The Stacks Project


Tag 00PD

Chapter 10: Commutative Algebra > Section 10.118: Around Krull-Akizuki

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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 27633–27647 (see updates for more information).

    \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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 00PD

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?