3
$\begingroup$

Fix a number field $K$ with ring of integers $O_K$; moreover $\sigma:K\to \mathbb C$ is an embedding of fields. Let $X\to\operatorname{Spec} O_K$ be an arithmetic surface ($X$ is regular and projective amd geometrically irreducible) and let $D=\sum_Yv_Y(D)Y$ be a divisor in $X$.

I was wondering what is the object usually indicated in the literature with the symbol with $D_\sigma$ (see for example Moriwaki's book about Arakelov geometry at page 101, line -6).

Here my guess: Usually $X_\sigma$ is the Riemann surface defined as the base change $X\times_{\operatorname{Spec} O_K}^\sigma \operatorname{Spec} \mathbb C$ through the morpshism $\operatorname{Spec}(\sigma):\operatorname{Spec} \mathbb C\to\operatorname{Spec} O_K$.The pullback of $D$ through the projection $p: X_\sigma\to X$ is a well defined divisor, since $p$ is flat. Therefore we put $D_\sigma:=p^\ast D$.

Is this argument correct?

$\endgroup$
2
$\begingroup$

We have the following two point of views of divisors on a nonsingular variety $X$.

  1. Weil divisor: linear combination of prime divisors.
  2. Cartier divisor: an element of $H^0(X, \text{Rat}(X)^\times/\mathcal{O}_X^\times)$, that is, data $\{(U_i, f_i)\}_{i = 1, \ldots, n}$ with the following properties.
    • $\bigcup U_i = X$ and $f_i$ is a rational function on $U_i$.
    • There is a unit function $u_{ij}$ on $U_i \cap U_j$ such that $f_i = u_{ij} f_j$ on $U_i \cap U_j$.

The two point of views are equivalent. The second point of view is better for understanding your question. $D_{\sigma}$ is just the pullback of $D$ by the morphism $X_{\sigma} \to X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.