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Suppose I have a hypersurface in $\mathbb{C}P^n$ given by some $f(z_1, \dots, z_{n+1}) = 0.$ Is there an algorithm which returns a rational parametrization if there is one, and "not rational" otherwise?

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I do not think so –  Francesco Polizzi Jul 11 '12 at 13:31
    
@Francesco: the lack of such algorithm would not shock me, but even heuristics would be nice... –  Igor Rivin Jul 11 '12 at 14:10
    
Do you want an algorithm, or just want to know whether one exists? i.e. whether or not the problem is decidable? –  Daniel Loughran Jul 11 '12 at 16:16
    
@Daniel to be honest, it did not occur to me that the problem was not decidable. Do you think it might be? –  Igor Rivin Jul 11 '12 at 17:06
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2 Answers

For smooth cubics in $\mathbb P^5$ this is unknown. That is, there are certain explicit families of such cubics that are known to be rational (those that admit a Pfaffian description, for example) but beyond these the problem of rationality for cubic $4$-folds is a famous unsolved problem.

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Here is my attempt at a heuristic as to why the problem should be undecidable.

Suppose we have a hypersurface $X$ of dimension $n$ and we wish to decide whether or not it is rational. I will assume that $n\geq2$. Then giving a rational map $\mathbb{P}^n \dashrightarrow X$ is the same as giving a $\mathbb{C}(t_1,\ldots,t_n)$-vauled point on $X$. However, "Hilbert's 10th problem" for such function fields is undecidable (see http://www.math.psu.edu/eisentra/varieties.pdf). Hence the problem you have asked for is undecidable.

Edit: As noted in the comments, this reasoning is not quite correct as for Hilbert's 10th problem we fix $m$ and a field $\mathbb{C}(t_1,\ldots,t_m)$, then allow the dimension $n$ to vary. Hence why it is only a heuristic!

Note that for Hilbert's 10th problem, the case $\mathbb{C}(t)$ is still open.

Edit: As remarked below, rationality for curves is decidable. One just needs to compute the genus of the normalisation of the projective closure of the curve.

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You can find an embedding of the normalization in projective space using blow-ups, say. This will give you a differential, say $dx/x - dy/y$. Counting the zeroes and poles of the differential computes the genus. I don't see any problems. –  Will Sawin Jul 11 '12 at 18:24
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Yes sorry I should have said that rationality is decidable for curves. I guess the thing which is not known to be decidable is whether or not a variety contains a rational curve. I will edit my answer. –  Daniel Loughran Jul 11 '12 at 18:47
    
Dear Daniel, maybe I am nitpicking a little bit, but to me undecidability of Hilbert's tenth problem for function fields does not seem to imply undecidability of the OP's question; HTP asks for an algorithm that works for all hypersurfaces, whereas in the OP's question the dimension of the hypersurface is fixed at the outset. (So the distinction is the same as that between decidability of rationality for curves, and decidability of existence of a rational curve in a variety.) –  Artie Prendergast-Smith Jul 11 '12 at 19:21
    
@Artie: Hmm I see your point. I will leave my answer up anyway as hopefully it at least gives a heuristic as to why the problem might be undecidable. –  Daniel Loughran Jul 11 '12 at 19:43
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