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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.

Put $K=\mathbb{C}((t))$. This is a complete discretely valued field, with valuation ring $\mathbb{C}[[t]]$ and residue field $\mathbb{C}$; its infinite residue field means it is not a locally compact local field. The finite-extension facts needed here are uniqueness of the extended valuation, completeness of the extension, and $[L:K]=ef$ for complete discretely valued fields with perfect residue field. They apply in this characteristic-zero setting.

Let $[L:K]=n$. The finite residue extension of $\mathbb{C}$ must be $\mathbb{C}$ itself, so $f=1$ and $e=n$. Normalize $v_L$ integrally and choose a uniformizer $\pi$. Since $v_L(t)=n$, write $t=u{\pi }^n$ with $u\in {\mathcal{O}}_L^{\times }$.

The residue of $u$ has an $n$th root $b_0\in {\mathbb{C}}^{\times }$. Apply Hensel lemma to $X^n-u$: its derivative at $b_0$ is the unit $n{b}_0^{n-1}$ because the residue characteristic is zero. Hence there is $v\in {\mathcal{O}}_L^{\times }$ with $v^n=u$. Setting $s=v\pi$ givesThe elements $1,s,\dots ,s^{n-1}$ are linearly independent over $K$. In a nontrivial relation ${\sum }_{i=0}^{n-1}{a}_i{s}^i=0$, the nonzero summands have valuations $n{v}_K(a_i)+i$, distinct modulo $n$, so exactly one has least valuation and cannot cancel. Since $L$ has dimension $n$, this proves $L=K(s)$.

The constants already embed $\mathbb{C}$ into ${\mathcal{O}}_L$ with residue field $\mathbb{C}$. Subtracting the residue coefficient and dividing by $s$ repeatedly, exactly as in Question 1(c), expresses every integral element uniquely as ${\sum }_{j\ge 0}{c}_j{s}^j$ with $c_j\in \mathbb{C}$. Completeness and the first-nonzero-coefficient valuation show that allowing finite negative powers givesThis proof of finite extensions of the complex Laurent series field did not need a Galois hypothesis. All $n$ distinct $n$th roots of unity lie in the constant field, and the substitutions $s\mapsto \zeta s$ preserve $s^n=t$. They give all $n$ automorphisms, so every such finite extension is cyclic Galois:

Choose compatible roots $s_n=t^{1/n}$ in an algebraic closure, first choosing them successively along the factorial-degree tower and then taking powers to define the other roots, with $s_n=s_m^{m/n}$ whenever $n\mid m$. The fields $K_n=K(s_n)$ are nested under divisibility. Every finite extension inside the closure equals one of them: its constructed root $s$ of $X^n-t$ differs from $s_n$ by a constant root of unity. Therefore their union is the entire algebraic closure. Explicitly, every algebraic element generates a finite extension, so lies in the union; any polynomial over the union has its coefficients in one finite extension, and its roots are still algebraic over $K$, hence also lie there.

For $n\mid m$, restriction of automorphisms sends ${\zeta }_m\in {\mu }_m$ to ${\zeta }_m^{m/n}\in {\mu }_n$. Using the compatible primitive roots $\exp (2\pi i/n)$, these are the reduction maps on exponent groups. Thus the absolute Galois group of the complex Laurent series field isThe inverse limit carries its profinite topology; choosing compatible roots gives the displayed identification with the profinite completion of the integers.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfyingA Non-Archimedean absolute value satisfies the stronger ultrametric inequality $|x+y|\le \max (|x|,|y|)$. Two equivalent absolute values induce the same topology, or equivalently differ by a positive real power. The trivial absolute value on a field takes value one on every nonzero element. The rational classification and the compactness criterion below concern nontrivial absolute values on a field; the trivial exceptions are given explicitly.

Here the additive valuation has real values. For any $c>1$, the mutually inverse constructions areMultiplicativity becomes $v(xy)=v(x)+v(y)$, and the ultrametric inequality becomes $v(x+y)\ge \min (v(x),v(y))$. Equivalent real-valued valuations differ by positive scaling, so these constructions give the required bijection on equivalence classes. Changing $c$ merely rescales the valuation.

To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to 0$. Thus two nontrivial absolute values on a field with the same topology give the same strict positivity relation on their additive valuations. Fix $t$ with $v_1(t)>0$. Comparing the signs of $m{v}_j(t)-n{v}_j(x)=v_j(t^m{x}^{-n})$, for integers $m$ and positive integers $n$, shows that $v_1(x)/v_1(t)$ and $v_2(x)/v_2(t)$ have identical rational cuts. They are equal, proving $v_2=c{v}_1$ for $c>0$. If one allows valuations in arbitrary ordered groups, the nontrivial classes arising this way are precisely rank-one valuations: higher-rank ordered value groups do not embed order-preservingly in $\mathbb{R}$.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, $|n|\le 1$ for every integer $n$, by repeatedly applying the ultrametric inequality to sums of ones. Some prime $p$ must have $|p|<1$, otherwise prime factorization and multiplicativity would make every nonzero rational have value one. There is at most one such prime: if both $|p|,|q|<1$, a Bezout identity $ap+bq=1$ contradicts the ultrametric inequality. If $p\nmid m$, another Bezout identity gives $|m|=1$. HenceThis proves the non-Archimedean part of the Ostrowski theorem. If the trivial absolute value on a field is admitted, it supplies one additional class and is not equivalent to any p-adic absolute value.

The valuation ring, its maximal ideal, and its residue field areSuppose the absolute value on a field is nontrivial and $R$ is compact. The ideal $\mathfrak{m}$ is an open additive subgroup of $R$, so $k$ is discrete; as a continuous image of a compact space it is finite. The subgroup $\mathfrak{m}$ is also closed, since all its cosets are open, and is therefore compact. The continuous function $|\cdot |$ attains a maximum $\rho$ on $\mathfrak{m}$, with $0<\rho <1$. Choose $\pi$ with $|\pi |=\rho$. Then the positive values of $v=-\log |\cdot |$ have least element $-\log \rho$. Division with remainder in this additive subgroup of $\mathbb{R}$ proves $v(K^{\times })=v(\pi )\mathbb{Z}$. Thus $K$ is a discretely valued field and $\pi$ is a uniformizer.

Conversely, normalize the discrete valuation by $v(\pi )=1$. If $k$ has $q$ elements, $R/{\pi }^{N}R$ has $q^N$ elements. Each quotient therefore supplies a finite cover by balls of radius tending to zero. The ring $R$ is a closed subset of the complete metric field $K$, hence complete and totally bounded, so it is compact. Equivalently,This is the local compactness criterion for a complete non-Archimedean field, with nontriviality understood. For the trivial absolute value on a field, $R=K$ has the discrete topology and is compact exactly when $K$ is a finite field; its value group is zero rather than a nonzero discrete cyclic group.

If $K$ is a complete discretely valued field whose residue field $k$ is a perfect field, and $\pi$ is a uniformizer, the multiplicative Teichmuller lift $[\cdot ]:k\to {\mathcal{O}}_K$ gives each $x\in {\mathcal{O}}_K$ a unique expansion

In a mixed-characteristic discretely valued field, a primitive $p^a$th root of unity reduces to one. For every $j$ prime to $p$, the ratio $({\zeta }^j-1)/(\zeta -1)$ is a unit, so all factors of ${\Phi }_{p^a}(1)=p$ have the same valuation. Taking valuations gives the formula. Because the left side is a positive integer for a normalized valuation, the existence of such a root forces $p^{a-1}(p-1)\mid v(p)$, bounding the possible $p$-power torsion.

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.