Let $S=k\left[x_1, \ldots, x_r\right]$, where $k$ is a field and (for simplicity) all the indeterminates have degree 1 . Let $\mathcal{C}$ be the category of finitely generated graded $S$-modules. The Hilbert series is an ไdditive function on $^{e "}$ with values in the (additive group of) formal power series $\mathbf{Z}[t]]$ in the sense that for each module $M \in \mathcal{C}$ we have the power series $h_M(t)=\sum_{d-0}^{\infty} \operatorname{dim}_k\left(M_d\right) \in \mathbf{Z}[[t]]$, and for each short exact sequence $0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{\prime \prime} \rightarrow 0$ in $\mathcal{C}$ we have $h_{M^{\prime \prime}}-h_M+h_{M^{\prime}}=0$ in $\mathbf{Z}[[t]]$. In fact, it is the universal additive function, in a certain sense. We prove this now:
We define the Grothendieck group $G_0(\mathcal{C})$ to be the additive group with a generator $[M]$ for each graded module $M$ in $\mathcal{C}$, and a relation $\left[M^2\right]- [M]+\left[M^T\right]$ for each short exact sequence $0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{\prime \prime} \rightarrow 0$ in $\mathcal{C}$.
Show that the map $M \mapsto[M]$ is the universal additive function on $\mathcal{C}$ in the sense that given any other additive function $h$ with values in a group $A$, there is a unique group homomorphism $h^{\prime}: G_0(\mathcal{C}) \rightarrow A$ such that $h(M)=h^{\prime}([M])$. In particular, the Hilbert series induces a function on $G_0(\mathcal{C})$. Show that this function is a monomorphism on $G_0(\mathcal{C})$, and maps $G_0(\mathcal{C})$ isomorphically to $\left.(1-t)^{-r}\left(\mathbf{Z}\left[t, t^{-1}\right]\right) \subset \mathbf{Z}[t]\right]$, using the following steps:
a. Define $K_0(\mathcal{C})$ to be the additive group with a generator $[F]$ for each graded free module $F$, and a relation $\left[F^m\right]-[F]+\left[F^m\right]$ for each short exact sequence $0 \rightarrow F^{\prime} \rightarrow F \rightarrow F^{\prime \prime} \rightarrow 0$ in $\mathcal{C}$. (Since every short exact sequence of free modules splits, we could instead have taken a relation $\left[F^{77}\right]-[F]+\left[F^{\prime \prime}\right]$ whenever $F \cong F^{\prime} \oplus F^{\prime \prime}$ as graded modules.) Show that regarding a free module as a module gives a map of groups $K_0(\mathcal{C}) \rightarrow G_0(\mathcal{C})$. Because of the existence of finite free resolutions of a graded module by graded free modules, this map has an inverse. Construct it.
b. Use part a to show that $G_0(\mathcal{C})=K_0(\mathcal{C})$ is the free group on the classes of modules $S(d)$ for $d \in \mathbf{Z}$. Now use Ex. 19.14a.