# Is a valuation domain PID when its maximal ideal is principal?

It is known that a valuation domain is a principal ideal ring if and only if its prime ideals are principal. Is it a principal ideal ring when its unique maximal ideal is principal?

• Yes, a proof can be found (for instance) in Serre's Local Fields. Aug 25, 2010 at 4:01
• @Torsten Ekedahl: could you tell me where can I find the proof in Serre's book? Thanks. Aug 25, 2010 at 6:38
• As Pete's example shows I forgot that the ring needs to be Noetherian. In any case if you can live with that restriction the result in Serre I:2.2. It is also true if the valuation is of rank 1 (which was I think was what really confused me). Aug 25, 2010 at 19:47

I assume that by a valuation domain you mean an integral domain $$R$$ with fraction field $$K$$ such that: for all $$x \in K^{\times}$$, at least one of $$x,x^{-1}$$ lies in $$R$$.

In this case, I believe the answer is no. Let $$R$$ be any valuation domain whose value group $$K^{\times}/R^{\times}$$ is isomorphic, as a totally ordered abelian group, to $$\mathbb{Z} \times \mathbb{Z}$$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $$(0,0)$$, but the valuation of any such element is at least $$(0,1)$$ and therefore any element of valuation $$(0,1)$$ gives a generator of the maximal ideal.

For some information on valuation rings, see e.g. Section 17 of

http://alpha.math.uga.edu/~pete/integral.pdf

$$\def\bbZ{\mathbb{Z}} \def\frm{\mathfrak{m}} \def\bbQ{\mathbb{Q}}$$(For me, a DVR is never a field.)

As a complement to Pete L. Clark's answer, here's a recipe to construct a valuation ring with $$\bbZ\times\bbZ$$ as value group (the following is taken from Matsumura's Commutative Ring Theory, Remark after Theorem 11.1): Let $$K$$ be a field, and suppose that $$A$$ is a DVR of $$K$$. Suppose $$\mathcal{R}$$ is a DVR of $$k=A/\frm_A$$, and let $$R$$ be the inverse image of $$\mathcal{R}$$ along $$\varphi:A\to k$$, which is a valuation ring of $$K$$. Let $$v_A$$ (resp., $$v_\mathcal{R}$$) be a valuation on $$K$$ (resp., on $$k$$) with ring $$A$$ (resp., $$R$$). Let $$f$$ be a uniformizer for $$A$$. Then \begin{align*} v:K^*&\to\bbZ\times\bbZ\\ x&\mapsto(n,m),\quad n=v_A(x),\;m=v_\mathcal{R}(\varphi(xf^{-n})), \end{align*} is an onto valuation on $$K$$ with ring $$A$$. (Hint: in Matsumura's book it is argued that if $$\overline{g}\in \mathcal{R}$$ is a uniformizer, then $$g$$ is a uniformizer for $$R$$ and $$g\in A\setminus\frm_A$$. It follows that $$v$$ is onto since $$x=f^ng^m$$ satisfies $$v(x)=(n,m)$$, and leveraging $$f$$ and $$g$$—plus writing elements of the quotient field of a DVR as a product of a unit of the DVR times a power of some uniformizer—it is an exercise to show that $$v$$ is a valuation.)

For an explicit example, consider $$K=\bbQ((x))$$, $$A=\bbQ[[x]]$$ and $$\mathcal{R}=\bbZ_{(p)}$$, where $$p\in\bbZ$$ is a prime.