If is an integral domain then it is a PID iff it has a Dedekind-Hasse valuation, that is, a function such that for any either

• 

  or

• 

Proof: First, let be a Dedekind-Hasse valuation and let be an ideal of an integral domain . Take some with minimal (this exists because the integers are well-ordered) and some such that . must contain both and , and since it is closed under addition, for any .

Since is minimal, the second possibility above is ruled out, so it follows that . But this holds for any , so , and therefore every ideal is princple.

For the converse, let be a PID. Then define for any unit. Any non-zero, non-unit can be factored into a finite product of irreducibles (since every PID is a UFD), and every such factorization of is of the same length, . So for , a non-zero non-unit, let . Obviously .

Then take any and suppose . Then take the ideal of elements of the form . Since this is a PID, it is a principal ideal for some , and since , there is some non-unit such that . Then . But since is not a unit, the factorization of must be longer than the factorization of , so , so is a Dedekind-Hasse valuation.