[proofplan] The proof has two parts. For the forward direction, we show that a uniformizer $\pi_L$ of a totally ramified extension $L/K$ generates $L$ over $K$ by a degree argument using the ramification index, and then verify that its minimal polynomial is Eisenstein by analysing the valuations of its coefficients via the extended valuation. For the converse, we show that if $\alpha$ has an Eisenstein minimal polynomial of degree $n$, then $w(\alpha) = 1/n$, which forces $e_{L/K} = n = [L:K]$, giving total ramification and proving $\alpha$ is a uniformizer. [/proofplan] [step:Show $L = K(\pi_L)$ when $L/K$ is totally ramified] Let $n = [L:K]$ and let $w$ denote the unique extension of the normalised valuation $v_K$ to $L$, so that $w|_K = v_K$. Since $L/K$ is totally ramified, the ramification index satisfies $e_{L/K} = n$ and the residue degree is $f_{L/K} = 1$. The normalisation of $w$ on $L$ gives $w(\pi_L) = 1/n$ (since $w(\pi_K) = 1 = e_{L/K} \cdot w(\pi_L)$). The subextension $K(\pi_L)/K$ satisfies $[K(\pi_L) : K] \leq [L : K] = n$. On the other hand, since $\pi_L \in \mathfrak{m}_L$, we have $\pi_L \in \mathfrak{m}_{K(\pi_L)}$, and therefore \begin{align*} e_{K(\pi_L)/K} \geq \frac{1}{w(\pi_L)} = n, \end{align*} because the ramification index satisfies $e_{K(\pi_L)/K} \cdot w(\pi_{K(\pi_L)}) = 1$ and $w(\pi_{K(\pi_L)}) \leq w(\pi_L) = 1/n$ (since $\pi_L \in \mathfrak{m}_{K(\pi_L)}$ and the uniformizer has the smallest positive valuation). Since $e_{K(\pi_L)/K} \leq [K(\pi_L) : K] \leq n$, we conclude $[K(\pi_L) : K] = n$, hence $L = K(\pi_L)$. [guided] The key inequality is $e_{K(\pi_L)/K} \geq n$. Why does this hold? The ramification index $e_{K(\pi_L)/K}$ is defined as $w(\pi_K)/w(\pi_{K(\pi_L)})$ where $\pi_{K(\pi_L)}$ is a uniformizer of $K(\pi_L)$. Since $\pi_L$ belongs to the maximal ideal $\mathfrak{m}_{K(\pi_L)}$ of $\mathcal{O}_{K(\pi_L)}$, we have $w(\pi_L) \geq w(\pi_{K(\pi_L)})$ (a uniformizer has the smallest positive valuation in a DVR). Therefore: \begin{align*} e_{K(\pi_L)/K} = \frac{w(\pi_K)}{w(\pi_{K(\pi_L)})} \geq \frac{w(\pi_K)}{w(\pi_L)} = \frac{1}{1/n} = n. \end{align*} Combined with the fundamental inequality $e_{K(\pi_L)/K} \cdot f_{K(\pi_L)/K} \leq [K(\pi_L) : K] \leq n$, we get $e_{K(\pi_L)/K} = n$ and $[K(\pi_L) : K] = n$. [/guided] [/step] [step:Show the minimal polynomial of $\pi_L$ over $K$ is Eisenstein] Let $f(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 \in K[x]$ be the minimal polynomial of $\pi_L$ over $K$ (monic of degree $n = [K(\pi_L) : K]$). Since $f(\pi_L) = 0$: \begin{align*} \pi_L^n + c_{n-1}\pi_L^{n-1} + \cdots + c_1\pi_L + c_0 = 0. \end{align*} Applying $w$ and using $w(\pi_L) = 1/n$, the term $\pi_L^n$ has valuation $w(\pi_L^n) = 1$, while $c_i \pi_L^i$ has valuation $w(c_i\pi_L^i) = v_K(c_i) + i/n$ for $0 \leq i \leq n-1$. For the sum to vanish, the minimum valuation must be achieved by at least two terms (by the ultrametric property: if one term had strictly smaller valuation than all others, the sum could not be zero). The term $\pi_L^n$ has valuation $1$. For $0 \leq i \leq n-1$, the valuations $v_K(c_i) + i/n$ take values in $(1/n)\mathbb{Z}$, and $v_K(c_i) \in \mathbb{Z}$. We show $v_K(c_i) \geq 1$ for $1 \leq i \leq n-1$ and $v_K(c_0) = 1$: - **$v_K(c_0) = 1$:** The $i = 0$ term has valuation $v_K(c_0)$, which is an integer. For $1 \leq i \leq n-1$, the valuation $v_K(c_i) + i/n$ has fractional part $i/n \neq 0$ (since $0 < i < n$). The term $\pi_L^n$ has valuation $1$. For the minimum to be achieved at $i = 0$ and matched by $\pi_L^n$, we need $v_K(c_0) = 1$. More precisely: suppose $v_K(c_0) \geq 2$. Then every term except $\pi_L^n$ has valuation $> 1$ (since $v_K(c_i) + i/n \geq 0 + 1/n > 1$ for $v_K(c_i) \geq 1$ and $i \geq 1$, and $v_K(c_0) \geq 2 > 1$). But then $\pi_L^n$ would be the unique minimum-valuation term, contradicting $f(\pi_L) = 0$. Suppose $v_K(c_0) = 0$. Then the $i = 0$ term has valuation $0 < 1$, and it must be matched by some other term. But for $1 \leq i \leq n-1$, $v_K(c_i) + i/n \geq i/n > 0$, and $w(\pi_L^n) = 1 > 0$, so no term can match valuation $0$. This contradicts $f(\pi_L) = 0$. Hence $v_K(c_0) = 1$. - **$v_K(c_i) \geq 1$ for $1 \leq i \leq n-1$:** If $v_K(c_j) = 0$ for some $1 \leq j \leq n-1$, then $w(c_j\pi_L^j) = j/n < 1$, which is strictly less than $w(\pi_L^n) = 1$ and $w(c_0) = v_K(c_0) = 1$, and has fractional part $j/n$. No other term can have the same fractional part (terms with $i \neq j$ and $v_K(c_i) = 0$ would have fractional part $i/n \neq j/n$). So $c_j\pi_L^j$ would be the unique minimum-valuation term, again contradicting $f(\pi_L) = 0$. Hence $v_K(c_i) \geq 1$ for all $1 \leq i \leq n-1$. Therefore $f$ is Eisenstein: $v_K(c_i) \geq 1$ for $0 \leq i \leq n-1$ and $v_K(c_0) = 1$. [guided] The argument relies on the following consequence of the ultrametric inequality: in a sum $b_1 + \cdots + b_m = 0$ in a non-archimedean valued field, the minimum valuation must be achieved by at least two terms. Indeed, if $w(b_1) < w(b_j)$ for all $j \geq 2$, then $w(b_2 + \cdots + b_m) \geq \min_{j \geq 2} w(b_j) > w(b_1)$, so $w(b_1 + \cdots + b_m) = w(b_1) > -\infty$, contradicting the sum being zero. The key structural observation is that the valuations $v_K(c_i) + i/n$ for $0 \leq i \leq n-1$ and the valuation $1$ of $\pi_L^n$ must collectively allow at least two terms to share the minimum. Since $v_K(c_i) \in \mathbb{Z}$ and the fractional parts $i/n$ for $0 \leq i \leq n-1$ are all distinct, two terms $c_i\pi_L^i$ and $c_j\pi_L^j$ with $i \neq j$ (both in $\{0, \ldots, n-1\}$) can only have equal valuation if $v_K(c_i) + i/n = v_K(c_j) + j/n$, which requires $v_K(c_i) - v_K(c_j) = (j-i)/n$, but the LHS is an integer and the RHS is not (since $0 < |j-i| < n$). So no two terms among $c_0, c_1\pi_L, \ldots, c_{n-1}\pi_L^{n-1}$ can share the minimum valuation. The minimum must therefore be shared between exactly one of these and $\pi_L^n$ (valuation $1$). Only the $i = 0$ term can match valuation $1$ (since it has integer valuation), forcing $v_K(c_0) = 1$. [/guided] [/step] [step:Prove the converse: Eisenstein minimal polynomial implies total ramification] Suppose $L = K(\alpha)$ and the minimal polynomial $g(x) = x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in K[x]$ of $\alpha$ over $K$ is Eisenstein, i.e., $v_K(b_i) \geq 1$ for $0 \leq i \leq n-1$ and $v_K(b_0) = 1$. Then $[L:K] = n$. From $g(\alpha) = 0$: \begin{align*} \alpha^n = -(b_{n-1}\alpha^{n-1} + \cdots + b_1\alpha + b_0). \end{align*} We determine $w(\alpha)$. The valuations of the terms on the RHS are $w(b_i\alpha^i) = v_K(b_i) + i \cdot w(\alpha)$ for $0 \leq i \leq n-1$. Using the same non-archimedean argument as above (the fractional parts of $i \cdot w(\alpha)$ modulo $\mathbb{Z}$ must distinguish terms), the minimum on the RHS is achieved by $b_0$ alone (which has valuation $v_K(b_0) = 1$), provided $w(\alpha) > 0$. Matching with the LHS: $n \cdot w(\alpha) = 1$, giving \begin{align*} w(\alpha) = \frac{1}{n}. \end{align*} (We verify $w(\alpha) > 0$: since $g$ is Eisenstein, $\alpha$ is integral over $\mathcal{O}_K$ but $\alpha \notin \mathcal{O}_K^\times$ because $\alpha^n + b_{n-1}\alpha^{n-1} + \cdots + b_0 = 0$ with $b_0 \in \mathfrak{m}_K$ would force $\alpha \in \mathfrak{m}_L$ by reducing modulo $\mathfrak{m}_L$.) Since $w(\alpha) = 1/n$, the element $\alpha$ lies in $\mathfrak{m}_L$ and generates values $w(\alpha^j) = j/n$ for $0 \leq j \leq n-1$, which are $n$ distinct elements of $[0,1) \cap (1/n)\mathbb{Z}$. This forces \begin{align*} e_{L/K} \geq n. \end{align*} Since $e_{L/K} \cdot f_{L/K} = [L:K] = n$ and $e_{L/K} \leq n$, we conclude $e_{L/K} = n$ and $f_{L/K} = 1$. Hence $L/K$ is totally ramified, and $\alpha$ is a uniformizer of $L$ (since $w(\alpha) = 1/e_{L/K} = 1/n$ is the smallest positive valuation in $L$). [/step]