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?

1$\begingroup$ Yes, a proof can be found (for instance) in Serre's Local Fields. $\endgroup$– Torsten EkedahlAug 25, 2010 at 4:01

$\begingroup$ @Torsten Ekedahl: could you tell me where can I find the proof in Serre's book? Thanks. $\endgroup$– TmobiusXAug 25, 2010 at 6:38

3$\begingroup$ 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). $\endgroup$– Torsten EkedahlAug 25, 2010 at 19:47
2 Answers
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
$\def\bbZ{\mathbb{Z}} \def\frm{\mathfrak{m}} \def\bbQ{\mathbb{Q}}$(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.