Computing $K_{X'/\mathbb{A}^2}$

Question

Let $k$ be an algebraically closed field of characteristic zero, $X=\mathbb{A}^2_k$ and $X'=\operatorname{Bl}_{(x,y)} X$. I'm looking to compute $K_{X'/X}$.

Note that $$X'=\operatorname{Proj} k[x,y,T_1,T_2]/(xT_2-yT_1).$$ On $D_{+}(T_1)$, say, we have $$D_{+}(T_1)\simeq \operatorname{Spec} k[x,y,T_2/T_1]/\left(x(T_2/T_1)-y\right).$$

To study $K_{X'/X}$ on this patch, it suffices to compute the Jacobian determinant associated to $$\pi_1:D_{+}(T_1)\longrightarrow \operatorname{Spec} k[x,y].$$

Call $v=T_2/T_1$ and $u=x$. I could be wrong here, but I believe $\pi_1$ is induced by the inclusion $$k[x,y]\longrightarrow k[x,y,T_2/T_1]/\left(x(T_2/T_1)-y\right),$$ sending $x\mapsto u$ and $y\mapsto uv$.

Knowing $K_{X'/X}$ beforehand, I'm pretty sure I need this inclusion to satisfy $f(u,v)=(u,uv)$. However, I'm having trouble seeing how we "change coordinates" with the assignments $x\mapsto u, y\mapsto uv$ to get this desired form.

Thank you for any help.

Answer 1

I want to provide an alternative answer using differential forms. The general process is justified in the appendix. Let $\mu: X' \to \mathbb{A}^2$ denote the blowup at $(x,y)$.

Since $\omega_{\mathbb{A}^2}$ is freely generated by $dx \wedge dy$, we can compute $\mu^*\omega_{\mathbb{A}^2}$ on each patch by pulling back $x$ and $y$, which you correctly described above. On the patch $U_1 = \mathbb{A}^2_{u, v}$, where $T_1 \neq 0$, we see that $\mu^*\omega_{\mathbb{A}^1}$ is trivialized by $$\mu^*(dx \wedge dy) = du \wedge d(uv) = u(du \wedge dv).$$

Hence, we may describe $K_{X'/\mathbb{A}^2}|_{U_1} = \{u = 0\} = E|_{U_1}$.

Similarly, on the other patch $U_0$, we'll see $E|_{U_0}$. (I'll leave it to you to check.)

Note: The complement of $U_1$ in $X'$ is a single point and, in particular, has codimension $\geq 2$ in $X'$. As such, the restriction map $\operatorname{Pic}(X') \to \operatorname{Pic}(U_1)$ is an isomorphism so checking the other patch is unnecessary.

Appendix: The Jacobian Description

Suppose $\mu: Y \to X$ is a birational map of smooth $n$-dimensional varieties. We can define $\omega_{Y/X}$ to be the invertible sheaf $\omega_{Y} \otimes \mu^*\omega_{X}^{-1}$. To describe this sheaf locally, fix $x \in X$ and a set of local coordinates $x_1, \dots, x_n$ of $X$. Then, $\omega_{X}$ is freely generated in a neighborhood $U$ by $dx_1 \wedge \cdots \wedge dx_n$. Hence in a neighborhood $V$ of $y \in \mu^{-1}(U)$, with coordinates $y_1 \dots, y_n$, we find that $\mu^*\omega_{X}$ to be freely generated by $$\mu^*(dx_1 \wedge \cdots \wedge dx_n) = d\mu^*x_1 \wedge \cdots \wedge d\mu^*x_n = \operatorname{Jac}_V(\mu) dy_1 \wedge \cdots \wedge dy_n.$$ As such, covering $Y$ by these $V$'s, we find trivializations $\mathcal{O}_V \to \omega_Y \otimes \mu^*\omega_{X}^{-1}|_V$ by sending $f \mapsto f \operatorname{Jac}_V(\mu)^{-1}$. Using the correspondence between line bundles and Cartier divisors (see, for example, Hartshorne's proposition II.6.13), we conclude that this line bundle can be associated with effective Cartier divisor $\{(V, \operatorname{Jac}_V(\mu))\}.$

Answer 2

For the sake of simplicity I will assume $(x,y)=(0,0)$. Let $\pi : \tilde{X}\rightarrow X$ be the blow-up morphism and $E=\pi^{-1}(0,0)$ the exceptional divisor. It is well known that $E\simeq \mathbb{P}^1$, but I think I won't need that fact. Nevertheless $E$ is a divisor in $\tilde{X}$, so it is associated to a line bundle $\mathcal{L}(E)$. An abstract formula gives $\omega_{\tilde{X}/X}\simeq \mathcal{L}(E) $. For your knowledge if $Y \subset X$ are non-singular varieties, $r=\text{Codim}(Y,X)$, $\tilde{X} = \text{Bl}_Y(X)$ then $\omega_{\tilde{X}/X}\simeq (r-1)\mathcal{L}(E)$.

This equality is important but here I think you would prefer a down to earth computation. Your guess for $\pi_1$ is right, indeed the morphism $\tilde{X} \rightarrow X$ is just the composition of the inclusion $\tilde{X}\hookrightarrow \mathbb{A}^2 \times \mathbb{P}^1$ with the canonical projection on $\mathbb{A}^2$, so you must send $y$ on $y$, but in your local chart $y=uv$ with your notation.

In the other chart, your map will be of the same kind, but to change coordinates you will have to inverse $v$ and use $d(\frac{1}{v})=\frac{-1}{v^2}\cdot dv$.

However I don't understand what you mean by "it suffices to compute the Jacobian determinant", what do you find as an answer ? For me the Jacobian gives informations about $\Omega^1_{\tilde{X}/X}$, but it is not what we are looking for.