# Exercise 11.5 from Atiyah-MacDonald: Hilbert-Serre theorem and Grothendieck group

I don't understand Exercise 11.5 of Atiyah & MacDonald, which demands one elaborate upon or rephrase the Hilbert–Serre Theorem (11.1) in terms of the Grothendieck group $K(A_0)$.

Here's the set-up in more detail. $A$ is a commutative Noetherian graded ring, finitely generated as an algebra over its degree-$0$ summand $A_0$ by finitely many elements $x_j$ of degrees $k_j > 0$. $\lambda$ is some additive function from the class of finitely generated $A_0$-modules to the integers $\mathbb{Z}$ — no field is presumed involved, so $\lambda$ is not assumed to be dimension.

For a finitely generated graded $A$-module $M$, the Poincaré series of $M$ with respect to $\lambda$ is the power series

$$P_\lambda(M,t) := \sum \lambda(M_n) t^n \in \mathbb{Z}[[t]].$$

If we write $q(t) = \prod (1 - t^{k_j})$ and compute the reciprocal $q^{-1}$ of $q$ in the power series ring $\mathbb{Z}[[t]]$, then the theorem is that

$$P_\lambda(M,t) \in q^{-1} \mathbb{Z}[t] \subset \mathbb{Z}[[t]];$$

that is, the Poincaré series is actually just a polynomial times the reciprocal of $q$.

Under the book's definition, the Grothendieck group $K(A_0)$ is the quotient of the free abelian group on the isomorphism classes of finitely generated $A_0$-modules by the subgroup generated by all elements $[N] - [M] + [P]$ for short exact sequences $0 \to N \to M \to P \to 0$ of such modules. No grading is invoked in this definition, and all finitely generated modules are used as generators, not merely projective or flat ones.

The question, again, is how to reformulate the result about $P_\lambda(M,t)$ in terms of $K(A_0)$.

My attempts to say something meaningful, which have turned out to be rather inadequate, follow.

First, if we define $K_{\mathrm{gr}}(A)$ to be the graded Grothendieck group of $A$, meaning we only admit isomorphism classes of graded $A$-modules as generators in the preceding definition and degree-preserving $A$-module homomorphisms in the exact sequences we use to generate the subgroup we quotient by, then it seems clear, using additivity of $\lambda$ on the summands $M_i$, that $M \mapsto P_\lambda(M,t)$ induces a homomorphism of additive groups $K_{\mathrm{gr}}(A) \to \mathbb{Z}[[t]]$ with values in the same subgroup $q^{-1} \mathbb{Z}[t]$ as before.

Further, one can pass from additive $\mathbb{Z}$-valued functions $\lambda$ on the class of finitely generated $A_0$-modules to group homomorphisms $l\colon K(A_0) \to \mathbb{Z}$, and say that for any such $l$, one can define an analogous homomorphism $Q_l(-,t)\colon K_{\mathrm{gr}}(A) \to \mathbb{Z}[[t]]$, with image as before.

This reformulation hardly seems enlightening, though. If we make $l$ a variable in this expression too, we get a function

$$Q\colon \mathrm{Hom}\big(K(A_0),\mathbb{Z}\big) \times K_{\mathrm{gr}}(A) \to q^{-1} \mathbb{Z}[t] \subset \mathbb{Z}[[t]].$$

This looks a little different than the original, but is not particularly more interesting.

Finally, another thing we can do is to factor all the $P_\lambda$ or $Q_l$ through the additive group $K(A_0)[[t]]$ to get a sort of "universal" Poincaré series, and note that for all $l \in \mathrm{Hom}\big(K(A_0),\mathbb{Z}\big)$, the image is contained in (well, I believe it is) $q^{-1} \mathbb{Z}[t]$. I would like to be able to lift the result about the image up to $K(A_0)[[t]]$, but since $K(A_0)$ doesn't usually have a ring structure, multiplication and hence $q^{-1}$ are not apparently defined in $K(A_0)[[t]]$. We can also make $l$ a variable as before.

I have trouble believing that the intended answer could be something so insubstantial.

Now, looking in Eisenbud, I did find that in the case where $A$ is a polynomial ring over a field $k$ and $\lambda = \dim_k$, the graded Poincaré series, if I am rephrasing correctly, gives an isomorphism from $K_{\mathrm{gr}}(A)$ to $q^{-1} \mathbb{Z}[t]$ (and hence implies an isomorphism with $\mathbb{Z}[t]$). That seems substantive enough. However, I don't see how hypotheses on $A$ and $\lambda$ as broad as ours could yield anything similarly interesting.

So what do you suppose they are looking for here?

• I remember thinking about this exercise a pretty long time and I think all they mean is the (somehow unspectacular) following. Every homomorphism $K(A_0)\to\mathbb{Z}$ gives rise to an additive function $\lambda$, so you can define $P_\lambda$ for such homomorphisms and conclude that, again, $P_\lambda\in q^{-1}\mathbb{Z}[t]$. – Ben Oct 21 '12 at 19:04
• Thank you. Speaking of unspectacular, is the amount of response to this question over this length of time typical for this site? Do you suppose I've done something wrong in formulating, titling, or submitting this question? Or is it perhaps inherently just not a very appealing one? I've been checking pretty regularly here and am wondering what I could have done wrong. – jdc Oct 23 '12 at 19:02
• Dear @jdc, no, I don't think so. I actually liked your question and how extensive you presented your ideas. My response may have caused some idleness, but I rather think that simply none of the users reading the question got a better idea (so far). However, supposed my idea is correct, it would be interesting to see if this has any $K$-theoretic consequences. Maybe someone knows some and takes the time to write an expanded answer? – Ben Oct 23 '12 at 20:16
• Dear @Ben A., thank you for liking it. My worry is that maybe too few (sixty-one, possibly counting myself multiple times, if the software does that) people read it to begin with. It would be great if someone who knows more finds this and could give an expanded answer. – jdc Oct 24 '12 at 4:03
• @navigetor23: Thank you, I will do that. – jdc Oct 25 '12 at 6:44

This is not a true answer, but I think it could be useful: the paper of W. Smoke Dimension and Multiplicity for Graded Algebras treats the case where $A_0$ is a field and obtains a homomorphism $\chi_R$ of $\mathbb{Z}[t]$-modules from $K_{\text{gr}}(A)$ to $\mathbb{Z}[[t]]$. Furthermore, if any finitely generated $A$-graded module have a finite graded free resolution, then $\chi_R$ is an isomorphism from $K_{\text{gr}}(A)$ to $\mathbb{Z}[t]$.
• Is this in Smoke? I don't seem to see it. Anyway, it generalizes the exercise in Eisenbud (19.15), where your $A$ is a finitely generated polynomial ring over a field. Thank you for this. – jdc Nov 2 '12 at 5:15