Lemma 10.58.1. Let $S$ be a graded ring. A set of homogeneous elements $f_ i \in S_{+}$ generates $S$ as an algebra over $S_0$ if and only if they generate $S_{+}$ as an ideal of $S$.
10.58 Noetherian graded rings
A bit of theory on Noetherian graded rings including some material on Hilbert polynomials.
Proof. If the $f_ i$ generate $S$ as an algebra over $S_0$ then every element in $S_{+}$ is a polynomial without constant term in the $f_ i$ and hence $S_{+}$ is generated by the $f_ i$ as an ideal. Conversely, suppose that $S_{+} = \sum Sf_ i$. We will prove that any element $f$ of $S$ can be written as a polynomial in the $f_ i$ with coefficients in $S_0$. It suffices to do this for homogeneous elements. Say $f$ has degree $d$. Then we may perform induction on $d$. The case $d = 0$ is immediate. If $d > 0$ then $f \in S_{+}$ hence we can write $f = \sum g_ i f_ i$ for some $g_ i \in S$. As $S$ is graded we can replace $g_ i$ by its homogeneous component of degree $d - \deg (f_ i)$. By induction we see that each $g_ i$ is a polynomial in the $f_ i$ and we win. $\square$
Lemma 10.58.2. A graded ring $S$ is Noetherian if and only if $S_0$ is Noetherian and $S_{+}$ is finitely generated as an ideal of $S$.
Proof. It is clear that if $S$ is Noetherian then $S_0 = S/S_{+}$ is Noetherian and $S_{+}$ is finitely generated. Conversely, assume $S_0$ is Noetherian and $S_{+}$ finitely generated as an ideal of $S$. Pick generators $S_{+} = (f_1, \ldots , f_ n)$. By decomposing the $f_ i$ into homogeneous pieces we may assume each $f_ i$ is homogeneous. By Lemma 10.58.1 we see that $S_0[X_1, \ldots X_ n] \to S$ sending $X_ i$ to $f_ i$ is surjective. Thus $S$ is Noetherian by Lemma 10.31.1. $\square$
Definition 10.58.3. Let $A$ be an abelian group. We say that a function $f : n \mapsto f(n) \in A$ defined for all sufficient large integers $n$ is a numerical polynomial if there exists $r \geq 0$, elements $a_0, \ldots , a_ r\in A$ such that for all $n \gg 0$.
\[ f(n) = \sum \nolimits _{i = 0}^ r \binom {n}{i} a_ i \]
The reason for using the binomial coefficients is the elementary fact that any polynomial $P \in \mathbf{Q}[T]$ all of whose values at integer points are integers, is equal to a sum $P(T) = \sum a_ i \binom {T}{i}$ with $a_ i \in \mathbf{Z}$. Note that in particular the expressions $\binom {T + 1}{i + 1}$ are of this form.
Lemma 10.58.4. If $A \to A'$ is a homomorphism of abelian groups and if $f : n \mapsto f(n) \in A$ is a numerical polynomial, then so is the composition.
Proof. This is immediate from the definitions. $\square$
Lemma 10.58.5. Suppose that $f: n \mapsto f(n) \in A$ is defined for all $n$ sufficiently large and suppose that $n \mapsto f(n) - f(n-1)$ is a numerical polynomial. Then $f$ is a numerical polynomial.
Proof. Let $f(n) - f(n-1) = \sum \nolimits _{i = 0}^ r \binom {n}{i} a_ i$ for all $n \gg 0$. Set $g(n) = f(n) - \sum \nolimits _{i = 0}^ r \binom {n + 1}{i + 1} a_ i$. Then $g(n) - g(n-1) = 0$ for all $n \gg 0$. Hence $g$ is eventually constant, say equal to $a_{-1}$. We leave it to the reader to show that $a_{-1} + \sum \nolimits _{i = 0}^ r \binom {n + 1}{i + 1} a_ i$ has the required shape (see remark above the lemma). $\square$
Lemma 10.58.6. If $M$ is a finitely generated graded $S$-module, and if $S$ is finitely generated over $S_0$, then each $M_ n$ is a finite $S_0$-module.
Proof. Suppose the generators of $M$ are $m_ i$ and the generators of $S$ are $f_ i$. By taking homogeneous components we may assume that the $m_ i$ and the $f_ i$ are homogeneous and we may assume $f_ i \in S_{+}$. In this case it is clear that each $M_ n$ is generated over $S_0$ by the “monomials” $\prod f_ i^{e_ i} m_ j$ whose degree is $n$. $\square$
Proposition 10.58.7. Suppose that $S$ is a Noetherian graded ring and $M$ a finite graded $S$-module. Consider the function see Lemma 10.58.6. If $S_{+}$ is generated by elements of degree $1$, then this function is a numerical polynomial.
\[ \mathbf{Z} \longrightarrow K'_0(S_0), \quad n \longmapsto [M_ n] \]
Proof. We prove this by induction on the minimal number of generators of $S_1$. If this number is $0$, then $M_ n = 0$ for all $n \gg 0$ and the result holds. To prove the induction step, let $x\in S_1$ be one of a minimal set of generators, such that the induction hypothesis applies to the graded ring $S/(x)$.
First we show the result holds if $x$ is nilpotent on $M$. This we do by induction on the minimal integer $r$ such that $x^ r M = 0$. If $r = 1$, then $M$ is a module over $S/xS$ and the result holds (by the other induction hypothesis). If $r > 1$, then we can find a short exact sequence $0 \to M' \to M \to M'' \to 0$ such that the integers $r', r''$ are strictly smaller than $r$. Thus we know the result for $M''$ and $M'$. Hence we get the result for $M$ because of the relation $ [M_ d] = [M'_ d] + [M''_ d] $ in $K'_0(S_0)$.
If $x$ is not nilpotent on $M$, let $M' \subset M$ be the largest submodule on which $x$ is nilpotent. Consider the exact sequence $0 \to M' \to M \to M/M' \to 0$ we see again it suffices to prove the result for $M/M'$. In other words we may assume that multiplication by $x$ is injective.
Let $\overline{M} = M/xM$. Note that the map $x : M \to M$ is not a map of graded $S$-modules, since it does not map $M_ d$ into $M_ d$. Namely, for each $d$ we have the following short exact sequence
\[ 0 \to M_ d \xrightarrow {x} M_{d + 1} \to \overline{M}_{d + 1} \to 0 \]
This proves that $[M_{d + 1}] - [M_ d] = [\overline{M}_{d + 1}]$. Hence we win by Lemma 10.58.5. $\square$
Remark 10.58.8. If $S$ is still Noetherian but $S$ is not generated in degree $1$, then the function associated to a graded $S$-module is a periodic polynomial (i.e., it is a numerical polynomial on the congruence classes of integers modulo $n$ for some $n$).
Example 10.58.9. Suppose that $S = k[X_1, \ldots , X_ d]$. By Example 10.55.2 we may identify $K_0(k) = K'_0(k) = \mathbf{Z}$. Hence any finitely generated graded $k[X_1, \ldots , X_ d]$-module gives rise to a numerical polynomial $n \mapsto \dim _ k(M_ n)$.
Lemma 10.58.10. Let $k$ be a field. Suppose that $I \subset k[X_1, \ldots , X_ d]$ is a nonzero graded ideal. Let $M = k[X_1, \ldots , X_ d]/I$. Then the numerical polynomial $n \mapsto \dim _ k(M_ n)$ (see Example 10.58.9) has degree $ < d - 1$ (or is zero if $d = 1$).
Proof. The numerical polynomial associated to the graded module $k[X_1, \ldots , X_ d]$ is $n \mapsto \binom {n - 1 + d}{d - 1}$. For any nonzero homogeneous $f \in I$ of degree $e$ and any degree $n >> e$ we have $I_ n \supset f \cdot k[X_1, \ldots , X_ d]_{n-e}$ and hence $\dim _ k(I_ n) \geq \binom {n - e - 1 + d}{d - 1}$. Hence $\dim _ k(M_ n) \leq \binom {n - 1 + d}{d - 1} - \binom {n - e - 1 + d}{d - 1}$. We win because the last expression has degree $ < d - 1$ (or is zero if $d = 1$). $\square$
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