Discrete Valuation Rings

Definition: A Discrete Valuation Ring (DVR) is an integral domain $R$ with the following properties: 1) $R$ is a Noetherian ring. 2) $R$ is a local ring. 3) The unique maximal ideal of $R$ is a principal ideal.

Recall that an ideal $I$ of a ring $R$ is a principal ideal if it is generated by a single element.

Theorem 1: Let $R$ be a discrete valuation ring. Then there exists an irreducible element $t \in R$ such that every $z \in R \setminus \{ 0 \}$ can be written in the form $z = ut^n$ where $u \in R$ is a unit and $n \geq 0$.

• Proof: Let $R$ be a discrete valuation ring. Then $R$ is a local ring. Let $m$ be the unique maximal ideal of $R$. Then $m$ is also a principal ideal, so $m = (t)$ for some generator $t \in R$.

• Now let $z \in R \setminus \{ 0 \}$. If $z$ is a unit, then we may write $z = zt^0$. Suppose that $z$ is NOT a unit. Then $z = z_1t$ for some $z_1 \in R$. If $z_1$ is a unit we are done. Otherwise, $z_1 = z_2t$ for some $z_2 \in R$. We continue this process. If at step $n$, $z_n$ is a unit then $z = z_nt^n$ and we are done. If this process never terminates, then we obtain an infinite chain of ideals:

(1)

\begin{align} \quad (z_1) \subseteq (z_2) \subseteq ... \subseteq (z_n) \subseteq ... \end{align}

• But $R$ is a Noetherian ring, so this chain of ideals must terminate, so $(z_n) = (z_{n+1})$ for some $n \in \mathbb{N}$. So then there exists a $v \in R$ such that $z_{n+1} = vz_n$. But $z_n = z_{n+1}t$, so:

(2)

\begin{align} \quad z_{n+1} = vtz_{n+1} \end{align}

• So $vt = 1$ which implies that $t$ is a unit. But this is a contradiction since $m = (t)$ is a maximal ideal and hence cannot contain any units. So the assumption that this process does not terminate is false. So every element $z \in R \setminus \{ 0 \}$ is of the form $z = ut^n$ where $u$ is a unit and $n \geq 0$.

• We lastly show that this representation is unique. Let $z = ut^n$ and $z = vt^m$ where $u, v \in R$ are units and $n, m \geq 0$. Without loss of generality, assume that $n \geq m$. Then:

(3)

\begin{align} \quad ut^n = vt^m \\ \quad ut^{n-m} = v \end{align}

• But $t$ is an irreducible element so $n - m = 0$, i.e., $n = m$, and so $u = v$. $\blacksquare$

Definition: Let $R$ be a discrete valuation ring so that (by Theorem 1) every element $z \in R \setminus \{ 0 \}$ can be uniquely written in the form $z = ut^n$ where $u \in R$ is a unit and $n \geq 0$. The element $t \in R$ is called a Uniformizing Parameter for $R$.