Valuation Rings and Discrete Valuation Rings - Examples

Every nonzero rational number can be written uniquely as $x=u{p}^n$ with $n\in \mathbb{Z}$ and $u=a\mathrm{/}b\in {\mathbb{Q}}^{\times }$ having both $a$ and $b$ prime to $p$: factor all powers of $p$ from the numerator and denominator and collect the difference in the exponent $n$. Define $v(0)=\mathrm{\infty }$ and $v(x)=n$ for $x\mathrm{\ne }0$. Then $v(xy)=v(x)+v(y)$, the ultrametric inequality follows by factoring out the smaller power of $p$, and $v$ is surjective because $v(p)=1$. Its nonnegative locus is exactly ${\mathbb{Z}}_{(p)}$, so [F1] makes ${\mathbb{Z}}_{(p)}$ a DVR with uniformiser $p$.

By [L1], the units are exactly the elements of valuation $0$, namely the fractions with numerator and denominator both prime to $p$. By [L2], every nonzero ideal is generated by $p^n$ for a unique $n\ge 0$.

Let $a\in R$ be nonzero. If $a\in \mathfrak{p}$, then $a=\pi a_1$ because $\mathfrak{p}=(\pi )$. If $a_1\in \mathfrak{p}$, divide by $\pi$ again, and continue. This process stops, for otherwise $(a)⊊(a_1)⊊(a_2)⊊\cdots$ would be a strict ascending chain of principal ideals, contradicting [L1]. Thus $a=u{\pi }^{n}b$ with $n\ge 0$, $u$ a unit of $R$, and $b\mathrm{\notin }\mathfrak{p}$.

Every nonzero element of $R_{\mathfrak{p}}$ is therefore $(a\mathrm{/}s)=((ub)\mathrm{/}s)(\pi \mathrm{/}1{)}^n$ with $s\mathrm{\notin }\mathfrak{p}$ and $b\mathrm{\notin }\mathfrak{p}$. Since both $b$ and $s$ are in the denominator set, $(ub)\mathrm{/}s$ is a unit of $R_{\mathfrak{p}}$. Hence $R_{\mathfrak{p}}$ is local with nonzero maximal ideal generated by $\pi \mathrm{/}1$, and every ideal is principal by the same minimum-exponent argument as in a DVR.

The ring $R_{\mathfrak{p}}$ is not a field because $\pi \mathrm{/}1$ lies in its maximal ideal and is not a unit. So [L2] applies and shows that $R_{\mathfrak{p}}$ is a discrete valuation ring with uniformiser $\pi \mathrm{/}1$.

The element $t\in k[t]$ is integral over $A_{\mathfrak{m}}$ because it satisfies the monic equation $T^2-t^2=0$ with coefficient $t^2\in A\subseteq A_{\mathfrak{m}}$.

The element $t$ does not belong to $A_{\mathfrak{m}}$. Indeed, if $t=a\mathrm{/}s$ with $a\in A$ and $s\in A\setminus \mathfrak{m}$, then $st=a\in A$. Since $s\mathrm{\notin }\mathfrak{m}$, its constant term is nonzero, so $s=c+t^{2}h(t)$ with $c\in k^{\times }$. Then $st=ct+t^{3}h(t)$. Every element of $A=k[t^2,t^3]$ is a $k$-linear combination of monomials $t^n$ with $n=0$ or $n\ge 2$, so no element of $A$ has a nonzero $t^1$ term. But $ct+t^{3}h(t)$ does, contradiction.

Thus $A_{\mathfrak{m}}$ contains an element of its fraction field integral over it that does not lie in the ring. By [L1], $A_{\mathfrak{m}}$ is not a DVR.

Form the group algebra $k[t^{\mathrm{\Gamma }}]=\{{\sum }_{\gamma \in \mathrm{\Gamma }}{a}_{\gamma }{t}^{\gamma }:\text{only finitely many }{a}_{\gamma }\mathrm{\ne }0\}$, and let $K$ be its fraction field. For a nonzero element $f=\sum a_{\gamma }{t}^{\gamma }$, define $v(f)$ to be the smallest $\gamma$ in its finite support. Because the order on $\mathrm{\Gamma }$ is translation-invariant by [F1], the product of the two lowest terms is the unique lowest term of a product, so $v(fg)=v(f)+v(g)$. For sums, the minimum support can only stay the same or move upward, so $v(f+g)\ge \min \{v(f),v(g)\}$. Extending by $v(f\mathrm{/}g)=v(f)-v(g)$ and $v(0)=\mathrm{\infty }$ gives a valuation on $K$ in the sense of [F2], with value group all of $\mathrm{\Gamma }$.

Let $V:=\{x\in K:v(x)\ge 0\}$. Then $V$ is a valuation ring and is not a field because $t^{(0,1)}$ has positive value. It is not a DVR, since its value group is $\mathrm{\Gamma }$ rather than $\mathbb{Z}$: any cyclic subgroup of $\mathrm{\Gamma }$ is generated by one pair, so it cannot contain both $(1,0)$ and $(0,1)$. Hence [L1] implies that $V$ is not Noetherian.

As in the previous example, form the group algebra $k[t^{\mathrm{\Gamma }}]$ and its fraction field $K$. For a nonzero finite sum $f={\sum }_{\gamma \in \mathrm{\Gamma }}{a}_{\gamma }{t}^{\gamma }$, let $v(f)$ be the least element of its support. The same minimum-support argument as before gives $v(fg)=v(f)+v(g)$ and $v(f+g)\ge \min \{v(f),v(g)\}$, so extending by $v(f\mathrm{/}g)=v(f)-v(g)$ and $v(0)=\mathrm{\infty }$ yields a valuation on $K$ with value group $\mathrm{\Gamma }$. Its nonnegative locus is therefore a valuation ring.

The ordered group $\mathrm{\Gamma }$ has no least positive element. Indeed, $0<\sqrt{2}-1<1$, so any least positive element $\epsilon$ would satisfy $0<\epsilon <1$. Choose $n=\lceil 1\mathrm{/}\epsilon \rceil$, so $n>1$ and $0\le n\epsilon -1<\epsilon$. Equality $n\epsilon -1=0$ would give $\epsilon =1\mathrm{/}n$. But $1\mathrm{/}n$ cannot lie in $\mathrm{\Gamma }$: if $1\mathrm{/}n=a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$, then $nb\sqrt{2}=1-na$ is rational, so $b=0$ and then $na=1$, impossible for $n>1$. Thus $0<n\epsilon -1<\epsilon$, contradicting minimality. Hence the valuation is not discrete.

Each successive quotient in the displayed filtration is generated by the class of ${\pi }^i$, so multiplication by ${\pi }^i$ identifies it with $V\mathrm{/}(\pi )$. Thus every factor has length $1$.

Adding the $n$ successive factors recovers the quotient $V\mathrm{/}({\pi }^n)$, and [L1] confirms that the total length is exactly $n$.

By [F1], both $\pi$ and ${\pi }^{\mathrm{\prime }}$ generate the maximal ideal, so $(\pi )=({\pi }^{\mathrm{\prime }})$. Hence ${\pi }^{\mathrm{\prime }}\in (\pi )$, say ${\pi }^{\mathrm{\prime }}=u\pi$, and similarly $\pi =v{\pi }^{\mathrm{\prime }}$. Multiplying gives $\pi =vu\pi$, so $vu=1$ in the domain $V$ and $u$ is a unit.

The sum $({\pi }^m)+({\pi }^n)$ is one of the two comparable ideals $({\pi }^m)$ and $({\pi }^n)$, namely the larger one, so [L1] makes it $({\pi }^{\min \{m,n\}})$. Likewise the intersection is the smaller one, namely $({\pi }^{\max \{m,n\}})$.

The ideal-comparability statement is exactly [L1]. For the displayed set $I$, closure under multiplication by elements of $V$ is immediate because $v(zx)=v(z)+v(x)\ge v(x)>1$ when $v(z)\ge 0$. If $x,y\in I$, then $v(x+y)\ge \min \{v(x),v(y)\}>1$, so $x+y\in I$. Thus $I$ is an ideal of $V$.

If $I=(a)$ were principal, then $v(a)>1$ and every element of $I$ would have value at least $v(a)$. But [L2] says there is no least element of the set $\{\gamma \in \mathrm{\Gamma }:\gamma >1\}$, so one can choose $1<\gamma <v(a)$ and then an element $x\in V$ with $v(x)=\gamma$. Such an $x$ lies in $I$ but not in $(a)$, contradiction. Hence $I$ is nonprincipal.