associated to the ideal generated by {e, af, bd} in C[a,b,c,d,e,f].
It's a solid square cut into four triangles with a,b,f,d at the corners and c in the middle.
7. Draw the solid abc triangle, and put a vertex d in the middle, with new edges connected to a,b,c. (The red, but not blue, lines in this picture.) What's the corresponding Stanley-Reisner ideal, and its Hilbert series?
This wasn't at all well specified, for which I apologize. If you didn't manage to read my mind, but made clear what complex you were working with, that's good enough.
What I'd intended to indicate was a union of three solid triangles, abd & bcd & cad. So the only faces missing are abcd and abc itself. In particular the S-R ideal should be generated by abc.
Then the Hilbert series is 1/(1-t)^4 * (1-t^3).
