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# implications of having divisor theory

The existence of a divisor theory restricts strongly the type 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, 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 factors.

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 factor, there corresponds a class of associate factors of $\xi$, and the elements of distinct classes are non-associates. Since $\xi$ has not other factors, the number of its non-associated factors is at most $2^{r}$.

## Mathematics Subject Classification

11A51*no label found*13A05

*no label found*

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