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Let $X$ be smooth variety defined over $\mathbb{Q}$. If we want to check that $X$ is locally soluble at a prime $p$, then it suffices to find a non-singular $\mathbb{F}_p$ point, which can be lifted to a $\mathbb{Q}_p$ point by Hensel's lemma.

However, it might happen that $X$ does not have any non-singular $\mathbb{F}_p$ points. For example $X$ could be given by a quadratic form and we are interested in the prime $p=2$, in which case the reduction mod $2$ of $X$ is a non-reduced scheme and hence every point is singular (at least if I understand the situation correctly).

What general methods are there, if any, to check local solubility in this kind of situation?

A nice toy example is the equation $x^2 + y^2 + z^2=0$. This is locally soluble for all primes $p>=3$ (by Chevalley–Warning) and clearly not soluble for $p=\infty$, hence it is not soluble at $p=2$ (by the Hilbert symbol formulation of quadratic reciprocity). Is there a simple way to see this using general methods?

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Well, yes, there is a simple and general method which can however be computationally intensive: you look modulo higher powers of the maximal ideal (i.e., mod $p^n$). It is a theorem that when n is sufficiently large, you will either be able to apply Hensel's Lemma to get a solution in the valuation ring, or you will find that there are no (primitive) solutions mod $p^n$. For the example you gave, I believe the minimal value of $n$ for which this occurs is $n = 3$. Try it! – Pete L. Clark Jul 5 '10 at 17:00
Thanks for the answer! – Daniel Loughran Jul 6 '10 at 10:04
Doesn't looking at the equation modulo 4 (i.e. with n = 2) suffice? The only solutions have x = y = z = 0 mod 2, and now descent shows that there is no nontrivial solution. – Franz Lemmermeyer Jul 6 '10 at 12:06
@Franz: Belatedly, yes, of course you're right: going mod $4$ is enough. I must have been thinking about $x^2 + y^2 + z^2 = 7$, which of course does need to be checked mod $8$. (Clearly I didn't follow my own advice...) – Pete L. Clark Dec 10 '10 at 17:42

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