What are the divisors of a function field?

Question

Sorry for this basic question.... I can't find the definition of divisors of a function field ? And how do they define a group? In a paper by Robinson and Roquette , is written : let $p$ be a non-archimedean prime divisor of $K$.among the valuations belonging to $p$ there is exactly one which is normalized such that its value group is $Z$. Also a few pages later it is said if $F$ is a function fiel of one variable over $K$ , then the equivalence classes of non-trivial valuations on $F$ which are trivial over $K$ is a free $Z$- module generated by the functional primes.

Answer 1

A prime divisor of a function field $F|K$ is a valuation $v$ of $F$ trivial on $K$ such that for the residue field $\overline{F}$ the equation

$\mathrm{trdeg}(F|K)=\mathrm{trdeg}(\overline{F}|K)-1$

holds -- see for example Zariski-Samuel, Volume 2.

For a function field of transcendence degree $1$ the prime divisors are discrete valuations, that is their value groups are order-isomorphic to $\mathbb{Z}$.

The divisor group is the free abelian group generated by the prime divisors.

For a number field a prime divisor is either one of the p-adic valuations or one of the extensions of the absolute value of $\mathbb{Q}$. The first type is more specifically called non-archimedian prime divisor.

This terminology was introduced to emphasize the similarity between the arithmetic theory of number fields on one side and the theory of function fields on the other. Arithmetic here is understood to be in contrast to geometry.