**Definition**. A commutative domain is called a **valuation ring** if its ideals are totally ordered. That means for any two ideals of of either or

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 be a proper ideal of a valuation ring Prove that is a prime ideal.

**Solution.** Suppose with I’ll show that Let be such that and So and because the ideals of are totally ordered. Obviously

Now, suppose that Then for some and hence which is impossible. Therefore which gives us

Hence and therefore

**Remark.** Note that if is any prime ideal of which is properly contained in then This is because if for some integer then because is prime. But that would be impossible because we’re given that is “properly” contained in

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.