[proofplan] The proof has two parts: irreducibility and the characterization of totally ramified extensions via Eisenstein polynomials. For irreducibility, we use the Newton polygon: an Eisenstein polynomial has a single segment of slope $-1/n$, so any factorisation would require roots of fractional valuation, forcing the polynomial to be irreducible. For the characterization, we show that in a totally ramified extension $L/K$ the uniformizer $\pi_L$ generates $L$ over $K$ and its minimal polynomial is Eisenstein; conversely, any root of an Eisenstein polynomial is a uniformizer of a totally ramified extension. [/proofplan] [step:Prove irreducibility of an Eisenstein polynomial via the Newton polygon] Let $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathcal{O}_K[x]$ be Eisenstein, so $v_K(a_i) \geq 1$ for $0 \leq i \leq n-1$ and $v_K(a_0) = 1$. The Newton polygon of $f$ has vertices at $(i, v_K(a_i))$ for $0 \leq i \leq n-1$ and $(n, 0)$. Since $v_K(a_0) = 1$ and $v_K(a_i) \geq 1$ for $1 \leq i \leq n-1$, all points $(i, v_K(a_i))$ lie on or above the line segment from $(0, 1)$ to $(n, 0)$. By the [Irreducibility Forces a Single Segment](/theorems/???), the Newton polygon consists of a single segment of slope $-1/n$ from $(0,1)$ to $(n,0)$. Suppose $f = gh$ with $g, h \in K[x]$ monic of degrees $d$ and $n - d$ respectively, with $1 \leq d \leq n-1$. By Gauss's lemma (applied to the DVR $\mathcal{O}_K$), $g, h \in \mathcal{O}_K[x]$. The Newton polygon of $f$ is determined by the Newton polygons of $g$ and $h$: the slopes of $f$ are the union (with multiplicity) of the slopes of $g$ and $h$. Since $f$ has a single slope $-1/n$, the factors $g$ and $h$ must also have all slopes equal to $-1/n$. But the slopes of $g$ are rational numbers with denominator dividing $\deg g = d < n$, and a slope of $-1/n$ with $\gcd(1,n) = 1$ requires the denominator to be a multiple of $n$. Since $d < n$, this is impossible. Therefore $f$ is irreducible. [guided] The Newton polygon provides a valuative factorisation constraint. The Eisenstein conditions $v_K(a_i) \geq 1$ for $i < n$ and $v_K(a_0) = 1$ force the Newton polygon of $f$ to be a single line segment from $(0, v_K(a_0)) = (0, 1)$ to $(n, v_K(\text{leading coeff})) = (n, 0)$, with slope $-1/n$. Why does a single-segment Newton polygon of slope $-1/n$ imply irreducibility? Suppose $f = gh$ in $K[x]$ with $\deg g = d$, $1 \leq d \leq n-1$. Since $\mathcal{O}_K$ is a DVR (hence a UFD), Gauss's lemma gives $g, h \in \mathcal{O}_K[x]$. The Newton polygon of $f$ is the lower convex hull of the Newton polygons of $g$ and $h$; more precisely, the multiset of slopes of $f$ is the union of the slopes of $g$ and $h$. Since $f$ has all slopes equal to $-1/n$, every slope of $g$ equals $-1/n$. Now $g$ is a polynomial of degree $d$ in $\mathcal{O}_K[x]$, so its Newton polygon has slopes of the form $-a/b$ where $b \leq d$. A slope of $-1/n$ in lowest terms has denominator $n$ (since $\gcd(1,n) = 1$), but $d < n$, so no segment of $g$ can have slope $-1/n$. This contradiction shows $f$ is irreducible over $K$. [/guided] [/step] [step:Show that a totally ramified extension $L/K$ is generated by its uniformizer] Let $L/K$ be totally ramified of degree $n$, so $e_{L/K} = n$ and $f_{L/K} = 1$. Let $w$ denote the unique extension of $v_K$ to $L$ normalised so that $w(\pi_K) = 1$. By the [Ramification Index and the Extended Valuation](/theorems/???), $w(\pi_L) = 1/e_{L/K} = 1/n$. Consider the subextension $K(\pi_L) \subseteq L$. The element $\pi_L \in \mathfrak{m}_{K(\pi_L)}$, so \begin{align*} e_{K(\pi_L)/K}^{-1} = \min\{w(x) : x \in \mathfrak{m}_{K(\pi_L)}\} \leq w(\pi_L) = \frac{1}{n}. \end{align*} Therefore $e_{K(\pi_L)/K} \geq n$. Since $[K(\pi_L):K] \leq [L:K] = n$ and $e_{K(\pi_L)/K} \leq [K(\pi_L):K]$, we get $[K(\pi_L):K] = n = [L:K]$, so $L = K(\pi_L)$. [/step] [step:Prove the minimal polynomial of $\pi_L$ over $K$ is Eisenstein] Write the minimal polynomial of $\pi_L$ over $K$ as $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathcal{O}_K[x]$. Since $f(\pi_L) = 0$: \begin{align*} \pi_L^n = -(a_{n-1}\pi_L^{n-1} + \cdots + a_1\pi_L + a_0). \end{align*} Applying $w$ to both sides, $w(\pi_L^n) = n \cdot w(\pi_L) = 1$. On the right-hand side, the term $a_i \pi_L^i$ has $w$-value $v_K(a_i) + i/n$ (since $a_i \in K$ and $w|_K = v_K$, and $w(\pi_L) = 1/n$). By the ultrametric property, \begin{align*} 1 = w(\pi_L^n) = w\!\left(\sum_{i=0}^{n-1} a_i \pi_L^i\right) \geq \min_{0 \leq i \leq n-1}\!\left(v_K(a_i) + \frac{i}{n}\right). \end{align*} Equality must hold (since the left side equals $1$, the minimum on the right must be at most $1$, and the ultrametric inequality is an equality when the minimum is achieved by a unique term or by the [Isoceles Triangle Principle](/theorems/???)). For $1 \leq i \leq n-1$: if $v_K(a_i) = 0$, then $v_K(a_i) + i/n = i/n < 1$, which would make the minimum less than $1$, contradicting $w(\pi_L^n) = 1$. So $v_K(a_i) \geq 1$ for all $1 \leq i \leq n-1$, giving $v_K(a_i) + i/n \geq 1 + 1/n > 1$. For $i = 0$: the minimum must equal $1$, and since $v_K(a_i) + i/n > 1$ for $i \geq 1$, the minimum is achieved at $i = 0$, so $v_K(a_0) = 1$. Therefore $f$ is Eisenstein: $\pi_K \mid a_i$ for $0 \leq i \leq n-1$ and $v_K(a_0) = 1$ (equivalently $\pi_K^2 \nmid a_0$). [guided] We want to determine the coefficients of the minimal polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0$ of $\pi_L$ over $K$. The strategy is to use the extended valuation $w$ to read off the valuations of the coefficients. Since $f(\pi_L) = 0$, we have $\pi_L^n = -\sum_{i=0}^{n-1} a_i \pi_L^i$. The left-hand side has $w$-value $n \cdot w(\pi_L) = n \cdot (1/n) = 1$. Each term $a_i \pi_L^i$ on the right has $w$-value $v_K(a_i) + i/n$, since $a_i \in K$ (so $w(a_i) = v_K(a_i)$) and $w(\pi_L^i) = i/n$. The ultrametric inequality gives $w(\sum_{i=0}^{n-1} a_i\pi_L^i) \geq \min_i (v_K(a_i) + i/n)$. For the sum to have $w$-value exactly $1$, the minimum of $\{v_K(a_i) + i/n : 0 \leq i \leq n-1\}$ must be at most $1$. Now suppose some coefficient $a_i$ with $1 \leq i \leq n-1$ had $v_K(a_i) = 0$. Then $v_K(a_i) + i/n = i/n$, which lies strictly between $0$ and $1$. This term would have the smallest $w$-value among all terms, making the right-hand side have $w$-value $i/n < 1$. But the left-hand side has $w$-value $1$. This contradiction shows $v_K(a_i) \geq 1$ for $1 \leq i \leq n-1$, hence $v_K(a_i) + i/n \geq 1 + 1/n > 1$ for these indices. The minimum must then be achieved by the $i = 0$ term: $v_K(a_0) + 0 = v_K(a_0)$, and this minimum equals $1$ (matching the left-hand side). So $v_K(a_0) = 1$, which means $\pi_K \mid a_0$ but $\pi_K^2 \nmid a_0$. Combined with $v_K(a_i) \geq 1$ ($\pi_K \mid a_i$) for $i \geq 1$, this shows $f$ is Eisenstein. [/guided] [/step] [step:Prove the converse: a root of an Eisenstein polynomial generates a totally ramified extension] Let $g(x) = x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in \mathcal{O}_K[x]$ be Eisenstein with $\alpha$ a root, and set $L = K(\alpha)$ with $[L:K] = n$ (since $g$ is irreducible by the first step). Let $w$ be the unique extension of $v_K$ to $L$. The constant term satisfies $b_0 = (-1)^n \prod_{j} \alpha_j$ where $\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_n$ are the roots of $g$ in $\bar{K}$. All roots are conjugate over $K$, so they all have the same $w$-value (any $K$-automorphism $\sigma$ of $\bar{K}$ preserves $w$ by the [Galois Automorphisms Preserve the Absolute Value](/theorems/???)). From $w(b_0) = v_K(b_0) = 1$ and $w(b_0) = \sum_{j=1}^n w(\alpha_j) = n \cdot w(\alpha)$, we get \begin{align*} w(\alpha) = \frac{1}{n}. \end{align*} Since $\alpha \in \mathfrak{m}_L$, the ramification index satisfies \begin{align*} e_{L/K}^{-1} = \min\{w(x) : x \in \mathfrak{m}_L\} \leq w(\alpha) = \frac{1}{n} = [L:K]^{-1}. \end{align*} Therefore $e_{L/K} \geq n$. The fundamental inequality $e_{L/K} \cdot f_{L/K} \leq [L:K] = n$ forces $e_{L/K} = n$ and $f_{L/K} = 1$. So $L/K$ is totally ramified and $\alpha$ is a uniformizer of $L$ (since $w(\alpha) = 1/n = e_{L/K}^{-1} = w(\pi_L)$, the element $\alpha$ has minimal positive $w$-value in $\mathfrak{m}_L$). [guided] Given an Eisenstein polynomial $g(x) = x^n + b_{n-1}x^{n-1} + \cdots + b_0$ with root $\alpha$ and $L = K(\alpha)$, we need to show $e_{L/K} = n$. The idea is to compute $w(\alpha)$ from the product formula for the constant term. The constant term of $g$ is $b_0 = (-1)^n \alpha_1 \cdots \alpha_n$ where $\alpha_1, \ldots, \alpha_n$ are all roots. Since these roots are $K$-conjugates of $\alpha$, the [Galois Automorphisms Preserve the Absolute Value](/theorems/???) shows $w(\alpha_j) = w(\alpha)$ for all $j$ (any $K$-automorphism sending $\alpha$ to $\alpha_j$ preserves $w$ since $w$ is the unique extension of $v_K$). Therefore $w(b_0) = n \cdot w(\alpha)$. The Eisenstein condition gives $v_K(b_0) = 1$, and $w(b_0) = v_K(b_0) = 1$ since $b_0 \in K$. So $w(\alpha) = 1/n$. Now $\alpha \in \mathfrak{m}_L$ (since $w(\alpha) > 0$), and $w(\alpha) = 1/n \leq e_{L/K}^{-1}$ would require $e_{L/K} \leq n$. But we also have $e_{L/K}^{-1} \leq w(\alpha) = 1/n$ (since the minimum of $w$ on $\mathfrak{m}_L$ is $e_{L/K}^{-1}$ and $\alpha \in \mathfrak{m}_L$), giving $e_{L/K} \geq n$. Combined with the fundamental inequality $e_{L/K} \leq [L:K] = n$, we get $e_{L/K} = n$, so $L/K$ is totally ramified. Finally, $w(\alpha) = 1/n = e_{L/K}^{-1}$, which is the minimal positive value of $w$ on $L$, so $\alpha$ is a uniformizer of $L$. [/guided] [/step]