[proofplan]
The proof combines two ingredients: the Galois equivariance of the absolute value and the transitivity of the Galois action on roots of an irreducible polynomial. Since $f$ is irreducible over $K$, all roots in a splitting field form a single orbit under $\operatorname{Aut}(L/K)$. By the theorem that [Galois Automorphisms Preserve the Absolute Value](/theorems/???), all roots have the same absolute value. A polynomial whose roots all have a single common valuation has a Newton polygon with a single line segment.
[/proofplan]
[step:Show all roots of $f$ have the same absolute value in the splitting field]
Let $L$ be a splitting field of $f$ over $K$, equipped with the unique absolute value $|\cdot|_L$ extending $|\cdot|$ on $K$ (which exists and is unique by the [Unique Extension of Absolute Values](/theorems/???), applied iteratively if needed to obtain a complete extension containing all roots).
Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f$ in $L$. Since $f$ is irreducible over $K$, for any two roots $\alpha_i, \alpha_j$ there exists a $K$-isomorphism $\sigma: K(\alpha_i) \to K(\alpha_j)$ with $\sigma(\alpha_i) = \alpha_j$ (this is the standard property of irreducible polynomials: $K(\alpha_i) \cong K[x]/(f) \cong K(\alpha_j)$). This isomorphism extends to an automorphism $\sigma \in \operatorname{Aut}(L/K)$ (since $L$ is a splitting field of $f$ over $K$, hence normal over $K$).
By [Galois Automorphisms Preserve the Absolute Value](/theorems/???), $|\sigma(\alpha_i)|_L = |\alpha_i|_L$. Since $\sigma(\alpha_i) = \alpha_j$:
\begin{align*}
|\alpha_j|_L = |\sigma(\alpha_i)|_L = |\alpha_i|_L.
\end{align*}
Since $i$ and $j$ were arbitrary, all roots $\alpha_1, \ldots, \alpha_n$ have the same absolute value.
[guided]
The argument uses two facts: (1) the Galois group acts transitively on the roots of an irreducible polynomial, and (2) Galois automorphisms preserve the absolute value.
**Fact (1):** If $f$ is irreducible over $K$, then $K[x]/(f)$ is a field, and the map $x \mapsto \alpha_i$ gives an isomorphism $K[x]/(f) \cong K(\alpha_i)$ for each root $\alpha_i$. Composing two such isomorphisms gives a $K$-isomorphism $K(\alpha_i) \cong K(\alpha_j)$ sending $\alpha_i$ to $\alpha_j$. Since $L$ is a splitting field (hence a normal extension of $K$), every $K$-embedding of $K(\alpha_i)$ into $\overline{K}$ lands inside $L$, so this isomorphism extends to $\sigma \in \operatorname{Aut}(L/K)$.
**Fact (2):** The [Galois Automorphisms Preserve the Absolute Value](/theorems/???) theorem states $|\sigma(\alpha)|_L = |\alpha|_L$ for all $\sigma \in \operatorname{Aut}(L/K)$ and $\alpha \in L$.
Combining: $|\alpha_j|_L = |\sigma(\alpha_i)|_L = |\alpha_i|_L$, so all roots have the same absolute value.
What if $f$ were reducible? Then the roots could split into multiple Galois orbits, potentially with different absolute values. For example, if $f = (x - p)(x - 1)$ over $\mathbb{Q}_p$, the roots $p$ and $1$ have $|p|_p = p^{-1} \neq 1 = |1|_p$. This corresponds to a Newton polygon with two segments of different slopes.
[/guided]
[/step]
[step:Deduce that the Newton polygon consists of a single segment]
Write $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ with $a_n \neq 0$ and $a_0 \neq 0$ (both are nonzero since $f$ is irreducible, hence has no root at $0$ unless $n = 1$ and $a_0 = 0$; but $a_0 a_n \neq 0$ follows from irreducibility when $n \geq 1$). WLOG $a_n = 1$ (dividing by $a_n$ shifts the Newton polygon vertically without changing the slopes or segment structure).
Let $v$ denote the valuation on $K$ (so $|\alpha| = c^{-v(\alpha)}$ for some constant $c > 1$). Let $\lambda := v(\alpha_i)$ be the common valuation of all roots (established in the previous step).
By Vieta's formulas, the coefficients of $f$ are elementary symmetric polynomials of the roots:
\begin{align*}
a_{n-k} = (-1)^k \sum_{1 \leq i_1 < \cdots < i_k \leq n} \alpha_{i_1} \cdots \alpha_{i_k}.
\end{align*}
Each product $\alpha_{i_1} \cdots \alpha_{i_k}$ has valuation $k\lambda$ (since all roots have the same valuation $\lambda$). The sum has $\binom{n}{k}$ terms, each of valuation $k\lambda$. By the strong triangle inequality:
\begin{align*}
v(a_{n-k}) \geq \min_{1 \leq i_1 < \cdots < i_k \leq n} v(\alpha_{i_1} \cdots \alpha_{i_k}) = k\lambda.
\end{align*}
Therefore the point $(n - k, v(a_{n-k}))$ in the Newton polygon satisfies $v(a_{n-k}) \geq k\lambda$, i.e., $(n-k, v(a_{n-k}))$ lies on or above the line through $(n, 0)$ with slope $-\lambda$. The endpoints of the Newton polygon are $(0, v(a_0))$ and $(n, v(a_n)) = (n, 0)$. For $k = n$: $v(a_0) = v((-1)^n \alpha_1 \cdots \alpha_n) = n\lambda$, so the point $(0, n\lambda)$ lies exactly on the line from $(n, 0)$ with slope $-\lambda$.
Since the Newton polygon is the lower convex hull of the points $(k, v(a_k))$, and every point lies on or above the line segment from $(0, n\lambda)$ to $(n, 0)$ (which has slope $-\lambda$), while both endpoints lie exactly on this line, the Newton polygon is exactly this single line segment with slope $-\lambda$ and horizontal length $n$.
[/step]
