Lemma 10.118.7. Let $A$ be a ring. The following are equivalent.

The ring $A$ is a discrete valuation ring.

The ring $A$ is a valuation ring and Noetherian.

The ring $A$ is a regular local ring of dimension $1$.

The ring $A$ is a Noetherian local domain with maximal ideal $\mathfrak m$ generated by a single nonzero element.

The ring $A$ is a Noetherian local normal domain of dimension $1$.

In this case if $\pi $ is a generator of the maximal ideal of $A$, then every element of $A$ can be uniquely written as $u\pi ^ n$, where $u \in A$ is a unit.

**Proof.**
The equivalence of (1) and (2) is Lemma 10.49.18. Moreover, in the proof of Lemma 10.49.18 we saw that if $A$ is a discrete valuation ring, then $A$ is a PID, hence (3). Note that a regular local ring is a domain (see Lemma 10.105.2). Using this the equivalence of (3) and (4) follows from dimension theory, see Section 10.59.

Assume (3) and let $\pi $ be a generator of the maximal ideal $\mathfrak m$. For all $n \geq 0$ we have $\dim _{A/\mathfrak m} \mathfrak m^ n/\mathfrak m^{n + 1} = 1$ because it is generated by $\pi ^ n$ (and it cannot be zero). In particular $\mathfrak m^ n = (\pi ^ n)$ and the graded ring $\bigoplus \mathfrak m^ n/\mathfrak m^{n + 1}$ is isomorphic to the polynomial ring $A/\mathfrak m[T]$. For $x \in A \setminus \{ 0\} $ define $v(x) = \max \{ n \mid x \in \mathfrak m^ n\} $. In other words $x = u \pi ^{v(x)}$ with $u \in A^*$. By the remarks above we have $v(xy) = v(x) + v(y)$ for all $x, y \in A \setminus \{ 0\} $. We extend this to the field of fractions $K$ of $A$ by setting $v(a/b) = v(a) - v(b)$ (well defined by multiplicativity shown above). Then it is clear that $A$ is the set of elements of $K$ which have valuation $\geq 0$. Hence we see that $A$ is a valuation ring by Lemma 10.49.16.

A valuation ring is a normal domain by Lemma 10.49.10. Hence we see that the equivalent conditions (1) – (3) imply (5). Assume (5). Suppose that $\mathfrak m$ cannot be generated by $1$ element to get a contradiction. Then Lemma 10.118.3 implies there is a finite ring map $A \to A'$ which is an isomorphism after inverting any nonzero element of $\mathfrak m$ but not an isomorphism. In particular we may identify $A'$ with a subset of the fraction field of $A$. Since $A \to A'$ is finite it is integral (see Lemma 10.35.3). Since $A$ is normal we get $A = A'$ a contradiction.
$\square$

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