divisor theory and exponent valuations

A divisor theory  ${𝒪}^{*}\to 𝔇$  of an integral domain $𝒪$ determines via its prime divisors a certain set $N$ of exponent valuations on the quotient field of $𝒪$.  Assume to be known this set of exponents (http://planetmath.org/ExponentValuation2) ${\nu }_{𝔭}$ corresponding the prime divisors $𝔭$.  There is a bijective correspondence between the elements of $N$ and of the set of all prime divisors.  The set of the prime divisors determines completely the of the free monoid $𝔇$ of all divisors in question. The homomorphism  ${𝒪}^{*}\to 𝔇$  is then defined by the condition

$\alpha \mapsto {\displaystyle \prod _i}𝔭_{𝔦}{}^{{\nu }_{{𝔭}_i}(\alpha )}=(\alpha ),Ã\mathrm{‚}Â\mathrm{´}$(1)

since for any element $\alpha$ of ${𝒪}^{*}$ there exists only a finite number of exponents ${\nu }_{{𝔭}_i}$ which do not vanish on $\alpha$ (corresponding the different prime divisor factors (http://planetmath.org/DivisibilityInRings) of the principal divisor $(\alpha )$).

One can take the concept of exponent as foundation for divisor theory:

Theorem.  Let $𝒪$ be an integral domain with quotient field $K$ and $N$ a given set of exponents (http://planetmath.org/ExponentValuation2) of $K$.  The exponents in $N$ determine, as in (1), a divisor theory of $𝒪$ iff the following three conditions are in :

• For every  $\alpha \in 𝒪$  there is at most a finite number of exponents  $\nu \in N$  such that  $\nu (\alpha )\ne 0$.

• An element  $\alpha \in K$  belongs to $𝒪$ if and only if  $\nu (\alpha )\geqq 0$  for each $\nu \in N$.

• For any finite set  ${\nu }_1,\mathrm{\dots },{\nu }_n$  of distinct exponents in $N$ and for the arbitrary set  $k_1,\mathrm{\dots },k_n$ of non-negative integers, there exists an element $\alpha$ of $𝒪$ such that

  $${\nu }_1(\alpha )=k_1,\mathrm{\dots },{\nu }_n(\alpha )=k_n.$$

For the proof of the theorem, we mention only how to construct the divisors when we have the exponent set $N$ fulfilling the three conditions of the theorem.  We choose a commutative monoid $𝔇$ that allows unique prime factorisation and that may be mapped bijectively onto $N$.  The exponent in $N$ which corresponds to arbitrary prime element $𝔭$ is denoted by ${\nu }_{𝔭}$.  Then we obtain the homomorphism

$$\alpha \mapsto \prod _{\nu }{𝔭}^{{\nu }_{𝔭}(\alpha )}:=(\alpha )$$

which can be seen to satisfy all required properties for a divisor theory  ${𝒪}^{*}\to 𝔇$.