## A property of valuation rings

Posted: December 1, 2009 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Definition. A commutative domain $D$ is called a valuation ring if its ideals are totally ordered. That means for any two ideals of $I,J$ of $D$ either $I \subseteq J$ or $J \subseteq I.$

It’s easy to see that every valuation ring is local, i.e. it has only one maximal ideal. What I would like to show here is one of several nice properties of valuation rings: the intersection of all powers of a proper ideal of a valuation ring is a prime ideal!

Problem. Let $I$ be a proper ideal of a valuation ring $D.$ Prove that $J=\bigcap_{n=1}^{\infty}I^n$ is a prime ideal.

Solution. Suppose $x,y \in D$ with $x \notin J, \ y \notin J.$ I’ll show that $xy \notin J.$ Let $m, n \geq 1$ be such that $x \notin I^m$ and $y \notin I^n.$ So $I^m \subseteq \langle x \rangle$ and $I^n \subseteq \langle y \rangle,$ because the ideals of $D$ are totally ordered. Obviously

$I^m \langle y \rangle \subseteq \langle x \rangle \langle y \rangle = \langle xy \rangle.$

Now, suppose that $I^m \langle y \rangle = \langle xy \rangle.$ Then $ry=xy,$ for some $r \in I^m$ and hence $x=r \in I^m,$ which is impossible. Therefore $I^m \langle y \rangle \subset \langle xy \rangle,$ which gives us

$I^{m+n} \subseteq I^m \langle y \rangle \subset \langle xy \rangle.$

Hence $xy \notin I^{m+n}$ and therefore $xy \notin J. \ \Box$

Remark. Note that if $P$ is any prime ideal of $D$ which is properly contained in $I,$ then $P \subseteq J.$ This is because if $I^k \subseteq P,$ for some integer $k \geq 1,$ then $I \subseteq P,$ because $P$ is prime. But that would be impossible because we’re given that $P$ is “properly” contained in $I.$