# Graded Grothendieck Group and Hilbert Polynomial

I was wondering if any of the arguments from elementary dimension theory of local noetherian rings could be simplified with knowledge of the Grothendieck group.

Let $$A$$ be a noetherian graded $$K$$-algebra over a ring $$K$$. We write $$C$$ for the category of noetherian $$A$$-modules. Let $$K_0 (C)$$ be the Grothendieck group of $$C$$. The translation functor on noetherian $$A$$-modules gives an action of $$\mathbb{Z}[x, x^{-1}]$$ on $$K_0 (C)$$.

If $$\lambda$$ is an additive function on the set of isomorphism classes of noetherian modules (edit: noetherian $$K$$-modules, not noetherian $$A$$-modules) taking values in $$\mathbb{Z}$$, then $$\lambda$$ arranges into an additive function $$\lambda_* = \sum_{i \in \mathbb{Z}} \lambda (-)$$ on the isomorphism classes of $$C$$, sending a graded module $$\oplus_{i \in \mathbb{Z}} M_i$$ to the formal sum $$\sum_{i \in \mathbb{Z}} \lambda( M_i)$$. $$\lambda_*: \text{iso}(C) \rightarrow \mathbb{Z}[[x]]$$ must then factor through $$K_0 (C)$$.

Usually we use $$\lambda_*$$ in the following theorem:

Theorem: (Hilbert, Serre) Let $$A$$ be a noetherian graded $$K$$-algebra, and let $$M$$ be a noetherian module. Then there is a $$m \in \mathbb{Z}$$ such that $$\lambda(M)(n) = f(n) \prod_{i = 1}^n (1 - n^{d_i})$$ for $$n > m$$, where $$d_i$$ occur as the degrees of generators of $$A$$ over $$K$$.

If generators of degree $$1$$ can be chosen, then this leads to a notion of dimension, where the dimension is the degree of the pole at $$1$$ minus $$1$$.

But I wonder whether the grothendieck group can be used here instead. Is this possible? I'm hoping someone can point me towards a more theoretical approach along these lines.

For instance, one might try to characterize the function $$K_0 (C) \rightarrow \mathbb{N}_{\geq 0}$$ which sends a class of modules to its dimension. The standard way of getting a function (but which does not characterize it) is by going through $$\mathbb{Z}[[t]]$$ and look at the degree of the pole at $$1$$, but maybe there are properties which define this uniquely. Another way of looking at it is to ask, "what sort of equivalence relation do we put on $$K_0(C)$$ to get $$\mathbb{N}_{\geq 0}$$, with the quotient map giving the dimension?"

• I'm not sure what you mean. How is $\sum_{i\in\mathbb{Z}} \lambda(M_i)\in\mathbb{Z}[[x]]$ well-defined? Where does the $x$ come into play? And more importantly: What is $\lambda(M_i)$ ? If $\lambda$ is defined on the set of isomorphism classes $C/\cong$, then $\lambda(M_i)$ makes no sense, because the graded piece $M_i$ isn't an $A$-module itself. It is only a $K$-module, i.e. a vector space in general. And if $\lambda$ is defined on $K\mathsf{-vect}/\cong$, then it is just equal to the dimension and you don't get anything more general than what you already have with the Hilbert polynomial. – Johannes Hahn Feb 18 at 20:38
• $\lambda : D / \cong \rightarrow \mathbb{Z}$ where $D$ is noetherian $K$-modules, not noetherian $A$-modules. The graded pieces are $K$-modules, and $\lambda$ assembles into a map on $C / \cong \rightarrow \mathbb{Z} [[t]]$. It is a theorem that, if $K$ is a field (more generally, if it is artinian), then noetherian $K$-modules produce the grothendieck group $\mathbb{Z}$. This is not hard to show when $K$ is a field. In the artinian case, the quotient map onto the grothendieck group is the same as the length function, up to isomorphism. – Dean Young Feb 19 at 0:13
• I think we agree that this no more general than the hilbert polynomial, as $\lambda$ is identically the length for a noetherian module over an artinian ring. Greater generality is not why I'm interested in this approach. – Dean Young Feb 19 at 0:21
• Alright, so I overlooked that $K$ can be any ring, not necessarily a field. And you meant to write $\lambda_\ast(M) = \sum_{i} \lambda(M_i)x^i$, right? In that case the target is not $\mathbb{Z}[[x]]$, but $\mathbb{Z}[[x]][x^{-1}]$, because you can have modules with (finitely many) non-zero components of negative degrees. – Johannes Hahn Feb 19 at 12:30
• I want to point out that the restriction to degree one generators is not necessary, i.e. the degree of the pole of the Poincare series of $M$ at x=1 equals the Krull dimension of $A/ann(M)$ (cf. Benson, Modular Representation Theory, 1.8.7), at least if $K$ is artinian. – tj_ Feb 19 at 21:31

Your construction is basically the natural homomorphism of $$\mathbb{Z}[x^{\pm 1}]$$-modules $$K_0(A\mathsf{-grMod}) \to \left\{\text{formal "Laurent series"} \sum_{i=k}^\infty a_i x^i : k\in\mathbb{Z}, a_i\in K_0(K\mathsf{-mod})\right\},$$ sending the class $$[M]$$ of a graded $$A$$-module $$M=\bigoplus_i M_i$$ to $$\sum_{i\in\mathbb{Z}} [M_i]x^i$$. You can now post-compose with any out-going morphism from $$K_0(K\mathsf{-mod})$$ to $$\mathbb{Z}$$ (or anywhere else) and get a corresponding morphism from $$K_0(A\mathsf{-grMod})$$ to the Laurent series ring over $$\mathbb{Z}$$ ( or some other coefficients).
If $$K$$ is a field, then $$K_0(K\mathsf{-mod})$$ is just isomorphic to $$\mathbb{Z}$$ itself via the dimension so that in this case your homomorphism just coincides with taking the Hilbert series of a module.
A generalisation of the Krull-dimension is surely possible, but tricky in the general setting. First of all, that dimension need not be constant across the spectrum. For example if $$K$$ is decomposable as $$K=K_1\times K_2$$, then every $$K$$-module, $$A$$ and every $$A$$-module also decomposes into a direct sum of a $$K_1$$-module and a $$K_2$$-module which are completely independent and can have vastly different dimensions. Even if $$K$$ is indecomposable, non-trivial behaviour is expected and does occur. At the very least we should aim for a dimension map defined on the spectrum which assigns to $$\mathfrak{p}\in Spec(K)$$ the "dimension" of the $$A_{\mathfrak{p}}$$-module $$M_\mathfrak{p}$$.
In other words: Characterising $$K$$-modules is contained in the problem of characterising $$A$$-modules and unless $$K$$ is very nice already, it may be useful to restrict attention to only those $$A$$-modules which have nicely behaved and well-understood underlying $$K$$-modules (at least locally). For example one could restrict to lattices, i.e. those $$A$$-modules $$M$$ for which all $$M_i$$ are f.g. projective $$K$$-modules. In that case localising at $$\mathfrak{p}$$ gives free modules so that we can speak of the dimension as a map $$K_0(K\mathsf{-proj})\to \mathbb{Z}^{Spec(K)}, [M] \mapsto (\mathfrak{p} \mapsto \dim_{K_\mathfrak{p}} M_\mathfrak{p})$$ and similarly $$K_0(A\mathsf{-grMod}) \to \mathbb{Q}((x))^{Spec(K)}, [M]\mapsto (\mathfrak{p} \mapsto \sum_{i\in\mathbb{Z}} \dim_{K_\mathfrak{p}}(M_i)_\mathfrak{p} x^i).$$
And for any given $$M$$ the map on the Spectrum is even continuous w.r.t. the Zariski topology if I'm not mistaken.