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Homedivisor theory and exponent valuations
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divisor theory and exponent valuations
A divisor theory $\mathcal{O}^{*}\to\mathfrak{D}$ of an integral domain $\mathcal{O}$ determines via its prime divisors a certain set $N$ of exponent valuations on the quotient field of $\mathcal{O}$. Assume to be known this set of exponents $\nu_{{\mathfrak{p}}}$ corresponding the prime divisors $\mathfrak{p}$. 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 structure of the free monoid $\mathfrak{D}$ of all divisors in question. The homomorphism $\mathcal{O}^{*}\to\mathfrak{D}$ is then defined by the condition
$\displaystyle\alpha\;\mapsto\;\prod_{i}\mathfrak{p_{i}}^{{\nu_{{\mathfrak{p}_{% i}}}(\alpha)}}=(\alpha),Â´$  (1) 
since for any element $\alpha$ of $\mathcal{O}^{*}$ there exists only a finite number of exponents $\nu_{{\mathfrak{p}_{i}}}$ which do not vanish on $\alpha$ (corresponding the different prime divisor factors of the principal divisor $(\alpha)$).
One can take the concept of exponent as foundation for divisor theory:
Theorem. Let $\mathcal{O}$ be an integral domain with quotient field $K$ and $N$ a given set of exponents of $K$. The exponents in $N$ determine, as in (1), a divisor theory of $\mathcal{O}$ iff the following three conditions are in force:

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

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

For any finite set $\nu_{1},\,\ldots,\,\nu_{n}$ of distinct exponents in $N$ and for the arbitrary set $k_{1},\,\ldots,\,k_{n}$ of nonnegative integers, there exists an element $\alpha$ of $\mathcal{O}$ such that
$\nu_{1}(\alpha)=k_{1},\,\;\ldots,\,\;\nu_{n}(\alpha)=k_{n}.$
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 $\mathfrak{D}$ that allows unique prime factorisation and that may be mapped bijectively onto $N$. The exponent in $N$ which corresponds to arbitrary prime element $\mathfrak{p}$ is denoted by $\nu_{\mathfrak{p}}$. Then we obtain the homomorphism
$\alpha\mapsto\prod_{\nu}\mathfrak{p}^{{\nu_{\mathfrak{p}}(\alpha)}}:=(\alpha)$ 
which can be seen to satisfy all required properties for a divisor theory $\mathcal{O}^{*}\to\mathfrak{D}$.
References
 1 S. Borewicz & I. Safarevic: Zahlentheorie. Birkhäuser Verlag. Basel und Stuttgart (1966).
Mathematics Subject Classification
13A18 no label found12J20 no label found13A05 no label found Forums
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