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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?"

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    $\begingroup$ 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. $\endgroup$ Commented Feb 18, 2019 at 20:38
  • $\begingroup$ $\lambda : D / \cong \rightarrow \mathbb{Z}$ 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. $\endgroup$
    – user30211
    Commented Feb 19, 2019 at 0:13
  • $\begingroup$ 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. $\endgroup$
    – user30211
    Commented Feb 19, 2019 at 0:21
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    $\begingroup$ 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. $\endgroup$ Commented Feb 19, 2019 at 12:30
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    $\begingroup$ 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. $\endgroup$
    – tj_
    Commented Feb 19, 2019 at 21:31

1 Answer 1

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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.

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