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Let $g\geq3$ be an integer, let $\{\Gamma_i|i \in I\}$ be the set of all stable graphs of genus $g$. (We say a graph is stable if it is the dual graph of a stable curve.)

Let $X$ be a curve defined over $\mathbb{Q}$, we say it has reduction type $\Gamma_i$ , if there is a model $\mathcal{X}_{p_i}$ over $\mathbb{Z}_{p_i}$, whose generic fiber is $X_{\mathbb{Q}_{p_i}}$ and special fiber is a stable curve with dual graph $\Gamma_i$.

Given $g\geq3$, does there always exist a smooth curve over $X/\mathbb{Q}$, (or some number field $K$?) such that every $\Gamma_i$ appear as the reduction of certain prime $p_i$?

(If we ask the same question over global function field (finite extension of $\mathbb{F}_p(t)$), I think the answer is "yes", as we can interpolate the loci in $\overline{\mathcal{M}}_g$ by curves. )

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  • $\begingroup$ A smooth curve over $\mathbb Q$ only has finitely many points of bad reduction. $\endgroup$ – Angelo Oct 12 '19 at 6:48
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    $\begingroup$ @Angelo Yes, but there are only finitely many stable graphs in fixed genus. $\endgroup$ – Olivier Benoist Oct 12 '19 at 10:10
  • $\begingroup$ Just to clarify, is the question asking for every $g \geq 3$ there exists such a curve, or just for some $g$ or possibly infinitely many such $g$? $\endgroup$ – Stanley Yao Xiao Oct 12 '19 at 11:59
  • $\begingroup$ @StanleyYaoXiao Thanks! I've edited the question $\endgroup$ – Qixiao Oct 12 '19 at 13:00
  • $\begingroup$ Oops, sorry, this was a really moronic comment. $\endgroup$ – Angelo Oct 12 '19 at 15:26
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For any list $p_i$ of primes, we can find such a curve over a number field that has reduction type $\Gamma_i$ at some prime lying over $p_i$.

We can iteratively blow up the strata of the stable graph stratification of $\overline{M}_g$ until they all have codimension $1$. Then by taking an intersection of general hyperplanes in the blow-up and applying Bertini, we can find an irreducible curve in $\overline{M}_g$ that meets every locus. Now map that curve to $\mathbb P^1$. Take a number field $K'$ where the images of the meeting points with every locus are defined, and pull back the $p_i$ to primes $\mathfrak p_i$ of $K'$. We can easily find a point of $\mathbb P^1(K')$ which is congruent mod $\mathfrak p_i$ to the image of the meeting point with the $\Gamma_i$ stratum, as this is just finitely many congruence conditions. We can now lift this point to a point of the curve, over a possibly larger number field, and then to a point of $\overline{\mathcal M}_g$ with stable reduction, over an even larger number field. At the point where it meets the $\Gamma_i$ locus in $\mathcal M_g$, becaue it has stable reduction, it must in fact have reduction type $\Gamma_i$.

I am sure the question over $\mathbb Q$ is open, as not much is known about the rational points on $M_g$ in general, except for very small $g$ which are unirational.

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Here is a stacky variant of Will Savin's answer, with a more precise result. The reference is my paper:

Problèmes de Skolem sur les champs algébriques, Compo. Math. 125 (2001), 1—30

(1) First, note the following: Let $F$ be a local field, $Y$ an $F$-scheme of finite type, $C\to Y$ a stable curve of genus $g$ over $Y$, and $\Gamma$ a stable graph of genus $g$. Let $U$ (resp. $U'\subset U$) be the set of points $y\in Y(F)$ such that $C_y$ has (potential) reduction type $\Gamma$ (resp. has stable reduction on $O_F$, with reduction type $\Gamma$). Then both $U$ and $U'$ are open in $Y(F)$, for the valuation topology. (Exercise).

(2) Now let $I'\subset I$ correspond to singular curves (i.e. we remove the one-point graph from $I$).
For each $i\in I'$, fix a prime $p_i$ (they have to be pairwise distinct!). I will make the assumption that
(*) there exists a smooth curve over $\mathbb{Q}_{p_i}$ with potential reduction graph $\Gamma_i$.
(I am pretty sure that this is always true; otherwise, choose $p_i$ accordingly).
Put $R:=\mathbb{Z}\left[(1/p_i)_{i\in I'}\right]$.

Theorem. There is a number field $K$, and a stable curve $\mathscr{X}$ over $O_K$ such that:
$\mathscr{X}$ is smooth over $R\otimes_{\mathbb{Z}}O_K$;
• for each $i\in I'$ and each prime $\mathfrak{p}$ of $O_K$ above $p_i$, $\mathscr{X}$ has reduction graph $\Gamma_i$ at $\mathfrak{p}$.
Assume further that for each $i\in I'$ there is a stable curve over $\mathbb{F}_{p_i}$ with graph $\Gamma_i$.
(This holds if $p_i$ is large enough). Then we can take $K$ totally split at each $p_i$.

Proof: Consider the moduli stack $\mathscr{M}_g$ (resp. $\overline{\mathscr{M}}_g$) of smooth (resp. stable) curves of genus $g$.
Let us prove the first claim. For each $i$, let $\Omega_i\subset{\mathscr{M}}_g(\mathbb{Q}_{p_i})$ be the subcategory of those smooth curves with (potential) reduction graph $\Gamma_i$. By (1), this is $p_i$-adically open in ${\mathscr{M}}_g(\mathbb{Q}_{p_i})$, in the sense of Definition 2.2 in the paper. Moreover, it is not empty, due to assumption (*).
It is straightforward to check that $\mathscr{M}_{g,R}\to\mathrm{Spec}(R)$ with the local data $(\Omega_i)_{i\in I'}$ constitutes a Skolem datum in the sense of Definition 0.6. (We use the fact that $\mathscr{M}_g$ is smooth with geometrically connected fibers over $\mathrm{Spec}(\mathbb{Z})$). Now apply Theorem 0.7: this almost gives the result (with $K$ totally split at each $p_i$) except that the curve may not have stable reduction outside $\mathrm{Spec}(R)$. To fix this, just enlarge $K$, possibly losing splitness.
For the second claim, we do the same, replacing $\Omega_i$ by $\Omega'_i$ consisting of curves with stable reduction of type $\Gamma_i$ over $\mathbb{Z}_{p_i}$: the extra assumption on $p_i$ guarantees that $\Omega'_i\neq\emptyset$. QED

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