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?