How to compute the transfer maps for G-theory of Noetherian schemes

Question

Let $k$ be a field and $R$ be the ring $k[x,xy,xy^2,xy^3]$. Let $X$ be $\operatorname{Spec}(R)$ and $\tilde{X}$ be the blow-up of $X$ along the maximal ideal $I$ of $R$ generated by $x,xy,xy^2,xy^3$.I have proved that $\tilde{X}$ is line bundle over $\mathbb{P}^1_k$, so that the $G$-theory groups of $\tilde{X}$ agree with the $G$-theory groups of $\mathbb{P}^1_k$. I am trying to compute the $G$-theory groups of $X$ by using the long exact sequence induced by the blow-up square associated to $X$ and $I$, which is $\cdots\rightarrow G_n(\mathbb{P}^1)\rightarrow G_n(k)\oplus G_n(\tilde{X})\rightarrow G_n(X)\rightarrow G_{n-1}(\mathbb{P}^1)\rightarrow\cdots$. So far, I only managed to show that the map $G_0(\mathbb{P}^1)\rightarrow G_0(k)$ sends the class of the structure sheaf of $\mathbb{P}^1$ to the class of $k$ and sends the class of the twisted sheaf $O(-1)$ to 0.Note that $G_0(\mathbb{P}^1)$ is the free abelian group based on the classes of $O$ and $O(-1)$.So I know the map $G_0(\mathbb{P}^1)\rightarrow G_0(k)$. But I don’t know how to compute the maps $G_n(\mathbb{P}^1)\rightarrow G_n(k)$ for all $n\geq 1$ and the maps $G_n(\mathbb{P}^1)\rightarrow G_n(\tilde{X})$ for all $n\geq 0$. Could someone help me with this problem? Thank you very much in advance.

Answer 1

If $\tilde{X}$ is the total space of a line bundle $L$ and $i \colon \mathbb{P}^1 \to \tilde{X}$ is the embedding of the zero section there is an exact sequence $$ 0 \to p^* L^\vee \to \mathcal{O}_{\tilde{X}} \to i_* \mathcal{O}_{\mathbb{P}^1} \to 0.\tag{*} $$ If you tensor it with $p^*F$ and observe that $$ p^*F \otimes i_* \mathcal{O}_{\mathbb{P}^1} \cong i_*i^*p^*F \cong i_*F $$ (by projection formula) you get an exact sequence $$ 0 \to p^*(F \otimes L^\vee) \to p^* F \to i_* F \to 0. $$ This gives you an expression for the map $i_*$ in terms of the map $p^*$.

EDIT. Let me explain the sequence $(*)$. The variety $\tilde{X}$ is smooth and $Z = i(\mathbb{P}^1) \subset \tilde{X}$ is a Cartier divisor. Therefore, the ideal $I_Z$ is a line bundle.

On the other hand, every line bundle on $\tilde{X}$ is the pullback of a line bundle via $p$. So, to identify $I_Z$ it is enough to understand its restriction to $Z$. We have $$ I_Z\vert_Z = \mathcal{O}_{\tilde{X}}(-Z)\vert_Z = \mathcal{O}_Z(-Z) = \mathcal{O}_Z(3), $$ because the normal bundle of $Z$ is $\mathcal{O}(-3)$. Thus $$ I_Z \cong p^*\mathcal{O}(3), $$ and the sequence $(*)$ is just the Koszul resolution of $\mathcal{O}_Z$.