Valuation Rings and DVRs

Valuations on Fields

Firstly, we give an example. Let $p$ be a prime number. For any non-zero rational number $x\in \mathbb{Q}$, we can write $x=p^n(a/b)$ where $a$ and $b$ are integers not divisible by $p$.

• The exponential $p$-adic valuation ${\nu }_p$ on $\mathbb{Q}$ is the function $v_p:\mathbb{Q}\to \mathbb{Z}\cup \{\infty \}$ defined as ${\nu }_p(x)=n$, and we also define ${\nu }_p(0)=\infty$
• The multiplicative $p$-adic valuation $|\cdot {|}_p$ on $\mathbb{Q}$ is defined as $|x{|}_p=c^{{\nu }_p(x)}$, and we also define $|0{|}_p=0$.

Exponential Valuations

Definition

Let $k$ be a field. An (exponential) valuation of $k$ is a real-valued function $\nu$ defined on $k^{\ast }=k\setminus \{0\}$ satisfying

• $\nu (ab)=\nu (a)+\nu (b)$ and
• $\nu (a+b)⩾\min \{\nu (a),\nu (b)\}$.

Note that $\nu :K^{\times }\to (\mathbb{R},+)$ is a group homomorphism. The image $\nu (k^{\times })$ is called the value group of $\nu$, and we set $\nu (0)=\infty$. We call $(k,\nu )$ valuation field.

Facts.

• $\nu (1)=\nu (-1)=0$, $\nu (a)=\nu (-a)$, $\nu (a^{-1})=-\nu (a)$
• $\nu (a)>\nu (b)$ yields $\nu (a+b)=\nu (b)$, see
  • Since $\nu (a+b)⩾\min \{\nu (a),\nu (b)\}=\nu (b)$, it suffices to show $\nu (a+b)>\nu (b)$ is impossible. Otherwise assume $\nu (a+b)>\nu (b)$ holds, then $\nu (b)=\nu (a+b+(-a))⩾\min \{\nu (a+b),\nu (a)\}>\nu (b)$, which is impossible. Hence $\nu (a+b)=\nu (b)$.

Definition

Let $(k,\nu )$ be a valuation field. Then $R=\{a\in k:\nu (a)⩾0\}$ is a subring of $k$, which is called valuation ring of $\nu$.

Let $R^{-1}=\{a^{-1}:a\in R\setminus \{0\}\}=\{a\in k:\nu (a)<0\}$. Then $k=R\cup R^{-1}$ and so $k$ is a fractional field of $R$.

The set $P=\{a\in l:\nu (a)>0\}$ is an ideal of $R$, which is called the valuation ideal of $\nu$.

The group of units of $R$ is $\{a\in k:\nu (a)=0\}$.

Note that $R$ is a local ring with $P=J(R)=\{\text{nonunits of }R\}$ by ^ysw57c, and $R/P$ is called residue field of $\nu$.

Definition

Let $\lambda \in R$ with $\lambda >0$. Define ${\nu }^{\prime }(a)=\lambda \nu (a)$ for all $a\in k^{\times }$. Then $\nu$ is also a valuation. We say $\nu$ is equivalent to ${\nu }^{\prime }$, and write $\nu \sim {\nu }^{\prime }$.

Definition

A valuation $\nu$ of $k$ is called discrete if its value group is isomorphic to $\mathbb{Z}$. In this situation, $R$ is called a discrete valuation ring.

In the case $\nu (k^{\times })=\mathbb{Z}$, $\nu$ is said to be normalized.

Example. $p$-adic on $\mathbb{Z}$ is a valuation.

Theorem

Let $(k,\nu )$ be a discrete normalized valuation field. Let $R$ be its valuation ring, and let $P$ be its valuation ideal.

• Let $\pi \in R$ with $\nu (\pi )=1$. Then any $\alpha \in k$ is expressed as $\alpha ={\pi }^{i}u$ where $i=\nu (\alpha )$, $u\in R\setminus P$.
• Any nonzero ideal of $R$ is of the form $p^i=({\pi }^i)$ for some $i⩾0$. In particular, $R$ is a $PID$.
• $R$ is integral closed.

\begin{proof} i) For $\alpha \in k$ with $\nu (\alpha )=i$, we have $\nu (\alpha {\pi }^{-i})=0$ and $\alpha {\pi }^{-i}\in R\setminus P$.

ii) Let $I$ be a nonzero ideal of $R$. Let $i=\min \{\nu (a):a\in I\}$. Then ${\pi }^i=a{u}^{-1}\in I$. Let $0\ne a\in I$ with $\nu (a)=j⩾i$. Then $a={\pi }^{j}u={\pi }^i({\pi }^{j-i}u)\in ({\pi }^i)$. So $I=({\pi }^i)$. In particular, $P=(\pi )$ and $I=P^i$.

iii) Take integral $\alpha \in k$, then there exists $a_i\in R$ such that

$${\alpha }^n+a_{n-1}{\alpha }^{n-1}+\cdots +a_0=0.$$

If $\alpha \notin R$, then $\alpha$ can be written as $\alpha ={\pi }^{-j}u$ with $j\in {\mathbb{N}}_{+}$. It deduces that

$$u^n+a_{n-1}{\pi }^j{u}^{n-1}+a_{n-2}{\pi }^{2j}{u}^{n-2}+\cdots +a_1{\pi }^{(n-1)j}u+a_0{\pi }^{nj}=0$$

and so

$$u^n=-{\pi }^j\left(a_{n-1}{u}^{n-1}+a_{n-2}{\pi }^j{u}^{n-2}+\cdots +a_1{\pi }^{(n-2)j}u+a_0{\pi }^{(n-1)j}\right).$$

Note that $\nu (\text{LHS})=n\nu (u)=0$ and $\nu (\text{RHS})=j+\nu (x)$ with $x\in R$, then we have $0=j+\nu (x)>0$, which is impossible. Therefore, $\alpha \in R$. \end{proof}

Corollary

Two discrete valuations $\nu$ and ${\nu }^{\prime }$ of $k$ are equivalent iff they have the same valuation ideal.

\begin{proof} Suppose that $\nu$ and ${\nu }^{\prime }$ have the same valuation ideal, say $P$. Then $R=k\setminus P^{-1}$ is the valuation ring, where $P^{-1}=\{x^{-1}:x\in P\setminus \{0\}\}$. Suppose $\nu$ and ${\nu }^{\prime }$ are normalized, then they have the same value in $R$ and so in $k$ by ^bb0319. Thus $\nu$ and ${\nu }^{\prime }$ are equivalent of $k$. \end{proof}

Theorem

Discrete valuation ring is a local PID.

\begin{proof} "←" Let $R$ be a local PID. Let $P=J(R)=(\pi )$. We claim that ${\cap }_{i=1}^{\infty }{P}^i=0$. Otherwise, if $0\ne a\in {\cap }_{i=1}^{\infty }{P}^i$, then $a=\pi a_1={\pi }^2{a}_2=\cdots$ with $a_i\in R$. It deduces that $a_i=\pi a_{i+1}$ and so we get a ascending series $(a_1)⊊(a_2)⊊\cdots$ Since $R$ is PID, the series terminate, leading to a contradiction.

For any $0\ne a\in R$, there exists $i\in {\mathbb{N}}_{+}$ such that $a\in P^i$ and $a\notin P^{i+1}$. Thus $a={\pi }^{i}u$ for some $u\in R\setminus P$. Let $k$ be the fractional field of $R$. Define $\nu (a/b)=\nu (a)-\nu (b)$ and $\nu$ is a discrete valuation.

"→" By ^bb0319. \end{proof}

Remark. In fact, DVR is a ED, where the Euclidean function is $\nu$. For any $a,b\in R$, if $\nu (a)⩾\nu (b)$, then $a/b\in R$ and so $q=b,r=0$; if $\nu (a)<\nu (b)$, then $q=1,r=b$.

Multiplicative Valuation

Definition

Let $\nu$ be an exponential valuation of a field $k$. Fix a real number $\gamma$ with $0<\gamma <1$. Define $\phi :k\to \mathbb{R},a\mapsto {\gamma }^{\nu (a)}$. Then

• $\phi (a)⩾0$ and $\phi (a)=0$ yields $a=0$;
• $\phi (ab)=\phi (a)\phi (b)$
• $\phi (a+b)⩽\max \{\phi (a),\phi (b)\}$
• $\phi (a^{-1})=\phi (a{)}^{-1}$
• $\phi (a)<\phi (b)$ yields $\phi (a+b)=\phi (b)$.

Define the valuation ring $R=\{a\in k:\phi (a)⩽1\}$ and valuation ideal $P=\{a\in k:\phi (a)<1\}$.

Two multiplicative valuation ${\phi }_1$ and ${\phi }_2$ are called equivalent if there exists $r\in {\mathbb{R}}_{⩾0}$ such that ${\phi }_2(a)={\phi }_1(a{)}^r$ for all $a\in k$.

Remark. We can get a multiplicative valuation from an exponential valuation. For a given a function $\phi :k\to R$ satisfying

• $\phi (a)⩾0$ and $\phi (a)=0$ yields $a=0$;
• $\phi (ab)=\phi (a)\phi (b)$
• $\phi (a+b)⩽\max \{\phi (a),\phi (b)\}$

define $\nu (a)={\log }_{\gamma }\phi (a)$, then $\nu$ is a valuation on $k$.

Definition

Define $d(a,b)=\phi (a-b)$, which is a metric on $k$. Then $k$ is a Hausdorff space and

• $k\times k\to k,(a,b)\mapsto a+b$
• $k\times k\to k,(a,b)\mapsto ab$
• $k^{\times }\times k^{\times }\to k,(a,b)\mapsto a{b}^{-1}$

are continuous.

Completion

Definition

Recall that if every Cauchy sequence in $k$ converges, then $k$ is called complete, or a complete field. The associated valuation ring is said to be a complete valuation ring.

Given a valuation field $k$, let $C$ denote the set of Cauchy sequence in $k$. Define $(a_n)+(b_n)=(a_n+b_n)$ and $(a_n)(b_n)=(a_n{b}_n)$. Then $C$ is a commutative ring. If $I$ denotes the subset of $C$ consisting of the sequences converge to $0$, then $I$ is a maximal ideal. Hence $\tilde{k}=C/I$ is a field.

If $(a_n)$ is a Cauchy sequence, then $|\phi (a_m)-\phi (a_n)|⩽\phi (a_m-a_n)$. Hence $(\phi (a_n))$ is a Cauchy sequence in $\mathbb{R}$ and ${\lim }_{n\to \infty }\phi (a_n)$ exists.

If $(a_n)-(b_n)\in I$, then ${\lim }_{n\to \infty }\phi (a_n)={\lim }_{n\to \infty }\phi (b_n)$. Define

$$\tilde{\phi }:\tilde{k}\to R,\widetilde{(a_n)}=\underset{n\to \infty }{\lim }\phi (a_n)$$

for $\widetilde{(a_n)}=(a_n)+I$. It is well-defined by ^147dxw. Furthermore, we can check $\tilde{\phi }$ is a multiplicative valuation of $\tilde{k}$.

For $a\in k$, the sequence $(a)=(a,a,a,\cdots )$ is a Cauchy sequence and $\widetilde{(a)}\ne \widetilde{(b)}$ if $a\ne b$. Thus we can regard $k$ as a subfield of $\tilde{k}$ by identifying $a$ with $\widetilde{(a)}$. So $\tilde{\phi }$ is an extension of $\phi$ and $k$ is dense in $\tilde{k}$ by definition. It deduces the following lemma.

Lemma

$\tilde{k}$ is a complete field.

The following theorem shows that completion preserves the residue field.

Theorem

Let $\tilde{R}$ and $\tilde{P}$ be the valuation ring and valuation ideal of $\tilde{\phi }$ respectively. Then $\tilde{R}/\tilde{P}\simeq R/P$.

\begin{proof} It suffices to show $R+\tilde{P}=\tilde{R}$. Note that $R+\tilde{P}\subseteq \tilde{R}$ is trivial.

Let $a\in \tilde{R}$ with $a\ne 0$. Then $0<\tilde{\phi }(a)⩽1$ and there exists $a_n\in k$ such that $a={\lim }_{n\to \infty }{a}_n$. Hence ${\lim }_{n\to \infty }\tilde{\phi }(a_n-a)=0$. So there exists $n$ such that $\tilde{\phi }(a_n-a)<\tilde{\phi }(a)$.

Notice that $\phi (a_n)=\tilde{\phi }(a_n-a+a)=\tilde{\phi }(a)⩽1$ and $a_n\in R$. And $\tilde{\phi }(a_n-a)<\tilde{\phi }(a)⩽1$ yields $a_n-a\in \tilde{P}$, Then $a=a_n+a-a_n\in R+\tilde{P}$. \end{proof}

Valuations on Vector Spaces and Field Extensions

Norms and Topology on Vector Spaces over Valued Fields

Let $\nu$ be a normalized discrete valuation of $k$ with valuation ring $R$ and valuation ideal $P=(\pi )$. Then $\{P,P^2,\cdots \}$ is a fundamental neighborhood system of $0$ by ${\cap }_{i=1}^{\infty }({\pi }^i)=0$. Therefore, $(a_n)$ is a Cauchy sequence iff for any $\ell \in {\mathbb{Z}}_{>0}$, there exists $N$ such that $a_m-a_n\in P^{\ell }$ for all $m,n⩾N$. Also ${\lim }_{n\to \infty }{a}_n=a$ iff given $\ell$, there exists $N$ such that $a_n-a\in P^{\ell }$ for all $n⩾N$.

Lemma

Let ${\phi }_1,{\phi }_2$ be discrete multiplicative valuations of $k$. TFAE:

• ${\phi }_1$ and ${\phi }_2$ are equivalent;
• $\{a\in k:{\phi }_1(a)<1\}=\{a\in k:{\phi }_2(a)<1\}$;
• the topologies on $k$ induced by ${\phi }_1$ and ${\phi }_2$ are same.

\begin{proof} i)←>ii) See ^ox907z.

ii)→iii) It suffices to show i)→iii), which is trivial because their fundamental neighborhood systems are equivalent.

iii)→ii) Suppose ${\phi }_1(a)<1$, then ${\lim }_{n\to \infty }{\phi }_1(a^n)={\lim }_{n\to \infty }{\phi }_1(a{)}^n=0$. So ${\lim }_{n\to \infty }{\phi }_2(a{)}^n=0$ and ${\phi }_2(a)<1$. Similarly, ${\phi }_2(a)<1$ yields ${\phi }_1(a)<1$. \end{proof}

Definition

Let $\phi$ be a multiplicative valuation of $k$. Let $V$ be a vector space over $k$. A real-valued function $||\cdot ||$ defined on $V$ is said to be a norm on $V$ if

• $||v||⩾0$ and $||v||=0⟺v=0$.
• $||av||=\phi (a)||v||$ for any $a\in k$ and $v\in V$.
• $||u+v||⩽\max \{||u||,||v||\}$.

If we set $d(u,v)=||u-v||$, then $V$ is a metric space. Note that $V\times V\to V,(u,v)\mapsto u+v$ and $k\times V\to V,(a,v)\mapsto av$ are continuous. Remark that norm induces metric, but not vice versa.

Lemma

Let $u_1,\cdots ,u_n$ be a basis of a finite-dimensional $k$-space $V$. For $v=a_1{u}_1+\cdots +a_n{u}_n\in V$, define $||v|{|}_0=\max \{\phi (a_i):1⩽i⩽n\}$. Then $||\cdot |{|}_0$ is a norm on $V$. The following holds for the metric indeed by $||\cdot |{|}_0$.

• Let ${\nu }_r=a_{r_1}{u}_1+\cdots +a_{r_n}{u}_n$ be given for $r=1,2,\cdots$. Then the sequence $\{v_r{\}}_r$ is a Cauchy sequence iff the sequence $(a_{r_i}{)}_r$ is a Cauchy sequence in $k$ for all $i=1,\cdots ,n$. Also ${\lim }_{r\to \infty }{v}_r={\sum }_{i=1}^n{a}_i{v}_i$ iff ${\lim }_{r\to \infty }{a}_{r_i}=a_i$ for all $i$.
• If $k$ is complete, then so is $V$.

\begin{proof} See homework. \end{proof}

all norms are equivalent

Suppose that $k$ is complete. Then for any norm $||v||$ on $V$, there exists constant $\lambda ,\mu$ such that $||v||⩽\lambda ||v|{|}_0$, $||v|{|}_0⩽\mu ||v||$ for all $v\in V$. Thus the topological induces by these two norms are the same.

\begin{proof} Let $\lambda =\max \{||u_i||:1⩽i⩽n\}$. Then for $v={\sum }_{i=1}^n{a}_i{u}_i$, $||v||⩽\max \{\phi (a_i)||u_i||\}⩽\lambda ||v|{|}_0$.

We will show the existence of $\mu$ by induction on $n$.

The case of $n=1$ is trivial. Let $n>1$. For $v=a_1{u}_1+\cdots +a_n{u}_n$, set ${\phi }_i(v)=\phi (a_i)$. Claim that for all $i$, there exists ${\mu }_i$ such that ${\phi }_i(v)⩽{\mu }_i||v||$. If it is true, then

$$||v|{|}_0=\max \{{\phi }_i(v)\}⩽\max \{{\mu }_i||v||\}=(\max \{{\mu }_i\})||v||$$

and $\mu =\max \{{\mu }_i\}$ is what we desired.

Let $V^{\prime }=k{u}_1\oplus \cdots \oplus k{u}_{n-1}$. By induction $||\cdot ||$ and $||\cdot |{|}_0$ induce the same topologies on $V$. In particular, $V^{\prime }$ is complete w.r.t. $||\cdot ||$ or $||\cdot |{|}_0$.

Suppose by way of contradiction that there is no ${\mu }_n$ with ${\phi }_n(v)⩽{\mu }_n||v||$ for all $v\in V$. Then for any $j\in {\mathbb{Z}}_{>0}$, there exists $v_j\in V\setminus V^{\prime }$ such that ${\phi }_n(v_j)>j||v_j||$.

This holds if $v_j$ is replaced by its nonzero scalar multiples, so we can assume that $v_j=a_{j,1}{u}_1+\cdots +a_{j,n-1}{u}_{n-1}+1\cdot u_n$. Then $||v_j||<1/j{\phi }_n(v_j)=1/j$, which yields ${\lim }_j{v}_j=0$ w.r.t. $||\cdot ||$. So ${\lim }_j(v_j-u_n)=-u_n$. But note that $v_j-u_n\in V^{\prime }$ for all $j$ and $V^{\prime }$ is closed in $V$, so $-u_n\in V^{\prime }$, leading to a contradiction. \end{proof}

Remark. The conclusion “all norms are equivalent” holds true for two very important classes of complete fields:

• Archimedean Fields: $\mathbb{R}$ and $\mathbb{C}$ (where the proof relies on local compactness).
• Non-Archimedean Fields: such as the complete valued fields defined here.

Extension of Valuations in Field Extensions

Let $K,L$ be discrete valuation fields with multiplicative valuation $\phi$ and $\Phi$, respectively. If $K\subseteq L$ and $\Phi$ is an extension of $\phi$, then we say $(L,\Phi )$ is an extension of $(K,\phi )$, and write $(K,\phi )\subseteq (L,\Phi )$. Consider $L$ as an $K$-space. Let $||x||=\Phi (x)$ for $x\in L$.

Theorem

Let $\phi$ be a complete discrete multiplicative valuation of $K$ and $L$ be a finite extension of $K$. If there exists an extension $(L,\Phi )$ of $(K,\phi )$, then $\Phi$ is unique up to equivalent. Furthermore, $(L,\Phi )$ is complete.

\begin{proof} By ^96mek7, since $K$ is complete and $[L:K]<\infty$, $L$ is also complete. Then by ^zs3z31 one can check $\Phi$ is unique up to equivalent. \end{proof}

$I$-adic Theory

Definition

Let $R$ be a ring, and let $V$ be an $R$-module. For an ideal $I\subseteq R$, assume that ${\cap }_{i=1}^{\infty }{I}^{i}V=0$. Fix $\gamma \in \mathbb{R}$ with $0<r<1$, define $||\cdot |{|}_I$ on $V$ by

• $||0|{|}_I=0$
• $||v|{|}_I={\gamma }^i$ for $v\in I^{i}V\setminus I^{i+1}V$.

Define $d_I(u,v):=||u-v|{|}_I$. Check: $d_I(u,v)⩽\max \{d_I(u,w),d_I(w,v)\}$. So $d_I$ is a metric on $V$.

In this situation, we say $d_I$ is defined on $V$.

Note that

• $v_n\in V$, $\{v_n{\}}_n$ is a Cauchy sequence iff for any integer $m>0$, there exists $N>0$ such that $v_i-v_j\in I^{n}V$ for all $i,j⩾N$.
• $v={\lim }_{n\to \infty }{v}_n$ iff for any integer $m>0$, there exists $N>0$ such that $v_i-v\in I^{m}V$ for all $i⩾N$.

Lemma

The maps

• $V\times V\to V$, $(a,b)\mapsto a+b$
• $R\times V\to V$, $(r,a)\mapsto ra$
• $f\in {\mathrm{Hom}}_R(V,W)$, where for $W$ we define $d_I$

are continuous.

Theorem

Assume that $V$ is finitely generated. If $R$ is complete (w.r.t. $d_I$), then $V$ is complete.

\begin{proof} Let $\{v_1,\cdots ,v_s\}$ be a set of generators of $V$, and let $\{w_i\}$ be a Cauchy sequence in $V$. There exists a sequence $\{m(n)\}$ of nonnegative integers such that $w_i-w_j\in I^{m(n)}V$ for all $i,j>n$. Note that $m(n)⩽m(n+1)$ and ${\lim }_{n\to \infty }m(n)=\infty$.

Let $w_1={\sum }_{t=1}^s{b}_{1t}{v}_t$ for $b_{1t}\in R$, and let

$$w_{n+1}-w_n=\sum _{t=1}^s{a}_{nt}{v}_t\text{ with }{a}_{nt}\in I^{m(n)}.$$

So $w_n={\sum }_{i=1}^s{b}_{nt}{v}_t$ with $b_{nt}=b_{1t}+{\sum }_{j=1}^{n-1}{a}_{jt}$. Notice that $b_{n+k,t}-b_{n,t}={\sum }_{j=n}^{n+k-1}{a}_{jt}\in I^{m(n)}$. Then $(b_{nt}{)}_n$ is a Cauchy sequence in $R$. It yields $b_t={\lim }_{n\to \infty }{b}_{nt}$ exists in $R$. Let $w={\sum }_{t=1}^s{b}_t{v}_t$. It is easy to check ${\lim }_{n\to \infty }{w}_t=w$ and $V$ is complete. \end{proof}

Remark. It is similar to ^96mek7.

Lemma

Suppose that $d_I$ is defined for all finitely generated $R$-modules. If $V$ is a finitely generated $R$-module, then every submodule of $V$ is closed.

\begin{proof} Let $W$ be a submodule of $V$. Since $f:V\to V/W$ is continuous by ^mccaaw, we know $W=f^{-1}(0)$ is closed. \end{proof}

Remark. It is worth noting that a complete discrete valuation ring $R$ provides the tools to induce a metric on its finitely generated modules. Let $R$ be complete discrete valuation ring with valuation $\nu$, and let $P=(\pi )$ and $F=R/P$ be the valuation ideal and residue field of $\nu$ respectively. Let $V$ be a finitely generated $R$-module. Then by ^353e4e， $V=R{u}_1\oplus \cdots \oplus R{u}_r$ is a direct sum of cyclic submodule. Since $({\pi }^n)V=({\pi }^n)u_1\oplus \cdots \oplus ({\pi }^n)u_r$, there is ${\cap }_{n=1}^{\infty }{\pi }^{n}V=0$. Then $d_P$ is defined for all finitely generated $R$-module. Since $R$ is complete, the module $V$ is also complete by ^ffpm0c.