# The Stacks Project

## Tag 00PD

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

1. The ring $A$ is a discrete valuation ring.
2. The ring $A$ is a valuation ring and Noetherian.
3. The ring $A$ is a regular local ring of dimension $1$.
4. The ring $A$ is a Noetherian local domain with maximal ideal $\mathfrak m$ generated by a single nonzero element.
5. 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 $A' \subset f.f.(A)$. 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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\begin{lemma}
\label{lemma-characterize-dvr}
Let $A$ be a ring. The following are equivalent.
\begin{enumerate}
\item The ring $A$ is a discrete valuation ring.
\item The ring $A$ is a valuation ring and Noetherian.
\item The ring $A$ is a regular local ring of dimension $1$.
\item The ring $A$ is a Noetherian local domain with maximal ideal
$\mathfrak m$ generated by a single nonzero element.
\item The ring $A$ is a Noetherian local normal domain of dimension $1$.
\end{enumerate}
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.
\end{lemma}

\begin{proof}
The equivalence of (1) and (2) is
Lemma \ref{lemma-valuation-ring-Noetherian-discrete}.
Moreover, in the proof of Lemma \ref{lemma-valuation-ring-Noetherian-discrete}
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 \ref{lemma-regular-domain}). Using this the equivalence of (3) and (4)
follows from dimension theory, see Section \ref{section-dimension}.

\medskip\noindent
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 \ref{lemma-valuation-valuation-ring}.

\medskip\noindent
A valuation ring is a normal domain by Lemma \ref{lemma-valuation-ring-normal}.
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 \ref{lemma-nonregular-dimension-one} 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 $A' \subset f.f.(A)$.
Since $A \to A'$ is finite it is integral (see
Lemma \ref{lemma-finite-is-integral}).
Since $A$ is normal we get $A = A'$ a contradiction.
\end{proof}

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