ac.commutative algebra - Is a valuation domain PID when its maximal ideal is principal?

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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