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Let $S$ be Spec $O_K$ with $O_K$ the ring of integers of a number field $K$.

Let $X\to S $ be an arithmetic variety, i.e., an integral smooth quasi-projective $S$-scheme with generic fibre $X_\eta$ geometrically connected.

Suppose that I highly suspect that $X\to O_K$ has no sections. Now, how could I prove that this is really the case? (This is equivalent to saying that $X_\eta$ has no integral points with respect to the model $X\to S$, I believe.)

Is there a "computable" necessary criterion $X\to S$ must satisfy for $X(S)$ to be non-empty? What about a sufficient criterion?

If $S$ is Spec $\mathbf{Q}$ one could try to check whether $X(\mathbf{Q}_v)$ is empty for example.

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up vote 10 down vote accepted

It sounds as if you are asking for an algorithm which, in particular, would be able to tell whether a given set of polynomials over $\mathbb{Z}$ admits an integer solution. This is Hilbert's 10th problem and it is known that no such algorithm exists in general.

For specific classes of variety, it's certainly possible to look at necessary conditions from the local-global point of view: taking $S = \textbf{Spec}\;\mathbb{Z}$, a necessary condition for solubility is existence of $\mathbb{Z}_p$-points for all $p$. Even for quadrics, this condition is known not to be sufficient, but one can define a Brauer-Manin obstruction: for a nice account of this, see Colliot-Thélène and Xu's article "Brauer-Manin obstruction for integral points of homogeneous spaces and representation of integral quadratic forms".

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