Let $X$ be an arithemetic surface. Let $(f)$ be a principal divisor. Let $D_{i}$ be the horizontal divisors showed up in $(f)$ and $d_i$ are their coefficients. I am trying to prove $\sum d_{i} m_i=0$, where $m_i$ is the degree of the quotient field of the horizontal divisor over the base. But I do not know how to do it.
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1$\begingroup$ Look on the generic divisor. (In down-to-earth language: it comes down to "#zeros = #poles" on an algebraic curve.) $\endgroup$– Noam D. ElkiesCommented Jun 10, 2018 at 2:57
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$\begingroup$ (Which means that it's actually $\sum_i d_i m_i = 0$, where $m_i$ is the "intersection multiplicity of $D_i$ with the generic fiber"; in more concrete language, $X$ is a curve over some number field $K$, and each $D_i$ is a point defined over some degree $m_i$ extension of $K$.) $\endgroup$– Noam D. ElkiesCommented Jun 10, 2018 at 19:43
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1$\begingroup$ Thanks! I proved this fact in a strange way; I use base change to $\mathbb{C}$ by making use of embeddings $K\rightarrow \mathbb{C}$. Then $\sum d_{i}dm_{i}$ is actually the sum of zeros and poles of $f$ on $X\otimes_{\mathbb{C}}$. Since $f\in K(X)$ the sum must be zero. I never heard of the notation of "generic divisor". Thanks for the hint and the help. $\endgroup$– GuestCommented Jun 10, 2018 at 23:43
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