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If a monic degree-$n$ polynomial over a discretely valued field with normalized valuation has all its roots of valuation $a/n$, where $\gcd (a,n)=1$, then it is irreducible. Each irreducible factor of degree $d$ has constant term of valuation $da/n\in \mathbb{Z}$, forcing $n\mid d$. Equivalently, a single Newton polygon segment of relatively prime horizontal and vertical lengths gives irreducibility.

This polynomial is Eisenstein at $2$. Over ${\mathbb{Q}}_3$, its Newton polygon has a single slope $-2/5$, so it is irreducible and a root generates a totally ramified extension of degree five. Over ${\mathbb{Q}}_7$, its factor degrees are $1,1,3$: the simple residue roots $1,-1$ lift, and the remaining roots have valuation $1/3$. Thus the degree-five number field generated by a root has one prime above $3$ and three above $7$.

For a polynomial with nonzero constant coefficient over a complete discretely valued field, plot coefficient exponents against their valuations. A lower Newton polygon segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity in an algebraic closure. Zero coefficients are omitted. Reflecting the horizontal axis changes the sign of every slope, so slope conventions must be fixed before reading root valuations.

Normalize $v_K$ by $v_K({\pi }_K)=1$ and put $v_K(0)=+\infty$. Write $f(X)={\sum }_{j=0}^d{a}_j{X}^j$, with $a_d=1$ and $a_0\ne 0$. In the usual coefficient-exponent convention, the Newton polygon $N_K(f)$ is the lower boundary of the convex hull of the upward vertical rays starting atIt runs from $(0,v_K(a_0))$ to $(d,0)$, with nondecreasing slopes from left to right. Zero coefficients contribute no finite point. Equivalently it is the largest convex piecewise-linear function lying below all these coefficient points.

The Newton polygon root valuation theorem says that a segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity, using the extended valuation on an algebraic closure. The nonzero constant coefficient excludes a zero root.

For a short justification, use the weighted Gauss valuation $w_t(P)={\min }_j(v(c_j)+jt)$ for $P=\sum c_j{X}^j$. It is multiplicative: after scaling $X$ by an element of valuation $t$ in a valued extension, the initial nonzero residue polynomials multiply without vanishing. Rational $t$ suffice. Factoring a monic polynomial into linear factors gives $w_t(f)={\sum }_r\min (t,v(r))$. This supporting-line function changes derivative at precisely the root valuations; its derivative drops count their multiplicities. Supporting lines to the lower polygon have slope $-t$, proving the sign and horizontal-length assertion.

Slope convention for part (b). Reflecting the horizontal axis, by plotting $(d-j,v_K(a_j))$, produces the reflected Newton polygon convention; its slopes are the root valuations themselves. Part (b)'s wording with $m$ instead of $-m$ is correct under this positive-root-valuation convention. The proof below gives both forms explicitly, so the result does not depend on silently changing the sign.

The polynomial is Eisenstein at $2$: every nonleading coefficient is even, and its constant term $126$ is not divisible by $4$. Therefore it is irreducible over $\mathbb{Q}$ andBy part (a), the number of primes above a rational prime is the number of irreducible factors over the corresponding $p$-adic field.

We use the Newton polygon theorem in this form: a segment of slope $-r$ and horizontal length $d$ accounts for $d$ roots of valuation $r$, counted with multiplicity, in an algebraic closure with $v_p(p)=1$. We also use the denominator criterion for Newton polygon irreducibility: if all roots have valuation $a/n$ with $\gcd (a,n)=1$, any monic factor of degree $d$ has constant-term valuation $da/n\in \mathbb{Z}$, hence $n\mid d$.

At $3$, the nonzero coefficient points areThe lower hull is the single segment from $(0,2)$ to $(5,0)$, with slope $-2/5$. All five roots therefore have valuation $2/5$. A proper factor of degree $d$ would require $5\mid d$, which is impossible for $1\le d<5$. Thus $g$ is irreducible over ${\mathbb{Q}}_3$, givingThe completion has ramification index $5$ and residue degree $1$: the value $2/5$ forces its ramification index to be divisible by $5$, and its degree is $5$.

At $7$, the nonzero coefficient points areThe lower hull consists of a length-three segment of slope $-1/3$ and a length-two horizontal segment. Thus three roots have valuation $1/3$, and two have valuation zero. Reduction givesBoth $1$ and $-1$ are simple roots, so the Hensel lemma gives two distinct linear factors over ${\mathbb{Q}}_7$. The remaining cubic accounts for all the valuation-$1/3$ roots and is irreducible by the same denominator argument. Hence the local factor degrees are $1,1,3$, andThe two linear factors give $(e,f)=(1,1)$, and the cubic gives $(e,f)=(3,1)$. These data sum to degree five, as required. This computes the local prime decomposition of T5 plus 6T3 plus 252T2 plus 126. Counting distinct factors merely modulo $3$ would not justify the result, since that reduction is $T^5$.

LetOver ${\mathbb{Q}}_2$, the Newton polygon has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from Hensel lemma. There is one unit root because $F(1)$ is even and $F^{\prime }(1)$ is odd. For the two remaining roots, putwhose parenthesized polynomial has a simple root $Y\equiv 1(\mathrm{mod}2)$, andwhose parenthesized polynomial has a simple root $Z\equiv 1(\mathrm{mod}2)$. Thus $F$ splits completely over ${\mathbb{Q}}_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one

Over ${\mathbb{Q}}_5$, $F$ is an Eisenstein polynomial. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root thenwith $e=3$ and $f=1$.

The polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $5$. Its polynomial discriminant iswhich is not a square number, so the Galois group of an irreducible cubic isAt $2$, all three roots already lie in ${\mathbb{Q}}_2$, so the local splitting field is trivial andAt $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv 3(\mathrm{mod}5)$, a nonsquare unit, so the quadratic resolvent field of a cubic is the unramified quadratic extension of ${\mathbb{Q}}_5$. The local splitting field therefore has degree six, withThese calculations are summarized by local factorization of X3 plus 25X2 minus 50X plus 40.

Define the Lubin–Tate torsion byThe scalar endomorphisms make this a module over ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$.

The Newton polygon or Weierstrass preparation theorem applied to the Lubin–Tate congruences shows that $[{\pi }^n{]}_f$ has exactly $q^n$ distinct roots in $\overline{\mathfrak{m}}$. More precisely, the quotient of the distinguished factors for $[{\pi }^n{]}_f$ and $[{\pi }^{n-1}{]}_f$ has degree $q^{n-1}(q-1)$, and its roots are precisely the points killed by ${\pi }^n$ but not by ${\pi }^{n-1}$.

Choose such a point ${\omega }_n$. If $a\in {\mathcal{O}}_K$, write $a={\pi }^{r}u$ with $u$ a unit. Since $[u{]}_f$ is an automorphism, $[a{]}_f({\omega }_n)=0$ exactly when $r\ge n$. Thusis injective. Both sides have $q^n$ elements, so it is an isomorphism of modules. Therefore ${\mu }_{f,n}$ is a free module of rank one.

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

The ramification polynomial records the normalised differences of conjugates of a uniformizer, including the zero difference from the identity automorphism. Its quotient by $X$ is the reduced ramification polynomial. In the usual coefficient-index convention, the finite slopes of its lower Newton polygon are negatives of the valuations of its nonzero roots; the coefficient at the zero root is treated as having infinite valuation.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

For a polynomial over a discretely valued field and a rational weight $t$, this coefficient-minimum function is multiplicative. Scale the variable by an element of valuation $t$ in a valued extension and reduce the initial terms; nonzero residue polynomials have nonzero product. For a monic polynomial factored into roots, $w_t(f)={\sum }_r\min (t,v(r))$. Its supporting lines determine the lower Newton polygon, giving the Newton polygon root valuation theorem and its multiplicities.