Here is exercise (10), chapter 3, from Beauville's Complex Algebraic Surfaces:
Let $S\subset\Bbb{P}^3$ be a surface, not necessarily smooth, such that every point of $S$ lies on a line contained in $S$. Show that $S$ is birationally isomorphic to a ruled surface. Conversely show that every ruled surface is birational to a surface of the preceding type (and more precisely, we can take this to be a cone in $\Bbb{P}^3$)
My questions are:
I'm bothered by the way he phrased it. By Beauville's definition, a ruled surface is a surface birationally equivalent to $C\times\Bbb{P}^1$. Therefore, to say that $S$ is birationally equivalent to a ruled surface wouldn't be equivalent to say that $S$ is a ruled surface? Why didn't he simply say "show that $S$ is ruled"?
Isn't the converse obvious? I mean, a ruled surface is birational to $C\times\Bbb{P}^1$, and it's easy (at least easy to imagine how) to construct a cone in $\Bbb{P}^3$ birational to $C\times\Bbb{P}^1$ which has the property described in the question.
If $S$ has the mentioned property, I guess in order to prove it is birational to $C\times\Bbb{P}^1$ we should be able to find something that looks like a fibration $S\to C$ with fibres that look like $\Bbb{P}^1$, but I'm not sure how. It would be nice, for example, to find a curve $C\subset S$ such that through each $P\in C$ there is only one line $L\subset S$ passing through $P$. But it obviously doesn't work always.
Any help will be appreciated. Thank you!