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Let $X \subset \mathbb{P}^n$ be a complete intersection of two quadrics. It is classical that, if $X$ contains a line, then it is rational. The proof is very simple and basically it is given by taking the projection off the line.

It is also stated in a paper by Colliot-Thélène, Sansuc and Swinnerton-Dyer that if $X$ contains a curve of odd degree, then it is rational as well. Is there a geometric proof of this?

Are there known examples of complete intersections of two quadrics that contain a curve of odd degree and not a line?

And finally: are there other sufficient conditions for the rationality of such a variety?

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up vote 3 down vote accepted

I've thought about this and discussed about this with someone else. A big part from what will follow is not my own contribution (in particular I hadn't thought about invoking Amer's theorem).

In fact it suffices to prove that any complete intersection of two quadrics containing a curve of odd degree must contain a line (hence the answer to your second question is affirmative).

By a theorem of Amer, an intersection of quadrics $f = g = 0$ contains a linear space of dimension $r$ over $k$ iff the quadric given by $f + tg = 0$ contains a linear space of dimension $r$ over $k(t)$. Hence it suffices to prove that any quadric which contains a curve of odd degree actually contains a line.

To do this, note that such a quadric must have a rational point: if you cut by a generic hyperplane, you will get a finite set of points, at least one of which has odd degree. But Springer's theorem says that if a quadric contains a point in an odd degree extension of the base field, then it must have a rational point. So the quadric is isotropic and we can split off a hyperbolic plane. With the "rest" of the quadric you can repeat the argument, although you probably have to be careful with configuration issues - I'd have to write down this part more carefully. Hence you will get two hyperbolic planes, hence a line.

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