Discrete Valuation Rings – The Math


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 $R$.



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