Graded modules over K[S, T]
In this section we set R = K[S, T]. We think of S, T as homogeneous of degree 1 and of the constants as homogeneous of degree 0. Then we see that R is a graded K-algebra
R = ⊕ R_d
with R_d the polynomials which are homogeneous of degree d. Hence we again have
R_d ⋅ R_{d'} ⊂ R_{d + d'}
as before. As before a graded module M over the ring R is an R-module M which comes equipped with a direct sum decomposition
M = ⊕ M_n
where now the integers n are allowed to be negative as well. The condition imposed on this decomposition is that
R_d ⋅ M_n ⊂ M_{d + n}.
As before we say an element x ∈ M is homogeneous if x ∈ M_n for some integer n. In this case we say that x is homogeneous of degree n. Again it is true that if M is a finitely generated R-module, then
Please make sure you agree with this before reading on. The Hilbert function of a finitely generated graded graded R-module M is the map
H_M : Z —> Z_{≥ 0}, n |—> dim_K(M_n)
Let's prove a lemma.
Lemma: There exist a, b ∈ Z such that H_M(n) = an + b for all n » 0.
Proof. Consider the map
M — S —> M
given by multiplication by S from M to M. This is an R-linear map which shifts degrees by 1. Let K = Ker(M — S —> M) and Q = Coker(M — S —> M). Then K and Q are also graded modules. Because R is a Noetherian ring, we see that K and Q are finitely generated R-modules too. But multiplication by S is zero on K and Q so we can think of K and Q as modules over K[T] and the results of Exercise 30 apply! OK, now consider the exact sequences
0 —> K_n —> M_n — S —> M_{n + 1} —> Q_{n + 1} —> 0
We see that
H_M(n + 1) - H_M(n) = H_Q(n + 1) - H_K(n)
If you now use what we know about H_K and H_Q then you can easily see why the lemma must be true. EndProof
Given a finitely generated graded R-module M the linear function an + b is called the Hilbert Polynomial of M. Maybe we'll denote it P_M sometimes…?
Exercise 31: Which pairs a, b can occur in the lemma? In other words: What are all possible Hilbert polynomials of finitely generated graded R-modules?
Continue reading about free graded modules. Back to the start page.