PIDAndUFDAreEquivalentInADedekindRing 2008-03-09 16:12:51 2008-04-30 23:39:18 Theorem PID % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here This article shows that if $A$ is a Dedekind domain, then $A$ is a UFD if and only if it is a PID. Note that this result implies the more specific result given in the article unique factorization and ideals in ring of integers. Since any PID is a UFD, we need only prove the other direction. So assume $A$ is a UFD, let $\mathfrak{p}$ be a nonzero (proper) prime ideal, and choose $0\neq x\in\mathfrak{p}$. Note that $x$ is a nonunit since $\mathfrak{p}$ is a proper ideal. Since $A$ is a UFD, we may write $x$ uniquely (up to units) as $x=p_1^{a_1}\cdots p_k^{a_k}$ where the $p_i$ are distinct irreducibles in $A$, the $a_i$ are positive integers, and $k>0$ since $x$ is not a unit. Since $\mathfrak{p}$ is prime and $x\in\mathfrak{p}$, it follows that some $p_i$, say $p_1$, is in $\mathfrak{p}$. Then $(p_1)\subset\mathfrak{p}$. But $(p_1)$ is prime since clearly in a UFD any ideal generated by an irreducible is prime. Since $A$ is Dedekind and thus has Krull dimension 1, it must be that $(p_1)=\mathfrak{p}$ and thus $\mathfrak{p}$ is principal.