Equivalent characterizations of a DVR

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.