implications of having divisor theory

The existence of a divisor theory restricts strongly the of an integral domain, as is seen from the following propositions.

Proposition 1. An integral domain $\mathcal{O}$ which has a divisor theory $\mathcal{O}^{*}\to\mathfrak{D}$ , is integrally closed in its quotient field.

Proof. Let $\xi$ be an element of the quotient field of $\mathcal{O}$ which is integral over $\mathcal{O}$ . Then $\xi$ satisfies an equation

\displaystyle\xi^{n}+\alpha_{1}\xi^{n-1}+\ldots+\alpha_{n}=0

(1)

where $\alpha_{1},\,\ldots,\,\alpha_{n}\in\mathcal{O}$ . Now, we can write $\displaystyle\xi=\frac{\varkappa}{\lambda}$ with $\varkappa,\,\lambda\in\mathcal{O}$ , whence (1) may be written

\displaystyle\varkappa^{n}=-\alpha_{1}\lambda\varkappa^{n-1}-\alpha_{2}\lambda% ^{2}\varkappa^{n-2}-\ldots-\alpha_{n}\lambda^{n}.

(2)

Let us make the antithesis that $\xi$ does not belong to $\mathcal{O}$ itself. Then $\lambda\nmid\varkappa$ and therefore we have for the corresponding principal divisors $(\lambda)\nmid(\varkappa)$ . We infer that there is a prime divisor factor $\mathfrak{p}$ of $(\lambda)$ and an integer $k\geqq 0$ such that

\mathfrak{p}^{k}\mid(\varkappa),\quad\mathfrak{p}^{k+1}\nmid(\varkappa),\quad% \mathfrak{p}^{k+1}\mid(\lambda).

By the condition 2 of the definition of divisor theory (http://planetmath.org/DivisorTheory), the right hand side of the equation (2) is divisible by

\mathfrak{p}^{(k+1)+(n-1)k}=\mathfrak{p}^{kn+1}.

On the other side, the highest power of $\mathfrak{p}$ , by which the divisor $(\varkappa^{n})$ is divisible, is $\mathfrak{p}^{kn}$ . Accordingly, the different sides of (2) show different divisibility by powers of $\mathfrak{p}$ . This contradictory situation means that the antithesis was wrong and thus the proposition has been proven.

Proposition 2. When an integral domain $\mathcal{O}$ has a divisor theory $\mathcal{O}^{*}\to\mathfrak{D}$ , then each element of $\mathcal{O}^{*}$ has only a finite number of non-associated (http://planetmath.org/Associates) factors (http://planetmath.org/DivisibilityInRings).

Proof. Let $\xi$ be an arbitrary non-zero element of $\mathcal{O}$ . We form the prime factor presentation of the corresponding principal divisor $(\xi)$ :

(\xi)=\mathfrak{p}_{1}\mathfrak{p}_{2}\cdots\mathfrak{p}_{r}

This is unique up to the ordering of the factors; $r\geqq 0$ . Then we form of the prime divisors $\mathfrak{p}_{i}$ all products having $k$ factors ( $0\leqq k\leqq r$ ) and choose from the products those which are principal divisors. Thus we obtain a set of factors of $(\xi)$ containing at most $\displaystyle r\choose k$ elements. All different principal divisor factors of $(\xi)$ are gotten, as $k$ runs all integers from 0 to $r$ , and their number is at most equal to

\sum_{k=0}^{r}{r\choose k}=2^{r}

(see 5. in the binomial coefficients). To every principal divisor , there corresponds a class (http://planetmath.org/EquivalenceClass) of associate factors of $\xi$ , and the elements of distinct classes (http://planetmath.org/EquivalenceClass) are non-associates. Since $\xi$ has not other factors, the number of its non-associated factors is at most $2^{r}$ .

Title	implications of having divisor theory
Canonical name	ImplicationsOfHavingDivisorTheory
Date of creation	2013-03-22 17:59:13
Last modified on	2013-03-22 17:59:13
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Topic
Classification	msc 11A51
Classification	msc 13A05
Synonym	properties of rings having a divisor theory
Related topic	DivisorTheoryAndExponentValuations