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Following Shafarevich "Algebraic Geometry II", I found this example.

Let $X_3\subset\mathbb{P}^3$ a smooth cubic surface. To prove that $X_3$ is rational he claims that there is a birational map $\phi: X_3\dashrightarrow \mathbb{P}^2$ given by $\phi(x)=\mathit{l}_x\cdot L$, where $L=\mathbb{P}^2\subset\mathbb{P}^3$ and $\mathit{l}_x$ is a line through the point $x\in X_3$ that intersects two fixed skew lines on $X_3$, $m$ and $m^\prime$.

I don't see why this map is birational and why the point $\mathit{l}_x$ should intersect two skew lines on $X_3$. Could someone elaborate a bit? I think i don't know very well the geometry of a cubic.

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First of all you say "why the point $x$ should intersect two skew lines", but that's not what Shafarevich means: he means that the line $l_x$ intersects those lines.

OK, let's check that $\phi$ really is a well-defined rational map.

Fix a point $x \in X_3$ that doesn't lie on either line $m$ or $m'$. Now think of all the lines through $x$ that intersect $m$: the union of these fills up a plane $\Pi \subset \mathbf P^3$. The plane $\Pi$ intersects the line $m'$ in precisely one point (if it didn't, we would have $m' \subset \Pi$, but then $m$ and $m'$ wouldn't be skew). So there is exactly one line through $x$ that intersects both $m$ and $m'$; this is the line $l_x$ we want. It intersects the plane $L$ in one point, and that point is $\phi(x)$.

Next, why is $\phi$ birational? That is, why does it have an inverse rational map? Well, we just reverse the argument above. Fix a point $p \in L$ that doesn't lie on either $m$ or $m'$. Then just as before, there will be a unique line, call it $\lambda_p$, that passes through $p$ and intersects the two lines $m$ and $m'$.

Now the crux: because $X_3$ is a cubic surface, $\lambda_p$ will intersect $X_3$ in exactly three points (counted with multiplicities). One each of these points will be on $m$ and $m'$; define $\psi(p)$ to be the third point. (Notice: this doesn't work as stated if $\lambda_p$ is tangent to $X_3$ at a point on $m$ or $m'$, but that's ok, because that won't happen for most $p$.) Finally, observe that $\phi$ and $\psi$ are inverse rational maps.

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  • $\begingroup$ thank you very much, I edited my question and corrected the error. $\endgroup$ – idioteca Feb 4 '14 at 13:34
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    $\begingroup$ Dear idioteca, note that to give a rational map, it is enough to specify it on any nonempty open set. It is then a theorem that a rational map between projective varieties can be extended to codimension 1. But you don't need to prove that. $\endgroup$ – user64687 Feb 4 '14 at 15:07
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    $\begingroup$ To answer your question, note that the lines $m$ and $m'$ each intersect $L$ in a single point, so there is only one line $l$ intersecting $m$ and $m'$ and contained in $L$. Moreover, this line $l$ intersects $X_3$ in three points: one each lying on $m$ and $m'$, and one other point $x$. This $x$ is the only point such that $l_x$ is contained in $L$. $\endgroup$ – user64687 Feb 4 '14 at 15:10
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    $\begingroup$ Note also that the given defintion of $\phi$ does not make sense when $x$ lies on either $m$ or $m'$. It's a good exercise to try to visualise $\phi$ and understand how it extends to (open sets in) those lines. $\endgroup$ – user64687 Feb 4 '14 at 15:12
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    $\begingroup$ Dear idioteca, what I said was that the given definition of $\phi$ does not make sense for those points. But indeed, the map $\phi$ must extend to almost all points on $m$ and $m'$. $\endgroup$ – user64687 Feb 4 '14 at 15:40

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