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Let $X$ be a smooth projective geometrically connected variety over $\mathbb{Q}$. It is said to have good reduction at a prime $p$ is there is a smooth projective $\mathcal{X}\to \mathrm{Spec}\:\mathbb{Z}_{(p)}$ with $\mathcal{X}_{\mathbb{Q}}\approx X$.

If I understand correctly, to say that $X$ has good reduction at $p$ is naively a $\Sigma^0_2$-statement. Gro-Tsen's answer shows it can in fact be decided by a halting oracle for ordinary Turing machines. Can it be decided by an ordinary Turing machine?

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    $\begingroup$ In the case of abelian varieties, by Neron-ogg-shafarevich, it is a $\Pi_1^0$ statement if I understand correctly. $\endgroup$
    – Asvin
    Commented Aug 11, 2020 at 21:43
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    $\begingroup$ -1 for changing the question after it received a complete answer $\endgroup$
    – Piotr Achinger
    Commented Aug 12, 2020 at 8:57

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Consider a Turing machine which enumerates all possible projective $\mathcal{X} \to \operatorname{Spec}\mathbb{Z}_{(p)}$, all possible morphisms $\varphi\colon X \to \mathcal{X}_\eta$ (where $\mathcal{X}_\eta$ is the generic fiber) and all possible morphisms $\psi\colon \mathcal{X}_\eta \to X$, and, for each $(\mathcal{X},\varphi,\psi)$, checks whether $\mathcal{X}$ is in fact smooth, and $\varphi$ and $\psi$ are inverse to each other: if so, it halts, otherwise, it proceeds to the next candidate. This machine stops iff $X$ has good reduction at $p$, so this can be decided using the halting oracle (for ordinary Turing machines), i.e., “$X$ has good reduction at $p$” is, in fact (equivalent to) a $\Sigma^0_1$ statement of arithmetic (namely the statement that the machine just described halts).

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  • $\begingroup$ Is it obvious that all those collections are enumerable? $\endgroup$
    – LSpice
    Commented Aug 11, 2020 at 23:08
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    $\begingroup$ I think so. We have to iterate over possible dimensions and for each dimension, the base ring is countable so there are countably many projective schemes and given $X_\eta,X$, morphism between them are another countable family. $\endgroup$
    – Asvin
    Commented Aug 11, 2020 at 23:18
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    $\begingroup$ @LSpice Well, yes and no. Once it is checked that these various algebraico-geometric objects can be represented by finitistic data that can be manipulated by a Turing machine, it's straightforward to just enumerate all possible such data. What isn't obvious is that they can indeed be represented (well, for $\mathscr{X}$ it is by the definition of “projective”). For this, see §16 of my joint paper with F. Orgogozo (not claiming originality here, it was probably considered folklore). $\endgroup$
    – Gro-Tsen
    Commented Aug 11, 2020 at 23:20
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    $\begingroup$ @Asvin Right, although the emphasis isn't so much on “countable” (although this is certainly a prerequisite) as on the fact that the relevant data (schemes, morphisms between them, etc.) can be indeed represented by computer. $\endgroup$
    – Gro-Tsen
    Commented Aug 11, 2020 at 23:23

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