A property of valuation rings

Posted: December 1, 2009 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

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.

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Comments
  1. Abdulrasool Azizi says:

    Let R be a totally ordered ring, r a non-unit element of R, and I=\bigcap_{n\in N}Rr^n.
    Then
    (i) If for some k ∈ N, r^k ∈ I, then r^k = 0.
    (ii) I is a prime ideal of R, or r is a nilpotent element of R.
    Proof. (i) Note that r^k ∈ I ⊆ Rr^{k+1}. So for some c ∈ R we have, r^k = cr^k+1, that is,
    r^k(1 −cr) = 0. Since 1− cr is a unit, r^k = 0.
    (ii) Let ab ∈ I, where a, b ∈ R, a \not\in I and b \not\in I. Then a \not\in Rrn and b \not\in Rrm
    for some n,m ∈ N. Every two ideals of R are comparable, so Rrn ⊆ Ra and
    Rrm ⊆ Rb. Thus r^{n+m} ∈ Rr^{n+m }⊆ Rab ⊆ I. Consequently by part (i), r^{n+m }= 0.

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