----- $K$ is a [[field]]. > [!proposition] Proposition. ([[every DVR is a Noetherian local domain of dimension 1]]) > Let $\nu:K^{*} \to \mathbb{Z}$ be a [[discrete valuation|discrete valuation]], and $A=\{ x \in K: \nu(x) \geq 0 \}$ the [[valuation ring]] of $\nu$ (necessarily $A$ is a [[DVR]]). Then: >1. An element $x \in A$ belongs to $A^{*}$ iff $\nu(x)=0$; >2. $A$ is a [[integral domain|domain]]; >3. Take any $\pi \in A$ with $\nu(\pi)=1$. Then the nonzero [[ideal|ideals]] are $A$ are precisely $\underbrace{ \langle \pi^{0} \rangle }_{ =A } \supsetneq\langle \pi^{1} \rangle \supsetneq \langle \pi^{2} \rangle \supsetneq \cdots,$ and so $\big( A, \langle \pi \rangle \big)$ is a [[Noetherian ring|Noetherian]] [[local ring|local]] [[integral domain|domain]]. Explicitly, $\langle \pi^{n} \rangle=\{ x \in A: \nu(x) \geq n \}$. >4. $\text{Spec }A=\{ (0), \langle \pi \rangle \}$, and so $\text{dim }A=1$. > This shows, in particular, that: > > Every [[DVR]] is a [[Noetherian ring|Noetherian]] [[local ring|local]] [[integral domain|domain]] of [[Krull dimension|dimension]] $1$. ^proposition > [!proposition] Corollary. > Together, $(1)$ and $(3, \text{locality})$ imply that $x \in A$ satisfies $\nu(x) > 0$ iff $x \in \mathfrak{m}$, where $\mathfrak{m}$ is the unique [[maximal ideal]] of $A$. ^proposition > [!NOTE] Remark. > What conditions need to be added for the converse to hold? This is analyzed in [[characterization of DVRs]]. One condition is [[regular local ring|regularity]]: $\text{DVR} \iff \text{\textcolor{Thistle}{regular} Noetherian local domain of dimension 1} .$ > [!proof]- Proof. ([[every DVR is a Noetherian local domain of dimension 1]]) > **1.** If $x=0$ then $\nu(x)=\infty \neq 0$. Now, take $x \neq 0$. Then $x \in A^{*} \iff x ^{-1} \in A \iff \underbrace{ \nu(x ^{-1}) }_{ =-\nu(x) } \geq 0 \iff \nu(x) \leq 0 \iff \nu(x)=0.$ **2.** $A$ is a [[subring]] of the [[field]] $K$ > > **3.** Let $\mathfrak{a} \neq 0$ be an [[ideal]] of $A$. Put $k=\min\{ \nu(x): x \in \mathfrak{a} \}$. We'll show $\mathfrak{a}=\langle \pi^{k} \rangle$. Since $\mathfrak{a} \neq 0$, $k < \infty$. For $x \in \mathfrak{a}$, we have $\nu(x \pi^{-k})=\nu(x)-k\nu(\pi)=\nu(x)-k$. $\nu(x) - k \geq 0$, and so $x \pi^{-k} \in A$, thus $x=x \pi^{-k} \cdot \pi^{k} \in \langle \pi^{k} \rangle$. So $\mathfrak{a} \subset \langle \pi^{k} \rangle$. On the other hand, for $x_{0} \in \mathfrak{a}$ such that $\nu(x_{0})=k$, we have $\nu(\pi^{k} x_{0} ^{-1})=k-k=0$, meaning that $\pi^{k}=\pi^{k} x_{0}^{-1} x_{0} \in \mathfrak{a}$. Thus $\langle \pi^{k}\rangle \subset \mathfrak{a}$. So $\mathfrak{a}=\langle \pi^{k} \rangle$. > > Explicitly, $\langle \pi^{n} \rangle=\{ x \in A : \nu(x) \geq n \}$. To see this, it is enough to check that $x \in \langle \pi^{n} \rangle \iff x \pi^{-n} \in A$. This is clear: if $x \in \langle \pi^{n} \rangle$, say, $x=a \pi^{n}$ for some $a \in A$, then $x \pi^{-n}=a \pi^{n}\pi^{-n}=a \in A$. Conversely, suppose $x \pi^{-n} \in A$. Now, $\langle x \rangle=\langle \pi^{m} \rangle$ for some $m \geq 0$, and so since $\langle \pi^{m} \rangle x \pi^{-n} \subset \langle \pi^{m} \rangle$, we have $x \pi^{m} \pi^{-n}=a \pi^{m}$ for some $a \in A$. Then $x=a \pi^{n} \in \langle \pi^{n} \rangle$. (That's a lot of writing for something rather clear.) > > > The explicit definition of $\langle \pi^{n} \rangle$ shows the inclusions are strict, and we are done. > > **4.** $(0)$ is prime and $\langle \pi \rangle$ is maximal by the points above. So certainly $\text{Spec }A \supset \{ (0), \langle \pi \rangle \}$. In fact it contains nothing else: the other ideals are $\pi^{k}$ for $k \geq 2$, but none of these are prime because e.g. $\pi^{k} \in \langle \pi^{k} \rangle$ while $\pi \notin \langle \pi^{k} \rangle$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```