Return to Answer

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

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

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 13 of

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

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

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 with 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 13 of

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

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 13 of

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