Strict Transform of a Line in a Blow Up 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. 


*

*Is this correct?

*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 $\mathbb{P}^1 \times \mathbb{A}^2$?

*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. 
 A: 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.  
