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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 $0$?

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 $\mathfrak p$.

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 $0$?

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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manifold
manifold
  • 321
  • 6
  • 15

Control on the locus of bad reduction for divisors

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 $\mathfrak p$.

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 $0$?