[proofplan] We prove both directions of the equivalence. The forward direction shows that $\mathcal{O}_K$ for a discretely valued field $K$ is a DVR by verifying it is a PID with a unique maximal ideal generated by a uniformizer. The converse constructs a discrete valuation on the fraction field of a DVR by using the unique factorisation $x = \pi^n u$ (with $u$ a unit) to define $v(x) = n$, then verifies that the valuation ring of this valuation recovers $R$. [/proofplan] [step:Show that $\mathcal{O}_K$ for a DVF is a DVR] Let $(K, v)$ be a discretely valued field with valuation ring $\mathcal{O}_K = \{x \in K : v(x) \geq 0\}$ and uniformizer $\pi$ (so $v(\pi) = 1$). We verify that $\mathcal{O}_K$ is a DVR. Every nonzero $x \in \mathcal{O}_K$ can be written as $x = \pi^n u$ where $n = v(x) \geq 0$ and $u = x\pi^{-n}$ satisfies $v(u) = 0$, hence $u \in \mathcal{O}_K^\times$. This factorisation is unique: if $\pi^n u = \pi^m u'$ with $u, u' \in \mathcal{O}_K^\times$, then $v(\pi^n u) = n = m = v(\pi^m u')$, so $n = m$ and $u = u'$. Every ideal $I \subseteq \mathcal{O}_K$ is principal: if $I \neq (0)$, let $n_0 = \min\{v(x) : x \in I\}$ (this minimum exists since $v(x) \in \mathbb{Z}_{\geq 0}$ for $x \in \mathcal{O}_K \setminus \{0\}$). Then $I = (\pi^{n_0})$, because every $x \in I$ with $v(x) = n_0 + k$ ($k \geq 0$) satisfies $x = \pi^{n_0} \cdot \pi^k u \in (\pi^{n_0})$, and conversely any element achieving the minimum valuation generates $\pi^{n_0}$ up to a unit. The unique maximal ideal is $\mathfrak{m} = (\pi) = \{x \in \mathcal{O}_K : v(x) \geq 1\}$, and $\pi$ is the unique prime element up to units. Thus $\mathcal{O}_K$ is a PID with a unique prime, i.e., a DVR. [guided] Recall that a DVR is a principal ideal domain with exactly one nonzero prime ideal (equivalently, a local PID that is not a field). We verify each property: **$\mathcal{O}_K$ is an integral domain:** It is a subring of the field $K$, so it has no zero divisors. **Every ideal is principal:** Let $I \neq (0)$ be an ideal. Since $v$ takes values in $\mathbb{Z}_{\geq 0}$ on $\mathcal{O}_K \setminus \{0\}$, the set $\{v(x) : x \in I, x \neq 0\}$ is a nonempty subset of $\mathbb{Z}_{\geq 0}$ and hence has a minimum $n_0$. Pick any $y \in I$ with $v(y) = n_0$. Then $y = \pi^{n_0} u$ for some $u \in \mathcal{O}_K^\times$, so $\pi^{n_0} = yu^{-1} \in I$ and $(\pi^{n_0}) \subseteq I$. Conversely, for any $x \in I$ with $v(x) = n_0 + k$ ($k \geq 0$), we have $x = \pi^{n_0 + k} u' = \pi^{n_0}(\pi^k u')$ with $\pi^k u' \in \mathcal{O}_K$, so $x \in (\pi^{n_0})$. Hence $I = (\pi^{n_0})$. **Unique maximal ideal:** The ideals of $\mathcal{O}_K$ are exactly $\{(\pi^n) : n \geq 0\} \cup \{(0)\}$, forming a chain $(0) \subsetneq \cdots \subsetneq (\pi^2) \subsetneq (\pi) \subsetneq \mathcal{O}_K$. The unique maximal ideal is $(\pi)$, and $(\pi)$ is the only nonzero prime ideal. [/guided] [/step] [step:Construct a discrete valuation on the fraction field of a DVR] Let $R$ be a DVR with prime element $\pi$. Since $R$ is a PID (hence a UFD), every nonzero element $x \in R$ has a unique factorisation $x = \pi^n u$ with $n \geq 0$ and $u \in R^\times$. Define \begin{align*} v: R \setminus \{0\} &\to \mathbb{Z}_{\geq 0} \\ \pi^n u &\mapsto n. \end{align*} Let $K = \operatorname{Frac}(R)$ be the field of fractions of $R$. Every nonzero element of $K$ can be written as $\pi^n u$ with $n \in \mathbb{Z}$ and $u \in R^\times$: if $a/b \in K^\times$ with $a = \pi^{n_1} u_1$ and $b = \pi^{n_2} u_2$, then $a/b = \pi^{n_1 - n_2} (u_1 u_2^{-1})$. Extend $v$ to $K^\times$ by setting $v(\pi^n u) = n$ for $n \in \mathbb{Z}$, $u \in R^\times$, and $v(0) = \infty$. [guided] Why does $K = R[\pi^{-1}]$? Every element of $\operatorname{Frac}(R)$ has the form $a/b$ with $a, b \in R$, $b \neq 0$. Writing $b = \pi^m u$ with $u \in R^\times$, we get $a/b = a \cdot u^{-1} \cdot \pi^{-m}$. Since $a \in R$ and $u^{-1} \in R^\times \subset R$, the product $a u^{-1} \in R$, so $a/b \in R[\pi^{-1}]$. The other inclusion $R[\pi^{-1}] \subseteq \operatorname{Frac}(R)$ is immediate. Hence $K = R[\pi^{-1}] = \operatorname{Frac}(R)$. [/guided] [/step] [step:Verify that $v$ is a discrete valuation on $K$] We check the valuation axioms for $v: K^\times \to \mathbb{Z}$: **Multiplicativity:** For $x = \pi^m u$ and $y = \pi^n u'$ (with $u, u' \in R^\times$), we have $xy = \pi^{m+n}(uu')$ and $uu' \in R^\times$, so $v(xy) = m + n = v(x) + v(y)$. **Ultrametric inequality:** Let $x = \pi^m u$ and $y = \pi^n u'$ with $m \leq n$. Then $x + y = \pi^m(u + \pi^{n-m} u')$. The element $u + \pi^{n-m} u'$ lies in $R$ (since $u \in R^\times \subset R$ and $\pi^{n-m} u' \in R$). If $u + \pi^{n-m}u' = 0$, then $v(x+y) = \infty \geq \min(m,n)$. Otherwise, writing $u + \pi^{n-m}u' = \pi^k w$ with $k \geq 0$ and $w \in R^\times$, we get $v(x+y) = m + k \geq m = \min(v(x), v(y))$. **Discreteness:** The value group is $v(K^\times) = \mathbb{Z}$, generated by $v(\pi) = 1$. [guided] The ultrametric inequality deserves a closer look. When $m = n$, we have $x + y = \pi^m(u + u')$. The element $u + u'$ might be zero (if $u' = -u$), in which case $v(x + y) = \infty > m$. If $u + u' \neq 0$, then $v(u + u') \geq 0$ since $u, u' \in R$ and $R$ is closed under addition, so $v(x + y) = m + v(u + u') \geq m = \min(v(x), v(y))$. When $m < n$, we have $u + \pi^{n-m}u'$ with $n - m \geq 1$, so $\pi^{n-m}u' \in \mathfrak{m}_R$ and $u + \pi^{n-m}u' \equiv u \pmod{\mathfrak{m}_R}$. Since $u \in R^\times$ and $\mathfrak{m}_R$ does not contain any unit, $u + \pi^{n-m}u' \notin \mathfrak{m}_R$, hence $v(u + \pi^{n-m}u') = 0$ and $v(x + y) = m = \min(v(x), v(y))$. [/guided] [/step] [step:Show that $R = \mathcal{O}_K$ and conclude] The valuation ring of $(K, v)$ is \begin{align*} \mathcal{O}_K = \{x \in K : v(x) \geq 0\} = \{\pi^n u : n \geq 0,\, u \in R^\times\} \cup \{0\} = R, \end{align*} where the last equality holds because every element of $R$ has the form $\pi^n u$ with $n \geq 0$ and $u \in R^\times$ (by the unique factorisation in the DVR), and conversely every such element belongs to $R$. This shows $R \cong \mathcal{O}_K$ where $K = \operatorname{Frac}(R)$ is a discretely valued field with valuation $v$, completing the proof of the converse. [/step]