9 DVR & Dedekind Domain

Definition

Let $k$ be a field. A discrete valuation on $k$ is a map $\nu :k^{\ast }\to \mathbb{Z}$ such that

• $\nu (xy)=\nu (x)+\nu (y)$;
• $\nu (x+y)⩾\min \{\nu (x),\nu (y)\}$ (non-Archimedean valuation)

(Sometimes define $\nu (0)=+\infty$.)

Definition

Let $R=\{x\in k:\nu (x)⩾0\}$ which contains $0$, then $R$ is a ring, which is called the valuation ring of $\nu$. It is a local ring with $m_R=\{x\in R:r(x)>0\}$.

\begin{proof} It is trivial to show $R$ is a ring. To see it is local with the unique maximal ideal $m_R$, it suffices to check $x\in R\setminus m_R$ is a unit. But in this case $\nu (x)=0$ iff $\nu (1/x)=\nu (1)-\nu (x)=0$ and so $1/x\in R$. \end{proof}

Remark. Given a valuation as above, one can check $||x||={ϵ}^{\nu (x)}$ with $0<ϵ<1$ a fixed element defines a norm in $k$. That is to say, $d(x,y)=||x-y||={ϵ}^{\nu (x-y)}$ is a distance. To show it, it suffices to show it satisfies the triangle inequality.

“在这个世界里，所有的三角形都是等腰三角形。”

For a triangle $\Delta ABC$, we have

$$d(a-c)=d((a-b)+(b-c))={ϵ}^{\nu (a-b+b-c)}={ϵ}^{\nu (a-b)}=d(a-b)$$

if $\nu (a-b)>\nu (b-c)$. See here.

Examples.

• $p$-adic valuation on $\mathbb{Q}$.

  • Fix a prime $p$, let $x=a/b\in \mathbb{Q}$ with $(a,b)=1$. Define

    $${\nu }_p(x)=\{\begin{array}{ll}i,& \text{if }a=p^i\cdot a^{\prime },\\ -j,& \text{if }b=p^j\cdot b^{\prime }.\end{array}$$

  • To see it is a valuation, it suffices to check ${\nu }_p(a/b+c/d)⩾\min \{{\nu }_p(a/b),{\nu }_p(c/d)\}$.

    • Suppose $p\nmid b$ and $p\nmid d$, then

      $${\nu }_p(a/b+c/d)={\nu }_p((ad+bc)/bd)={\nu }_p(ad+bc)⩾\min \{{\nu }_p(a),{\nu }_p(c)\}.$$

    • For general case, multiply $p^n$ to $a/b$ and $c/d$ for a large enough $n$. Then we get back to the first case.

  • For ${\nu }_p$ on $\mathbb{Q}$, the valuation ring is $\frac{\mathbb{Z}}{\mathbb{Z}\setminus p\mathbb{Z}}={\mathbb{Z}}_{(p)}$, whose unique maximal ideal is $p{\mathbb{Z}}_{(p)}$ and residue field is ${\mathbb{Z}}_{(p)}/p{\mathbb{Z}}_{(p)}\simeq {\mathbb{F}}_p$.

• $k=k(x)=\mathrm{Frac}(k[x])$ is a field.
  • Take an irreducible polynomial $r(x)\in k[x]$. Then we define $\nu (f(x)/g(x))=\nu (f(x))-\nu (g(x))$. For $f(x)=r(x{)}^a{\prod }_{i=1}^n{r}_i(x)$, let $\nu (f(x))=a$.
  • It is valuation, similar to ${\nu }_p$ on $\mathbb{Q}$.

Remark. Valuation ring and ED are different, as a norm function $N$ does not satisfies $N(a+b)⩾\min \{N(a),N(b)\}$.

Definition

We say an integral is a DVR (discrete valuation ring) if there exists a discrete valuation $\mathrm{Frac}A\stackrel{\nu }{\twoheadrightarrow }\mathbb{Z}$ such that $A$ is a valuation ring.

Fact. DVR is a PID. It is also proved in ^bb0319.

• Since $\mathrm{Frac}A\stackrel{\nu }{\twoheadrightarrow }\mathbb{Z}$ is surjective, there exists $x\in \mathrm{Frac}A$ such that $\nu (x)=1$ (such $x$ is called a uniformizer).
  • Claim 1: the unique maximal ideal is $m_k=(x)$.
    • Indeed, given any $y\in m_k$, notice that $y\in m_k$ iff $\nu (y)⩾1$. Then $\nu (y/x)⩾1-1⩾0$ and $y/x\in A$. Thus $y\in (x)$ and $m_k\subseteq (x)$ yield $m_k=(x)$.
  • Claim 2: given any ideal $I\subseteq A$, $I=(x^k)$ for some $k⩾0$.
    • Let $k=\min \{\nu (z):z\in I\}$. There exists $y\in I$ such that $\nu (y)=k$. Then $\nu (y/x^k)=k-k=0$ and $I\supseteq (y)=(x^k)$.
    • For any $z\in I$, we have $\nu (z)⩾k$ and $\nu (z/x^k)⩾0$. Thus $z\in (x^k)$ for all $z\in I$ and $I\subseteq (x^k)$.
    • Now we prove that $I=(x^k)$.

Proposition

Let $A$ be a Noetherian local domain with dimension $1$. Let $m$ be the maximal ideal of $A$, and let $k=A/m$. TFAE:

• $A$ is a DVR;
• $A$ is integral closed;
• $m$ is a principal ideal;
• ${\dim }_k(m/m^2)=1$;
• any nonzero ideal of $R$ is a power of $m$;
• there exists $x\in A$ such that any non-zero ideal is of form $(x^k)$ with $k⩾0$.

\begin{proof} We starts with $2$ facts on $A$.

• Fact A: for ideal $\alpha$ such that $0⊊\alpha ⊊A$, then $\alpha$ is $m$-primary and $\alpha \supseteq m^n$ for some $n$.
  • proof. Since $\dim A=1$ and $A$ is a domain, we know the only primes of $A$ are $(0),m$. Hence $r(\alpha )={\cap }_{p\supseteq \alpha }p=m$. By ^s2eg5f $\alpha$ is $m$-primary and $\alpha \supseteq m^n$.
• Fact B: $m^n\ne m^{n+1}$ for all $n⩾0$.
  • proof. Otherwise $A$ is Artinian local and so $\dim A=0$ by ^myy86x, which is impossible.

i)→ii) See here.

ii)→iii) Choose any nonzero $a\in m$. By Fact A, there exists $n$ such that $m^n\subseteq (a)$ and $m^{n-1}⊈(a)$. Pick some $b\in m^{n-1}$ such that $b\notin (a)$. Let $x=a/b\in K=\mathrm{Frac}(A)$. We aim to show $x$ is a “uniformizer”, that is, $m=(x)$. Note $x^{-1}=b/a\notin A$, then $x^{-1}$ is not integral over $A$ by $A$ integral closed. Claim $x^{-1}m=A$ as subsets of $K$. (Remark that we don’t know whether $x\in A$.) If $x^{-1}m=A$, then $m=(x)$ and iii) holds.

Now we prove the claim. Notice that $x^{-1}m=(b/a)m\subseteq A$, because for any $y\in m$, $by\in m^n\subseteq (a)$. Also note that $x^{-1}m$ is an ideal of $A$. So it remains to prove $x^{-1}m⊈m$. Suppose otherwise $x^{-1}m\subseteq m$, then this defines an $A[x^{-1}]$-module structure on $m$. This is a faithful $A[x^{-1}]$-module, because all multiplications happen inside $K$. By ^jy6o7s, $x^{-1}$ is integral over $A$, leading to a contradiction. Now we proved the claim.

iii)→iv) Assume $m=(x)$, then $m^2=(x^2)$. Then $m/m^2\simeq k\overline{x}$ as a $1$-dimensional vector space.

iv)→v) For $\alpha \subseteq A$, by Fact A $\alpha \supseteq m^n$ for some $n$. Consider $A/m^n$, which is a Noetherian local ring with dimension $0$. Hence $A/m^n$ is Artinian. By ^f166vm $\overline{\alpha }$ is principal and so $\overline{\alpha }={\overline{m}}^k$. Hence $\alpha =m^{k+n}$.

v)→vi) By Fact B, $m⊋m^2$, then there exists $x\in m\setminus m^2$. Since $(x)=m^k$ for some $k$, we have $k=1$ and $(x)=m$.

vi)→i) By vi), $m=(x)$. By Fact B, $(x^k)⊋(x^{k+1})$. Now for any $a\in A$, $(a)=(x^{\alpha })$ with a unique $\alpha$. Now define $\nu (a)=\alpha$. It is easy to check $\nu$ is a valuation. \end{proof}