# 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)