Valuation Rings and Discrete Valuation Rings

Let $\mathfrak{m}:=\{x\in V:x=0\text{ or }{x}^{-1}\mathrm{\notin }V\}$. By [A1], an element of $V$ lies outside $\mathfrak{m}$ exactly when it is a unit, so $1\mathrm{\notin }\mathfrak{m}$ and $\mathfrak{m}$ is proper.

If $r\in V$ and $x\in \mathfrak{m}$, then $rx\in \mathfrak{m}$: if $rx\mathrm{\ne }0$ and $(rx{)}^{-1}\in V$, then $x^{-1}=r(rx{)}^{-1}\in V$, contradicting $x\in \mathfrak{m}$.

Let $x,y\in \mathfrak{m}$. If $x+y\mathrm{\notin }\mathfrak{m}$, then $x+y$ is a unit by step 1.1. If $x=0$ or $y=0$ this contradicts $x,y\in \mathfrak{m}$, so assume $x,y\mathrm{\ne }0$. By [F1], either $y\mathrm{/}x\in V$ or $x\mathrm{/}y\in V$; in the first case $x^{-1}=(x+y{)}^{-1}(1+y\mathrm{/}x)\in V$, and in the second case $y^{-1}=(x+y{)}^{-1}(1+x\mathrm{/}y)\in V$, again a contradiction. Thus $x+y\in \mathfrak{m}$.

Steps 2.1 and 2.2 show that $\mathfrak{m}$ is an ideal. Every proper ideal contains no unit, so every proper ideal is contained in $\mathfrak{m}$. Hence $\mathfrak{m}$ is the unique maximal ideal of $V$, and $V$ is local.

Assume condition 1. Let $a,b\in V$. If $a=0$ or $b=0$, divisibility is trivial. If $a,b\mathrm{\ne }0$, apply [F1] to $a\mathrm{/}b\in K^{\times }$: if $a\mathrm{/}b\in V$, then $a=(a\mathrm{/}b)b$, so $b$ divides $a$; if $b\mathrm{/}a\in V$, then $a$ divides $b$. Thus condition 2 holds.

Assume condition 2. Let $I$ and $J$ be ideals of $V$. If $I\subseteq ̸J$, choose $a\in I\setminus J$. For any $b\in J$, condition 2 says either $a$ divides $b$ or $b$ divides $a$; the second option would put $a$ in $J$, so $b=ca$ for some $c\in V$ and hence $b\in I$. Therefore $J\subseteq I$. By symmetry, any two ideals are comparable, so condition 3 holds.

Assume condition 3. Let $x\in K^{\times }$, and choose $a,b\in V$ with $b\mathrm{\ne }0$ and $x=a\mathrm{/}b$ by [A1]. The principal ideals $(a)$ and $(b)$ are comparable. If $(a)\subseteq (b)$, then $a=bc$ for some $c\in V$, so $x=c\in V$. If $(b)\subseteq (a)$, then $b=ad$ for some $d\in V$, so $x^{-1}=d\in V$. Thus condition 1 holds.

Under condition 3, a finitely generated ideal $I=(a_1,\dots ,a_n)$ is principal: among the finitely many comparable principal ideals $(a_i)$, choose a largest one, say $(a_j)$. Then every $a_i$ lies in $(a_j)$, so $I=(a_j)$. The zero ideal is $(0)$, and step 1.3 now identifies $V$ as a valuation ring, so [L1] records the local consequence for nonunits.

The order relation of [L2] is well defined on cosets: if $x^{\mathrm{\prime }}=ux$ and $y^{\mathrm{\prime }}=vy$ with $u,v\in V^{\times }$, then $y^{\mathrm{\prime }}\mathrm{/}{x}^{\mathrm{\prime }}=(v\mathrm{/}u)(y\mathrm{/}x)$, and [A1] shows that $y^{\mathrm{\prime }}\mathrm{/}{x}^{\mathrm{\prime }}\in V$ exactly when $y\mathrm{/}x\in V$.

The order is reflexive because $x\mathrm{/}x=1\in V$. It is antisymmetric because if $\bar{x}\le \bar{y}$ and $\bar{y}\le \bar{x}$, then both $y\mathrm{/}x$ and $x\mathrm{/}y$ lie in $V$, so $y\mathrm{/}x$ is a unit of $V$. Hence $x$ and $y$ represent the same coset in $\mathrm{\Gamma }$.

The order is transitive because $y\mathrm{/}x\in V$ and $z\mathrm{/}y\in V$ imply $z\mathrm{/}x=(z\mathrm{/}y)(y\mathrm{/}x)\in V$. It is total because $V$ is a valuation ring: for any $x,y\in K^{\times }$, the quotient $y\mathrm{/}x$ either lies in $V$ or has inverse $x\mathrm{/}y$ in $V$. It is translation-invariant because $\bar{x}+\bar{z}\le \bar{y}+\bar{z}$ is equivalent to $(yz)\mathrm{/}(xz)=y\mathrm{/}x\in V$. Thus $\mathrm{\Gamma }$ is a totally ordered abelian group.

Define $v(0)=\mathrm{\infty }$ and $v(x)=\bar{x}$ for $x\mathrm{\ne }0$. Then $v(x)=\mathrm{\infty }$ exactly when $x=0$, and for $x,y\mathrm{\ne }0$ one has $v(xy)=\overline{xy}=\bar{x}+\bar{y}=v(x)+v(y)$. If $x+y=0$, then $v(x+y)=\mathrm{\infty }\ge \min \{v(x),v(y)\}$. Otherwise, after swapping $x$ and $y$ if needed, step 1.3 gives $v(x)\le v(y)$, so $y\mathrm{/}x\in V$ and $x+y=x(1+y\mathrm{/}x)$ with $1+y\mathrm{/}x\in V$; hence $v(x+y)\ge v(x)=\min \{v(x),v(y)\}$. Therefore $v$ satisfies the valuation axioms of [L1].

The nonnegative locus of $v$ is exactly $V$: for $x\in K^{\times }$, the condition $0\le v(x)$ means $\bar{1}\le \bar{x}$, which by [L2] is equivalent to $x\in V$. Since $0\in V$ as well, the valuation ring of $v$ is precisely $V$.

Let $x$ be an element of the field of fractions of $V$ that is integral over $V$. By [A1], regard $x$ as an element of $K$. If $x\mathrm{\notin }V$, then [F1] gives $x^{-1}\in V$. This element is not a unit of $V$, because a unit inverse would put $x$ back in $V$. Hence [L2] places $x^{-1}$ in the maximal ideal $\mathfrak{m}$ of $V$.

Choose a monic equation $x^n+a_{n-1}{x}^{n-1}+\cdots +a_0=0$ with $a_i\in V$. Multiplying by $x^{-n}$ gives $1+a_{n-1}{x}^{-1}+\cdots +a_0{x}^{-n}=0$. Since $x^{-1}\in \mathfrak{m}$ and $\mathfrak{m}$ is an ideal, every term except $1$ lies in $\mathfrak{m}$. Therefore $1\in \mathfrak{m}$, contradicting maximality.

So $x\in V$. By [L1], this proves that $V$ is integrally closed; by [A1], it is also a domain.

Let $x\in K^{\times }$ and put $n:=v(x)\in \mathbb{Z}$. Since $v(\pi )=1$ by [F1], one has $v(x{\pi }^{-n})=v(x)-nv(\pi )=0$. Setting $u:=x{\pi }^{-n}$, [F2] gives $u\in V$ and $u^{-1}\in V$, hence $u\in V^{\times }$. Therefore $x=u{\pi }^n$.

Suppose also that $x=u^{\mathrm{\prime }}{\pi }^m$ with $u^{\mathrm{\prime }}\in V^{\times }$ and $m\in \mathbb{Z}$. Applying $v$ gives $n=v(x)=v(u)+n=v(u^{\mathrm{\prime }})+m=m$ because units have valuation $0$. Then $u=x{\pi }^{-n}=x{\pi }^{-m}=u^{\mathrm{\prime }}$. So the expression is unique.

Let $I\mathrm{\ne }0$ be an ideal of $V$. Because $I\subseteq V$, every nonzero element of $I$ has valuation in ${\mathbb{Z}}_{\ge 0}$. Choose $x\in I\setminus \{0\}$ with minimal valuation $n$. By [L1], $x=u{\pi }^n$ for a unit $u$, so ${\pi }^n=u^{-1}x\in I$.

If $y\in I$ is nonzero, then [L1] gives $y=u^{\mathrm{\prime }}{\pi }^m$ for some unit $u^{\mathrm{\prime }}$ and $m\ge 0$. Minimality of $n$ yields $m\ge n$, so $y=u^{\mathrm{\prime }}{\pi }^{m-n}{\pi }^n\in ({\pi }^n)$. Thus $I\subseteq ({\pi }^n)$, while step 1.1 gave $({\pi }^n)\subseteq I$. Hence $I=({\pi }^n)$. Since [F1] gives $\mathfrak{m}=(\pi )$, this is also ${\mathfrak{m}}^n$.

If $({\pi }^n)=({\pi }^m)$, then ${\pi }^n\in ({\pi }^m)$ and ${\pi }^m\in ({\pi }^n)$, so $n\ge m$ and $m\ge n$. Therefore $n=m$, and the exponent is unique.

Let $\mathfrak{p}$ be a nonzero prime ideal of $V$. Choose $0\mathrm{\ne }x\in \mathfrak{p}$. By [L1], $x=u{\pi }^n$ for some unit $u$ and some $n\ge 1$, so ${\pi }^n\in \mathfrak{p}$. Since $\mathfrak{p}$ is prime, $\pi \in \mathfrak{p}$. Therefore $\mathfrak{m}=(\pi )\subseteq \mathfrak{p}$, and maximality of $\mathfrak{m}$ forces $\mathfrak{p}=\mathfrak{m}$.

By [A1], $(0)$ is prime, and step 1.1 shows there are no other nonzero primes besides $\mathfrak{m}$. Therefore the prime spectrum has exactly the strict chain $(0)⊊\mathfrak{m}$. Its length is $1$, and no longer chain exists. Hence [F1] gives $\dim V=1$.

Assume condition 1. Then [L3] makes $R$ local. By [L4], every ideal of $R$ is principal, so [F3] shows that $R$ is a PID; in particular it is condition 4. Because every ideal is principal, [F2] makes $R$ Noetherian, so condition 2 holds as well. By [L1] and [L5], $R$ is integrally closed and one-dimensional, so condition 3 also holds.

Assume condition 2. By [L3], the ring is local with maximal ideal $\mathfrak{m}$. Because $R$ is not a field, $\mathfrak{m}\mathrm{\ne }0$. By [F2], every ideal of $R$ is finitely generated, so choose generators $a_1,\dots ,a_r$ of $\mathfrak{m}$. By [L2], one of them divides all the others; rename it $\pi$. Then $\mathfrak{m}=(\pi )$.

Assume condition 3. Let $\mathfrak{m}$ be the unique maximal ideal. It is nonzero because $R$ is not a field. Choose $0\mathrm{\ne }x\in \mathfrak{m}$. The quotient $R\mathrm{/}xR$ is Noetherian by [L6]. Every prime ideal of $R$ containing $xR$ is nonzero, hence equals $\mathfrak{m}$ because $\dim R=1$. Therefore the nilradical of $R\mathrm{/}xR$ is $\mathfrak{m}\mathrm{/}xR$, so [L7] yields an integer $n\ge 1$ with ${\mathfrak{m}}^n\subseteq xR$.

Assume condition 4. Then the ring is local with nonzero maximal ideal $\mathfrak{m}=(\pi )$. Because every ideal is principal, [F2] makes $R$ Noetherian. Every element outside $\mathfrak{m}$ is a unit: if $x\mathrm{\notin }\mathfrak{m}$, then $(x)$ is not contained in the unique maximal ideal, so $(x)=R$. Hence every nonunit is a multiple of $\pi$.

In the situations of steps 1.2 and 1.4, the ring is a Noetherian local domain with nonzero principal maximal ideal $\mathfrak{m}=(\pi )$. Let $0\mathrm{\ne }x\in R$. If $x$ is not a unit, the maximal-ideal description gives $x=\pi x_1$. If $x_1$ is not a unit, write $x_1=\pi x_2$, and continue. This process stops, for otherwise $(x)⊊(x_1)⊊(x_2)⊊\cdots$ would be a strict ascending chain of ideals, contradicting Noetherianity. Thus every nonzero element has the form $u{\pi }^n$ with $u$ a unit and $n\ge 0$. If $u{\pi }^n=u^{\mathrm{\prime }}{\pi }^m$ with units $u,u^{\mathrm{\prime }}$, then $n=m$, for otherwise a positive power of the nonunit $\pi$ would equal a unit.

In the same situations, every nonzero ideal $I$ of $R$ is $({\pi }^n)$ for a unique integer $n\ge 0$: choose $0\mathrm{\ne }x\in I$ whose exponent in step 2.1 is minimal, say $x=u{\pi }^n$. Then ${\pi }^n=u^{-1}x\in I$. For any nonzero $y\in I$, write $y=u^{\mathrm{\prime }}{\pi }^m$ by step 2.1; minimality gives $m\ge n$, so $y\in ({\pi }^n)$. Hence $I=({\pi }^n)$. Therefore step 1.2 already yields condition 4.

Under condition 4, let $x=a\mathrm{/}b\in K^{\times }$ with $a,b\in R\setminus \{0\}$. By step 2.1, write $a=u{\pi }^m$ and $b=v{\pi }^n$. Then $x=(u{v}^{-1}){\pi }^{m-n}$. If $m\ge n$, then $x\in R$; if $m<n$, then $x^{-1}\in R$. Thus $R$ is a valuation ring, and step 1.4 already makes it Noetherian. Therefore condition 2 holds.

Return to condition 3. Choose $n\ge 1$ minimal with ${\mathfrak{m}}^n\subseteq xR$. If $n=1$, then $\mathfrak{m}=(x)$, so steps 1.3 and 3.1 give condition 4. Suppose instead that $n>1$, and choose $y\in {\mathfrak{m}}^{n-1}\setminus xR$. Then $y\mathfrak{m}\subseteq {\mathfrak{m}}^n\subseteq xR$, so $z:=y\mathrm{/}x\in K$ satisfies $z\mathfrak{m}\subseteq R$ while $z\mathrm{\notin }R$.

Under condition 4, define $v:K\to \mathbb{Z}\cup \{\mathrm{\infty }\}$ by $v(0)=\mathrm{\infty }$ and $v(x)=m-n$ when $x=(u{v}^{-1}){\pi }^{m-n}$ as in step 3.2. The uniqueness part of step 2.1 makes this well defined. Multiplicativity is immediate from exponents. If $v(x)\le v(y)$, write $x=u{\pi }^m$ and $y=u^{\mathrm{\prime }}{\pi }^n$ with $m\le n$; then $x+y={\pi }^m(u+u^{\mathrm{\prime }}{\pi }^{n-m})$, and the bracket lies in $R$, so $v(x+y)\ge m=\min \{v(x),v(y)\}$. Because $v(\pi )=1$, this valuation is surjective, and its nonnegative locus is exactly $R$. Therefore condition 1 holds.

Still under condition 3, suppose also that $z\mathfrak{m}\subseteq \mathfrak{m}$. Because $R$ is Noetherian, [F2] lets us choose generators $b_1,\dots ,b_r$ of $\mathfrak{m}$. Write $z{b}_j={\sum }_{i=1}^r{c}_{ij}{b}_i$ with $c_{ij}\in R$. In matrix form this is $(z{I}_r-C)\mathbf{b}=0$. By [L8], $\det (z{I}_r-C)\mathbf{b}=0$. Choose an index $j$ with $b_j\mathrm{\ne }0$; since $R$ is a domain, the equality $\det (z{I}_r-C)b_j=0$ forces $\det (z{I}_r-C)=0$. This is a monic polynomial equation for $z$ with coefficients in $R$, so $z$ is integral over $R$. Because $z\in K$ and condition 3 says $R$ is integrally closed, [F4] then forces $z\in R$, a contradiction. Hence $z\mathfrak{m}⊈\mathfrak{m}$.

By step 5.1, choose $a\in \mathfrak{m}$ with $az\mathrm{\notin }\mathfrak{m}$. Since $z\mathfrak{m}\subseteq R$, the element $u:=az$ lies in $R$. Being outside the maximal ideal, $u$ is a unit. For every $b\in \mathfrak{m}$, the element $zb$ lies in $R$, so $b=u^{-1}a(zb)\in aR$. Thus $\mathfrak{m}\subseteq aR\subseteq \mathfrak{m}$, and $\mathfrak{m}=aR$ is principal. Together with step 1.3, step 3.1 now gives condition 4.

Step 1.1 proves $(1)\Rightarrow (2),(1)\Rightarrow (3),(1)\Rightarrow (4)$; steps 1.2 and 3.1 prove $(2)\Rightarrow (4)$; step 3.2 proves $(4)\Rightarrow (2)$; step 4.2 proves $(4)\Rightarrow (1)$; and steps 1.3 to 6.1 prove $(3)\Rightarrow (4)$. Therefore all four conditions are equivalent.

Suppose $V$ is Noetherian. A valuation ring is a domain, so if $V$ is not a field then [L2] applies and shows that $V$ is a DVR. Thus a Noetherian valuation ring is a field or a DVR.

Conversely, every field is Noetherian because its only ideals are $(0)$ and the whole ring. If $V$ is a DVR, then [L2] applied in the forward direction shows that it is a Noetherian valuation ring. Hence $V$ is Noetherian exactly in the two stated cases.

The case $n=0$ is $V\mathrm{/}(1)=0$, whose length is $0$ by [F1]. For $n=1$, the ideals of $V\mathrm{/}(\pi )$ correspond to the ideals of $V$ containing $(\pi )$. By [L2], those are only $(\pi )$ and $V$, so $V\mathrm{/}(\pi )$ is simple and has length $1$.

For each $n\ge 1$ there is a short exact sequence $0\to ({\pi }^n)\mathrm{/}({\pi }^{n+1})\to V\mathrm{/}({\pi }^{n+1})\to V\mathrm{/}({\pi }^n)\to 0$. Multiplication by ${\pi }^n$ induces an isomorphism $V\mathrm{/}(\pi )\cong ({\pi }^n)\mathrm{/}({\pi }^{n+1})$, so step 1.1 gives ${\mathrm{\ell }}_V(({\pi }^n)\mathrm{/}({\pi }^{n+1}))=1$. Therefore [L3] yields ${\mathrm{\ell }}_V(V\mathrm{/}({\pi }^{n+1}))={\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))+1$.

By induction on $n$, step 1.1 and step 2.1 give ${\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))=n$ for every $n\ge 0$.

Let $x\in V$ be nonzero, and write $x=u{\pi }^n$ as in [L1]. Multiplication by the unit $u^{-1}$ identifies the ideals $(x)$ and $({\pi }^n)$, so $V\mathrm{/}(x)\cong V\mathrm{/}({\pi }^n)$ as $V$-modules. Hence ${\mathrm{\ell }}_V(V\mathrm{/}(x))={\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))=n$.

By [L1] and [L3], the ring $R_{\mathfrak{p}}$ is a Noetherian local domain. By [F1], its Krull dimension is $\dim (R_{\mathfrak{p}})=\mathrm{ht}(\mathfrak{p})=1$.

The localisation remains integrally closed. Let $x$ lie in the fraction field of $R_{\mathfrak{p}}$ and be integral over $R_{\mathfrak{p}}$. Write a monic equation $x^n+\frac{a_{n-1}}{s_{n-1}}{x}^{n-1}+\cdots +\frac{a_0}{s_0}=0$ with each $s_i\mathrm{\notin }\mathfrak{p}$. Put $s=s_0\cdots s_{n-1}$. Then $sx$ is integral over $R$, so [F3] gives $sx\in R$. Since $s\mathrm{\notin }\mathfrak{p}$, we conclude $x=(sx)\mathrm{/}s\in R_{\mathfrak{p}}$. Thus $R_{\mathfrak{p}}$ is integrally closed.

The maximal ideal of $R_{\mathfrak{p}}$ is nonzero. Choose $0\mathrm{\ne }a\in \mathfrak{p}$. If $a\mathrm{/}1=0$ in $R_{\mathfrak{p}}$, then some $s\mathrm{\notin }\mathfrak{p}$ satisfies $sa=0$, impossible in the domain $R$. Hence $a\mathrm{/}1\in \mathfrak{p}{R}_{\mathfrak{p}}$ is nonzero, so $R_{\mathfrak{p}}$ is not a field.

Steps 1.1, 1.2, and 1.3 verify condition 3 of [L4] for the ring $R_{\mathfrak{p}}$. Therefore $R_{\mathfrak{p}}$ is a discrete valuation ring.

A discrete valuation ring is a domain. Its zero ideal is principal, and [L1] shows that every nonzero ideal is principal.

Therefore every ideal of the domain $V$ is principal, so [F1] makes $V$ a principal ideal domain.