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I am trying to understand blow-ups, so I'd like to get some advises. Let me show you how I started to blow up a line $L\subset \mathbb A^3$. Any hint/comment/correction is highly appreciated.

Suppose $L$ is given parametrically by $x=at,y=bt,z=ct$, so that $x/a=y/b=z/c$. The blow-up $\pi:\textrm{Bl}_L\mathbb A^3\to \mathbb A^3$ is the resolution of the rational map $$ \phi:\mathbb A^3\dashrightarrow \mathbb P^1 $$ sending $P\mapsto \lambda$, where $H_\lambda\subset \mathbb A^3$ is the unique plane through $L$ and $P$. Hence $$\textrm{Bl}_L\mathbb A^3=\{(P,\lambda)\in \mathbb A^3\times\mathbb P^1\,|\,P\in H_\lambda \}\subset \mathbb A^3\times\mathbb P^1.$$ If $\lambda=(u:v)$, then $H_\lambda$ has equation $u(xb-ya)+v(yc-zb)=0$, thus $$ \textrm{Bl}_L\mathbb A^3=\{((x,y,z);(u:v))\in \mathbb A^3\times\mathbb P^1\,|\,u(xb-ya)+v(yc-zb)=0 \}. $$

Question 1. Is this correct?

Question 2. Can you please give me a hint to find the equations of the exceptional divisor $E=\pi^{-1}(L)$?

(I am stuck on question 2, because I know I should add one more equation to $\textrm{Bl}_L\mathbb A^3$ but I cannot figure which: $L$ has two defining equations)

Thanks in advance.

Brenin
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1 Answers1

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Notice that for any $(x,y,z)\in L,$ we have $xb-ya=yc-zb=0,$ which implies that for every $[u:v]\in\mathbb P^1,$ we have $((x,y,z),[u:v])\in\operatorname{Bl}_L\mathbb A^3.$ These are all the points of the exceptional divisor. To get "an" equation of the exceptional divisor, we need to focus on the two charts of $\operatorname{Bl}_L\mathbb A^3$ separately. In the chart $u=1$ the blowup is cut out by $(xb-ya)+v'(yc-zb)\in\mathbb A^4 = \operatorname{Spec}(k[x,y,z,v']).$ In particular, in the coordinate ring of the blowup, which we write as $k[x',y',z',v']$ we have $x'b-y'a = -v'(y'c-z'b),$ and in this chart, the exceptional divisor is defined by $y'c-z'b=0.$ The same goes, mutatis mutandis, in the other chart.

I find it a little easier to understand as follows. The equations $f_1=xb-ya$ and $f_2=yc-zb$ cut out the line $L$, so we know that the blow up has two charts defined by their coordinate rings $R_1=k[x,y,z][f_2/f_1]\subseteq k(x,y,z)$ and $R_2=k[x,y,z][f_1/f_2]\subseteq k(x,y,z).$ The equations defining $L$ map, under $k[x,y,z]\to R_i,$ to $f_1,(f_2/f_1)f_1$ and $(f_1/f_2)f_2,f_2$ respectively, and the equation of the exceptional divisor in each chart is $f_1=0$ and $f_2=0.$ I encourage you to write this out explicitly in the case $L:y=0,z=0$ is the $x$-axis, so that it looks less like gobbledygook.

Andrew
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  • Thank you, @Andrew. I find your first paragraph very clear, but the second one remains a bit obscure to me. For instance why is it fine to consider $R_1$ and $R_2$ defined that way? (i.e. why do we look at $f_i/f_j$?) – Brenin Apr 18 '13 at 20:55
  • Dear @atricolf, this is a different way to view the construction of blowups, by the gluing contruction. Blowing up the plane at a point entails a very similar computation. So, for $\operatorname{Bl}_{(0,0)}\mathbb A^2$ we glue the charts with coordinate rings $k[x,y,y/x]=k[x,y/x]$ and $k[x,y,x/y]=k[x/y,y]$ in the obvious way, using $f_1=x,f_2=y.$ The $y/x$ and $x/y$ are actually our coordinates for the exceptional $\mathbb P^1$ in the charts. This works for other examples as well, including yours. – Andrew Apr 18 '13 at 21:22
  • I maybe should mention, it suffices to look at $R_1,R_2$ because $\mathbb A^3$ is integral. We can view $R_1,R_2\subseteq k(x,y,z),$ and it is not hard to see that $R_2\cong k[x',y',z',v']$, with $v'$ corresponding to $-(f_1/f_2).$ We also have $R_1\cong k[x',y',z',u']$ with $u'\leftrightarrow -(f_2/f_1)$ (and a similar relation holding in $k[x',y',z',u']$). – Andrew Apr 18 '13 at 22:08
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    Dear @Andrew, thank you for all your explanations. I should look again at "The Geometry of Schemes", where I read about this approach some time ago... – Brenin Apr 19 '13 at 17:39