# 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 ImplicationsOfHavingDivisorTheory 2013-03-22 17:59:13 2013-03-22 17:59:13 pahio (2872) pahio (2872) 6 pahio (2872) Topic msc 11A51 msc 13A05 properties of rings having a divisor theory DivisorTheoryAndExponentValuations