How to compute the integer corresponding to a class in $G_0(B_{\mathrm{red}})$ for a commutative noetherian ring $B$?

Question

$\newcommand{\red}{\mathrm{red}}$Let $k$ be an algebraically closed field of characteristic zero and $m$ be a positive integer. Let $R$ be the subring $k[x,xy,xy^2,…,xy^m]$ of the polynomial ring $k[x,y]$, and $B$ be the quotient ring $R/xR$. I am trying to compute the $G$-theory groups of the ring $R$.

So far, I have tried to use the $G$-theory localization sequence induced by the fibration sequence $G(R/xR)\rightarrow G(R)\rightarrow G(R_x)$ to do this computation. By using this localization sequence, it suffices to compute the image of the boundary map $\partial:G_1(R_x)\rightarrow G_0(R/xR)$. I have also computed that $B_{\red}\cong k[t]$. So by devissage, we have $$G_0(R/xR)\cong G_0((R/xR)_{\red})=G_0(B_{\red})\cong G_0(k[t])\cong\mathbb{Z}.$$ The isomorphism $G_0(R/xR)\rightarrow G_0((R/xR)_{\red})$ maps the class $[R/xR]$ in $G_0(R/xR)$ to the class $[B/I]+[I/I^2]+…+[I^{m-1}/I^m]$ in $G_0((R/xR)_{\red})$, where $I$ is the nilradical of $B$. This comes from the filtration $0=I^m\subseteq I^{m-1}\subseteq…\subseteq I\subseteq B$.

My question is, what integer does this class in $G_0((R/xR)_{\red})$ correspond to? How do I compute this integer?

Thank you so much for your kind help.

Answer 1

Here is an overkill answer.

Let $A=k[x,xy^m]\subset R$. Then, as I said in a previous answer for a previous question of yours, $R$ is integral over $A$ and thus it is a finite module over $A$. It is torsion free of rank $m$ as a module over $A$. Thus one has a resolution, $0\to P\to F\to R\to 0$ as $A$-modules, where $F$ is a free module of say rank $n$ over $A$. By Auslander-Buchsbaum, $P$ is projective (over $A$) of rank $n-m$ (in fact free by Seshadri's theorem). Tensoring by $A/xA$, we get an exact sequence $0\to P/xP\to F/xF\to B\to 0$. Since $P/xP$ is free (over $A/xA=k[xy^m]$), we see that $[B]=m\in G_0(A/xA)$. On the other hand, we have the natural map $G_0(B_{red})\to G_0(B)$, which as you noted is an isomorphism. Thus, the map $G_0(B)\to G_0(A/xA)$ is an isomorphism, since the composite $G_0(B_{red})\to G_0(A/xA)$ is. This shows the class of $B$ in $G_0(B_{red})$ is $m$.