Free graded modules
We continue to work over R = k[S, T]. Let M be a graded R-module. The shift or twist of M by an integer e is the graded R-module M(e) defined by the rules
In other words we just shift the grading but don't change anything else. Note that
H_{M(e)}(n) = H_M(e + n)
for the Hilbert functions. A special case is R(e) which is the R-module which has R_{e + n} in degree n. So the Hilbert polynomial of R(0) = R is n + 1 and the Hilbert polynomial of R(e) is n + e + 1. Given a homogeneous polynomial G ∈ R of degree d multiplication by G gives a graded R-module map
R(-d) — G —> R
which does not shift degrees.
A free graded module is a finitely generated graded R-module of the form
M = R(e_1) ⊕ R(e_2) ⊕ … ⊕ R(e_r)
Note that the Hilbert polynomial of M is rn + e_1 + … + e_r + r. We will always order the e_i such that
e_1 ≤ e_2 ≤ … ≤ e_r
The integers e_1, …, e_r themselves are invariants of the isomorphism class of M as a graded R-module. We will say that e_1, …, e_r is the splitting type of M. Namely, if M = R(e_1) ⊕ R(e_2) ⊕ … ⊕ R(e_r), then M has a minimal set of generators
x_i = (0, …, 0, 1, 0, …, 0), i = 1, …, r
sitting in degree -e_i. Computationally, still assuming M is a free graded module, you would find e_1, …, e_r as follows:
Dividing by a submodule is tricky. An alternative is to observe that the Hilbert function of M uniquely determines the integers e_1, …, e_r.
Exercise 32: Explain an algorithm that finds the splitting type of a graded free module M over R in terms of its Hilbert function.
Exercise 33 (optional): Show that the kernel of a map of graded free modules over R is graded free over R.
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