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This is exercise $4.4$ part (c) of Hartshorne's book. Let $Y$ be the nodal cubic curve $y^{2}z=x^{2}(x+z)$ in $\mathbb{P}^{2}$. Show that the projection $f$ from the point $(0,0,1)$ to the line $z=0$ induces a birational map from $Y$ to $\mathbb{P}^{1}$.


Consider the open subset of $Y$ given by $Y \setminus V(z)$ , that is we set $z=1$.

Define $f: Y \setminus V(z) \rightarrow \mathbb{P}^{1} \setminus \{[1:1],[1:-1]$ by:

$f([x : y : z]) = [x: y]$

Now define $g: \mathbb{P}^{1} \setminus \{[1 : 1],[1 : -1]\} \rightarrow Y \setminus V(z)$ by:

$g([x : y]) = [(y^{2}-x^{2})x : (y^{2}-x^{2})y : x^{3}]$

Question: what if $x=0$? then we get $y=0$ so $[x :y : 1] = [ 0 : 0 : 1]$ but the point $[x^{4} : x^{3}y : x^{3}]$ is not defined at $x=y=0$.

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$[ab:ac] = [b:c]$, right? –  Hurkyl May 17 '12 at 17:13
You can simplify your formula for $f$ (assuming it's right). p.s. your formula would be ill-defined were it not for the fact the simplification is possible: your functions aren't homogeneous, and so the required identity $f[x:y:z]=f[ax:ay:az]$ isn't immediately evident. –  Hurkyl May 17 '12 at 17:23
@Hurkl: just modified it, can you have a look please? –  user31509 May 17 '12 at 17:44
It's been a while since I've look at at this fine detail -- but as I recall, it's good enough for them to be inverses on an open set, right? You can restrict the domains further to make it work. –  Hurkyl May 18 '12 at 17:54

1 Answer 1

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Recall that the line in $\mathbb P^2$ joining $P=[a:b:c]$ to $P'=[a':b':c']$ is given parametrically by $[ua+u'a':ub+u'b':uc+u'c']$ where $[u:u'] \in \mathbb P^1$.

So, the line joining $P'=O=[0:0:1]\in Y$ to $P=[a:b:c]\neq O\in Y$ has parametric equation $[ua:ub:uc+u']$.
It intersects the line at infinity $z=0$ at the point $uc+u'=0$ i.e. at $[a:b:0]$.
So the required rational map $Y \cdots \to \mathbb P^1$ (not a morphism!) is $$\phi:U=Y\setminus \lbrace O\rbrace \to \mathbb P^1:[a:b:c]\mapsto [a:b] $$
The inverse rational map is the morphism ( you have actually already calculated this: your computation is correct) $$\psi=\phi^{-1}:\mathbb P^1\to Y: [c:d]\mapsto [c(d^2-c^2):d(d^2-c^2):c^3]$$
Notice that $\psi$ is the normalization of $Y$, a remark which enlightens the whole exercise.
In particular we see that the two points $[1:\pm1]\in \mathbb P^1$ are sent by $\psi$ to $O$ .
This is why the rational map $\phi$ we started with cannot be a morphim: a non-injective morphism like $\psi $ cannot be inverted as a morphism, only (sometimes!) as a rational map .

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