How to compute the $G$-theory groups of $k[x,y]/(xy)$ for any field $k$

Question

I am trying to compute the $G$-theory groups of the ring $k[x,y]/(xy)$ for any field $k$. What I have tried so far are two approaches.

Approach 1: Use the $G$-theory localization sequence for $k[x,y]/(xy)$. Set $A=k[x,y]/(xy)$ and consider $A/x\cong k[y]$ and $A_x\cong k[x,x^{-1}]$. There is the $G$-theory localization sequence

$$\dots\rightarrow G_{n+1}(A_x)\rightarrow G_n(A/x)\rightarrow G_n(A)\rightarrow G_n(A_x)\rightarrow …\rightarrow G_0(A_x)\rightarrow 0.$$

Since $A_x\cong k[x,x^{-1}]$, I obtained $G_0(A_x)\cong G_0(k)\cong \mathbb{Z}$. And the boundary map $G_1(A_x)\rightarrow G_0(A/x)$ is surjective, since the class $[A/x]\in G_0(A/x)$ is in the image of this boundary map. So by exactness of the localization sequence, I got $G_0(A)\cong G_0(A_x)\cong \mathbb{Z}$. I know that $G_n(A_x)\cong G_n(k)\oplus G_{n-1}(k)$ for any $n\geq 1$. And $G_n(A/x)\cong G_n(k)$ for all $n$. But I don’t know how to compute the boundary map $\partial:G_{n+1}(A_x)\rightarrow G_n(A/x)$. This is where I am stuck.

Approach 2: Use the coniveau spectral sequence. I know that $E_{\infty}^{0,-n}\cong F^0G_n(A)/F^1G_n(A)$, where $E_{\infty}^{0,-n}$ is the stable value at the $(0,-n)$ spot of the Coniveau spectral sequence. Now we have $F^0G_n(A)=G_n(A)$ and $F^1G_n(A)$ is defined to be the image of the map $K_nM^1(A)\rightarrow K_nM(A)$. Since the coniveau spectral sequence is a bounded fourth quadrant cohomological spectral sequence, I calculated that the $E_2$ page gives the stable values. So there are two difficulties here:how to compute $E_{\infty}^{0,-n}$ and how to compute $G_n(A)$ from the fact that $G_n(A)/F^1G_n(A)\cong E_{\infty}^{0,-n}$.

Any help will be greatly appreciated.

Answer 1

Consider the following square: $\require{AMScd}$ \begin{CD} pt \sqcup pt @>>> \mathbb{A}^1_k\sqcup \mathbb{A}^1_k\\ @V V V @VV V\\ pt @>>> \text{Spec}(k[x,y]/(xy)) \end{CD} This is a pullback square. The two affine lines are getting mapped to the two intersecting lines in the obvious manner. The fiber over the intersection point is just two disjoint points. This square is also a blow-up square since the right vertical map induces an isomorphism on the complement of the points. Such squares induce long exact sequence in the following manner: $$\ldots \rightarrow G_n(pt\sqcup pt)\rightarrow G_n(pt)\oplus G_n(\mathbb{A}^1_k\sqcup \mathbb{A}^1_k)\rightarrow \text{Spec}(A)\rightarrow \ldots$$ Looking at this it is easy to see that since $G_n(pt\sqcup pt) \rightarrow G_n(\mathbb{A}^1_k\sqcup \mathbb{A}^1_k)$ is an isomorphism the map $G_n(pt\sqcup pt)\rightarrow G_n(pt)\oplus G_n(\mathbb{A}^1_k\sqcup \mathbb{A}^1_k)$ is a split injection. Now you can see that $G_n(A)\cong G_n(pt)$.

It is also important to note that vertical maps are proper so the maps on the $G$-theory are defined via proper pushforward.