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Let $X$ be a smooth variety over a number field $K$ and let $\mathcal X$ be a normal, projective model of $X$ over the ring of integers $O_K$.

Now assume that $D\subset X$ is an irreducible divisor and take its closure $\mathcal D$ is $\mathcal X$. Then $\mathcal D$ is a model over $O_K$ of the variety $D$.

Let $f_\mathfrak p(D)$ be the number of irreducible components of the reduction $D_{\mathfrak p}$ at the prime $\mathfrak p$. In other words, $f_{\mathfrak p}(D)$ is the number of the irreducible components of the fibre over $\mathfrak p$ on the arithmetic variety $\mathcal D\to\operatorname{Spec} O_K$.

Is it possible estimate an upper bound (depending on $\mathcal X$ but not on $D$) for the sum: $$ S_D:=\sum_{\mathfrak p} (f_\mathfrak p(D)-1) $$ In other words, can we control the number of irreducible components for the fibres of models of divisors? For instance: is $S_D$ bounded from above by $S_X$? Can $S_D$ be $0$?

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    $\begingroup$ Do you want $D$ to be geometrically irreducible, rather than just irreducible? Otherwise $S_D$ can be infinite. $\endgroup$
    – Daniel Loughran
    Commented Jul 25, 2022 at 14:43
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    $\begingroup$ There is no upper bound depending only on $\mathcal{X}$ but not on $D$. So long as the dimension of $X$ is at least $2$, a union $\mathcal{D}_p$ of $s$ (pairwise distinct) general very ample divisors in a single fiber $\mathcal{X}_p$ will admit a "lift" $D$ that is geometrically irreducible. $\endgroup$
    – Jason Starr
    Commented Jul 25, 2022 at 15:41
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    $\begingroup$ I would guess that the best you can hope for is an upper bound which depends on the degree of $D$ measured with respect to a fixed ample line bundle on $\mathcal{X}$. Of course fixing the degree gives an upper bound on $f_\mathfrak{p}(D)$ but the trickier part is to use the fact that $D$ varies within a finite type Hilbert scheme to show there exists an upper bound on the number of primes of bad reduction. Also making this effective sounds very hard. $\endgroup$
    – Dori Bejleri
    Commented Jul 25, 2022 at 21:05
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    $\begingroup$ @DoriBejleri To get such an upper bound, you also need to bound the height of the "moduli point" $[D]$ inside its complete linear system (such a height is sometimes called a "moduli height"). $\endgroup$
    – Jason Starr
    Commented Jul 25, 2022 at 22:21
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    $\begingroup$ @JasonStarr Yes that makes sense thanks! $\endgroup$
    – Dori Bejleri
    Commented Jul 25, 2022 at 22:32

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