Consider the blow up $\pi:B \to \mathbb{A}^2$ of the origin in $\mathbb{A}^2$. Let $L=Z(ax+by)$ be a line through the origin in $\mathbb{A}^2$ and let $\widetilde{L}$ be the strict transform of $L$ in $B$. The exceptional divisor $E=\pi^{-1}((0,0))$ is covered by the affine open sets $U$ and $V$ with coordinate rings $k[U]=k[x,y/x]$ and $k[V]=k[y,x/y]$. We compute the ideal of $\widetilde{L} \cap U_1$ in $U_1$ to be $\langle a+b(y/x)\rangle$ and the ideal of $\widetilde{L} \cap U_2$ to in $U_2$ to be $\langle a(x/y)+b\rangle$. I have a few questions.

  1. Is this correct?
  2. How can I compute from this the intersection $\widetilde{L} \cap E$? It should be one point, correct? Is it equal to the point $((a/b:b/a),(0,0))$ of $\mathbb{P}^1 \times \mathbb{A}^2$?
  3. I want to show that the function $\mathbb{P}^1 \to E$ given by $(a:b) \mapsto \widetilde{L} \cap E$ is a bijection. From my guess for the intersection $\widetilde{L}\cap E$ in Question 2, it is not clear how this map can be surjective.

I have been struggling with this material and haven't found an adequate reference where these things are worked out in great enough detail. So I would be very appreciative if someone could help me understand this example. Thank you so much in advance for your time.


Let $L$ be a line through the origin. $L$ is defined up to multiplication by a nonzero scalar, so we may assume first that $L=Z(x+by)$, some $b \in k$. Let $x,y$ be coordinates on $\mathbb{A}^2$ and $u,v$ be homogeneous coordinates on $\mathbb{P}^1$. Then the exceptional divisor $E=\pi^{-1}((0,0))$ is covered by the affine open subsets $U_1=\mathbb{A}^2\times\mathbb{P}_u^1$ and $U_2=\mathbb{A}^2\times\mathbb{P}_v^1$. On $U_1$, we may as well set $u=1$. Then $x=yv$, and substituting this into the equation for $L$ gives $y(v+b)=0$. If $y=0$ then also $x=0$, and this is the component of the exceptional divisor in $U_1$. The other component, $v+b=0$ is the component of the strict transform in $U_1$. On $U_2$ we have $y=xu$, and so $x(1+bu)=0$. Again, $x=0,y=0$ is $E$. For the other component, $u \neq 0$, so in this case the strict transform is contained in the chart $U_1$, and is given by $u=1,v+b=0$. If $b \neq 0$, we similarly obtain that $L'$ is contained in the chart $U_2$ and is given by $v=1,u+a=0$. Thus, $L' \cap E=\{((0,0),(1:b))\}$ if $(a,b) \in D(x)$ and $L' \cap E=\{((0,0),(a:1))\}$ if $(a:b) \in D(y)$. Since $E=\{p\}\times\mathbb{P}^1$, it is now easy to see that the map $\mathbb{P}^1 \to E$ given by $(a:b) \mapsto L' \cap E$ is bijective.


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