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Let $X\rightarrow S$ be an arithmetic surface (this means, in the language of Qing Liu's book, $X$ is a regular integral scheme of dimension 2, projective and flat over a Dedekind scheme $S$ of dimension 1). Let $\eta$ be the generic point of $S$.

Let $\omega_{X/S}$ be the dualizing sheaf of $X/S$, which is invertible, hence there is a Cartier divisor $K_{X/S}$ such that $\omega_{X/S}\cong\mathcal{O}_X(K_{X/S})$.

At first, it seemed natural to me that since $\omega_{X/S}$ is a "relative object", $K_{X/S}$ should be linearly equivalent to a purely horizontal effective divisor - ie, is a nonnegative linear combination of horizontal prime divisors. However, this cannot always be the case, since if $X_\eta$ is genus $\ge 1$, then $K_{X/S}$ can be chosen to be effective, and if $X$ is not minimal (ie, contains an exceptional (vertical) divisor $E$), then $K_{X/S}\cdot E < 0$, which can only happen if $E$ is a component of $K_{X/S}$ (for any effective canonical divisor $K_{X/S}$).

Thus, the question is: If $X_\eta$ is say smooth of genus $\ge 1$ (or $\ge 2$ if $g = 1$ is special), and $X$ is minimal, then can we always find an effective canonical divisor $K_{X/S}$ which is a linear combination of horizontal prime divisors?

Is there any intuition for why this is/isn't/should/shouldn't be the case?

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  • $\begingroup$ For $X → S$ be a regular fibered arithmetic surface and $0 < E ≤ X_s$ we have $p_a(E) = 1 + 1/2(E^2 + K_{X/S} · E).$ called adjunction formula $\endgroup$
    – user21574
    Apr 28, 2017 at 23:50

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