Discrete valuation ring

Remarks

Equivalently, $R$ is an integral domain with field of fractions $K$ such that there is a surjective discrete valuation $v \colon K^{\times} \to \Z$ with $R = \set{x \in K^{\times} : v(x) \ge 0} \cup \set{0}$. A generator $\pi$ of the maximal ideal is a uniformizer; every nonzero $x \in K$ is $x = u \pi^{v(x)}$ with $u$ a unit of $R$. The nonzero ideals of $R$ are exactly $(\pi^n)$ for $n \ge 0$.

Example: for a prime $p$, the ring $\Z_{(p)} = \set{a/b : a, b \in \Z,\ p \nmid b}$ is a discrete valuation ring. Every nonzero element is $p^n u$ with $u$ a unit and $n \ge 0$, so every nonzero ideal is $(p^n)$, which is principal. The only nonzero prime among these is $(p)$. The ring $k[\![t]\!]$ of formal power series over a field $k$ and the ring $\Z_p$ of $p$-adic integers are other examples.

Non-example: $\Z$ is a principal ideal domain, but it has infinitely many nonzero prime ideals $(2), (3), (5), \ldots$, so it is not a discrete valuation ring. A field is not a discrete valuation ring either: it has no nonzero prime ideal.