Let I = < f_1, ..., f_k > where the f_i are each homogeneous polynomials.
1. Show that I has a Grobner basis consisting of homogeneous polynomials.
When the reduction algorithm is asked to reduce a homogeneous polynomial, using other homogeneous polynomials, the result is homogeneous.
The S-polynomial of two homogeneous polynomials is homogeneous (of degree = degree(that lcm)).
Hence Buchberger's algorithm creates not just a Gr\"obner basis, but one consisting of homogeneous polynomials.
2. Assume hereafter that the {f_i} are a reduced Grobner basis.
Recall that the Hilbert function h_I(n) is the dimension of R_n / I_n.
Call a monomial standard if it isn't divisible by any of the leading monomials of the {f_i}.
Show that h_I(n) is the number of standard monomials of degree n.
Let {x^e} denote the set of standard monomials of degree n (e is really a vector, here).
Then the claim is that their images {x^e + I_n} form a basis of R_n / I_n.
Proof: given an element r of R_n (not necessarily in I), when we run the reduction algorithm on r we get stuck at a new element r+i in the same coset r+I. But this r+i is a C-linear combination of the {x^e} (or we wouldn't be stuck). So r+I is a combination of {x^e + I}, i.e. the {x^e + I_n} span R_n / I_n.
Now look at a linear combination \sum_e c_e x^e. For this to give the zero element of R_n / I_n, it needs to be in I_n, i.e. to be in I. But the reduction algorithm will be stuck on this, unless it's already the zero polynomial, i.e. all the c_e were zero. So the {x^e + I_n} are linearly independent.
3. Show that h_I = h_{init I}, so H_I = H_{init I} (the Hilbert series).
I and init I have the same standard monomials (since they were defined using only the initial terms in the Gr\"obner basis), so by #2 their Hilbert functions are the same, and the Hilbert
series are derived from those.
4. The polynomial b^2 - ac has two possible leading terms (depending on term order), so I = < b^2 - ac > has two possible init I. Compute each of their Hilbert series (and show they are equal, as problem 3 predicts).
Either way we get (1 - t^2) / (1-t)^3; the 2 is 0+2+0 when b^2 is the leading term, and 1+0+1 when ac is.
5. Let I = < the entries of M^2, where M is a 2x2 matrix >. Compute the Hilbert series H_I.
We computed a Gr\"obner basis for I in class,
bc + d^2, ac + cd, ab + bd, a^2 - d^2, ad^2 + d^3
whose initial terms are {bc, ac, ab, a^2, ad^2}.
Then our formula for the Hilbert series of that monomial ideal is 1/(1-t)^4 times
1 - (4t^2 + t^3) + (5t^3 + 4t^4 + t^5) - (t^3 + 3t^4 + 5t^5 + t^6) + (t^4 + t^5 + 3t^6) - t^6
where I've grouped the 2^5 terms according to the sizes of the subsets of the generators,
= 1 - 4t^2 + 3t^3 + 2t^4 - 3t^5 + t^6.
6. Let J be the larger ideal containing I and also the generator Trace(M). Compute the Hilbert series H_J.
This has a much simpler reduced Gr\"obner basis: a+d, bc+d^2.
The initial terms are {a,bc}, and the numerator of the Hilbert series is 1 - t - t^2 + t^3.
Note that this factors as (1-t)(1-t^2), which is secretly a statement about the independence of these generators.
