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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?

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Yes, a proof can be found (for instance) in Serre's Local Fields. –  Torsten Ekedahl Aug 25 '10 at 4:01
    
@Torsten Ekedahl: could you tell me where can I find the proof in Serre's book? Thanks. –  TmobiusX Aug 25 '10 at 6:38
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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). –  Torsten Ekedahl Aug 25 '10 at 19:47
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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://math.uga.edu/~pete/integral.pdf

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