Decomposition Inertia and Frobenius

Use Cauchy-sequence classes with distance $\lim \mathrm{\mid }{x}_n-y_n\mathrm{\mid }$. Addition and multiplication are defined termwise: Cauchy sequences are bounded, and $\mathrm{\mid }{x}_n{y}_n-x_m{y}_m\mathrm{\mid }\le \mathrm{\mid }{x}_n\mathrm{\mid }\mathrm{\mid }{y}_n-y_m\mathrm{\mid }+\mathrm{\mid }{y}_m\mathrm{\mid }\mathrm{\mid }{x}_n-x_m\mathrm{\mid }$ proves that products are Cauchy and independent of representatives. Addition is treated by the triangle inequality. Field identities follow termwise, and $\mathrm{\mid }[x_n]\mathrm{\mid }=\lim \mathrm{\mid }{x}_n\mathrm{\mid }$ is multiplicative and positive definite.

For a nonzero class x, eventually $\mathrm{\mid }{x}_n\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2>0$. The tail reciprocals are Cauchy because $\mathrm{\mid }{x}_n^{-1}-x_m^{-1}\mathrm{\mid }=\mathrm{\mid }{x}_n-x_m\mathrm{\mid }\mathrm{/}(\mathrm{\mid }{x}_n\mathrm{\mid }\mathrm{\mid }{x}_m\mathrm{\mid })$; finitely many initial entries may be set to one. Their class is the inverse of x. Zero and one are the constant classes, so this proves the field structure on the complete metric space.

An isometric field map to a complete field extends uniquely as a continuous map by the metric universal property. Taking limits of sums and products shows that the extension is a field map; taking distance limits shows it is an isometry. Density forces uniqueness of all these operations and of the extending map. The general completion theorem is used with its usual countable choices of representatives; no choice-free assertion for arbitrary F is inferred.

In the nonarchimedean case the strong triangle inequality passes to limits. If $x\mathrm{\ne }0$ in the completion, choose $a\in F$ with $\mathrm{\mid }x-a\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }$; the strong inequality applied in both directions gives $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }x\mathrm{\mid }$. Thus no new nonzero values appear. If $\mathrm{\mid }x\mathrm{\mid }\le 1$, approximation with $\mathrm{\mid }x-a\mathrm{\mid }<1$ has $\mathrm{\mid }a\mathrm{\mid }\le 1$ and gives the same residue. The kernel of the map of original valuation rings on residues is exactly $\mathrm{\mid }a\mathrm{\mid }<1$, proving the residue-field isomorphism. A trivial value gives the discrete already-complete field.

The norm axioms imply $\mathrm{\parallel }\sum a_i{v}_i\mathrm{\parallel }\le (\sum \mathrm{\parallel }{v}_i\mathrm{\parallel })\max \mathrm{\mid }{a}_i\mathrm{\mid }$ when n is positive. In dimension zero completeness and comparison are immediate. In dimension one $\mathrm{\parallel }a{v}_1\mathrm{\parallel }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }{v}_1\mathrm{\parallel }$ gives both bounds and completeness.

Proceed by finite induction. Assume the result for smaller dimensions. Every coordinate hyperplane $H_i$ is complete in its restricted norm and therefore closed: a point in its closure is approached by a sequence within distance 1/n, a Cauchy sequence whose limit in $H_i$ equals that point. Put $c_i={\inf }_{h\in H_i}\mathrm{\parallel }{v}_i-h\mathrm{\parallel }>0$, since the complement of the closed hyperplane is open. Translation and scaling by a nonzero scalar give ${\inf }_{h\in H_i}\mathrm{\parallel }{a}_i{v}_i-h\mathrm{\parallel }=\mathrm{\mid }{a}_i\mathrm{\mid }{c}_i$; the assertion is also valid for $a_i=0$.

For $v=\sum a_j{v}_j$, subtract its other coordinates to obtain $\mathrm{\mid }{a}_i\mathrm{\mid }{c}_i\le \mathrm{\parallel }v\mathrm{\parallel }$. Thus $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\le ({\max }_i{c}_i^{-1})\mathrm{\parallel }v\mathrm{\parallel }$, completing the induction. A Cauchy sequence has coordinatewise limits in F and converges by the upper bound, so V is complete. Every subspace, being finite-dimensional, is complete by the same argument and hence closed. No compactness of a unit sphere or nontrivial scalar valuation was assumed.

Two extending absolute values are norms on the finite-dimensional F-vector space E. Norm equivalence gives constants $c,C>0$ with $c\mathrm{\mid }x{\mathrm{\mid }}_1\le \mathrm{\mid }x{\mathrm{\mid }}_2\le C\mathrm{\mid }x{\mathrm{\mid }}_1$ for every x. It also gives completeness for either norm.

For $x\mathrm{\ne }0$, apply the comparison to $x^n$ and take nth roots: $c^{1\mathrm{/}n}\mathrm{\mid }x{\mathrm{\mid }}_1\le \mathrm{\mid }x{\mathrm{\mid }}_2\le C^{1\mathrm{/}n}\mathrm{\mid }x{\mathrm{\mid }}_1$. Let n tend to infinity to obtain equality. Both values of zero are zero. This includes the trivial valuation and E=F, and asserts uniqueness only if an extension exists.

If the valuation is trivial, A=k=F and the original factorization suffices. If $h_0=1$, take h=1 and H=g. Otherwise put $m=\mathrm{deg}{h}_0>0$, $N=\mathrm{deg}g$, lift $h_0$ to a monic $h_1$ of degree m and $H_0$ to $H_1$ of degree at most N-m. Lift a Bezout relation to polynomials r,s with $h_{1}r+H_{1}s\equiv 1$ modulo the maximal ideal. Among the finitely many nonzero coefficients of $g-h_1{H}_1$ and $h_{1}r+H_{1}s-1$, choose one of maximum absolute value, or any element with value strictly between zero and one if both errors vanish. Denote it by pi. Then both errors lie in $I[T]$, where $I=(\pi )=\{x:\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }\pi \mathrm{\mid }\}$ and $0<\mathrm{\mid }\pi \mathrm{\mid }<1$.

Suppose $g-h_n{H}_n\in I^n[T]$, $h_n,H_n\equiv h_1,H_1$ modulo I, with the stated degree bounds. Put $e_n=(g-h_n{H}_n)\mathrm{/}{\pi }^n\in A[T]$. Over A/I we need $q{H}_1+Q{h}_1=e_n$. Multiply the fixed Bezout relation by $e_n$; then divide $e_{n}s$ by the monic $h_1$, writing $e_{n}s=h_{1}u+q$, $\mathrm{deg}q<m$. Take $Q=e_{n}r+u{H}_1$. Modulo I, monicity of $h_1$ and $\mathrm{deg}(e_n-q{H}_1)\le N$ imply $\mathrm{deg}Q\le N-m$. Delete higher coefficients of Q, which lie in I. This explicitly solves the congruence with bounded degrees.

Set $h_{n+1}=h_n+{\pi }^{n}q$ and $H_{n+1}=H_n+{\pi }^{n}Q$. Their product equals g modulo $I^{n+1}$, since $2n\ge n+1$, and all degree bounds persist. This deterministic correction uses only the initial finite lifts and polynomial division, so no new arbitrary residue representatives are chosen at successive stages. The finitely many coefficient sequences are Cauchy because $\mathrm{\mid }\pi {\mathrm{\mid }}^n\to 0$. Completeness gives limits h,H in A[T], with h monic of degree m, and continuity of finite multiplication gives g=hH and the prescribed reductions.

For a simple root $\bar{a}$ of monic $\bar{f}$, apply the factorization to $(T-\bar{a})H_0$ with $H_0(\bar{a})\mathrm{\ne }0$. Write $f=(T-a)H$. Then $H(a)$ is a unit, so $f^{\mathrm{\prime }}(a)\mathrm{\ne }0$. If b is another root with $\bar{b}=\bar{a}$, then $H(b)$ is a unit and $(b-a)H(b)=0$ forces b=a. This proves both existence and uniqueness of the simple-root lift.

Suppose some coefficient has value greater than one. Choose one of maximal value, say $a_j$, and put $g=f\mathrm{/}{a}_j$. All coefficients of g lie in A, its constant and leading coefficients reduce to zero, and some intermediate coefficient reduces to a nonzero element. Consequently $\bar{g}=T^r{H}_0$ with $1\le r\le \mathrm{deg}\bar{g}<\mathrm{deg}f$ and $H_0(0)\mathrm{\ne }0$.

The two residue factors are coprime, so nonmonic Hensel lifting gives $g=hH$ with h monic of degree r. Both factors have positive degree, since $0<r<\mathrm{deg}g$. Multiplying back by $a_j$ contradicts irreducibility. Thus no coefficient exceeds one. Degree one and f=T are included without contradiction to the zero constant-term case.

Put $n=[L:K]$ and $v(x)=\mathrm{\mid }{N}_{L\mathrm{/}K}(x){\mathrm{\mid }}^{1\mathrm{/}n}$. The determinant interpretation shows that v vanishes exactly at zero and $v(a)=\mathrm{\mid }a\mathrm{\mid }$ for $a\in K$, since multiplication by a is a scalar n by n matrix. Norm multiplicativity gives $v(xy)=v(x)v(y)$.

For $v(x)\le 1$, let $E=K(x)$ and m=[E:K]. The tower law and determinant of the scalar E-linear action give $N_{L\mathrm{/}K}(x)=N_{E\mathrm{/}K}(x{)}^{[L:E]}$. The companion matrix of multiplication by x shows that its norm is $(-1{)}^{m}f(0)$ for the monic minimal polynomial f. Thus $\mathrm{\mid }f(0)\mathrm{\mid }\le 1$, so all coefficients lie in the valuation ring. It follows that $\mathrm{\mid }f(-1)\mathrm{\mid }\le 1$, and the same companion-matrix calculation for x+1 gives $v(x+1)\le 1$.

For y nonzero and $v(x)\le v(y)$, apply the preceding step to x/y to obtain $v(x+y)\le v(y)$. If y=0 there is nothing to prove; interchange x,y when needed. This proves the strong triangle inequality. The uniqueness lemma gives uniqueness and completeness. Its proof applies in arbitrary characteristic; no conjugate count or separability was used. For the trivial base value the norm formula is identically one on nonzero elements.

A nontrivial absolute value on K cannot restrict trivially to the rationals. If it did, for each algebraic x and all n, reduction by its fixed minimal equation would express $x^n$ in a fixed finite list of powers of x with rational coefficients of value at most one. Thus $\mathrm{\mid }x{\mathrm{\mid }}^n$ is bounded independently of n and $\mathrm{\mid }x\mathrm{\mid }\le 1$. Apply the same argument to $x^{-1}$ to obtain $\mathrm{\mid }x\mathrm{\mid }=1$ for nonzero x. Ostrowski and the equivalence characterization therefore leave precisely a p-adic restriction up to positive power, or the usual archimedean restriction up to positive power.

If the restriction is a positive power of the $p$-adic value, then all integers have value at most one. The binomial theorem gives $\mathrm{\mid }x+y{\mathrm{\mid }}^n\le (n+1)\max (\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }{)}^n$; taking $n$th roots and letting $n\to \mathrm{\infty }$ proves the ultrametric inequality on $K$. For this nonarchimedean value, every algebraic integer has value at most one: otherwise its leading term in a monic integral equation would have strictly greater value than the sum of the others. Then $P=\{a\in {\mathcal{O}}_K:\mathrm{\mid }a\mathrm{\mid }<1\}$ is a proper prime ideal, and it is nonzero since it contains the rational prime p. By [F10], ${\mathcal{O}}_K$ is Dedekind, so [F6] applies and $({\mathcal{O}}_K{)}_P$ is a DVR. Elements outside P have value one; every nonzero element of K is $u{\varpi }^m$ with u a local unit, so its value is $\mathrm{\mid }\varpi {\mathrm{\mid }}^m$, $0<\mathrm{\mid }\varpi \mathrm{\mid }<1$. Hence its place is exactly P's valuation place. The elements of ${\mathcal{O}}_K$ with value less than one recover P, proving uniqueness. Conversely a DVR valuation defines that nontrivial absolute-value place.

In the archimedean case write the restriction as the ordinary value to a power $0<a\le 1$; a>1 is excluded by the triangle inequality on positive integers. The completion contains the complete real field with this powered value. The image V of $\mathbb{R}{\otimes }_{\mathbb{Q}}K$ in the completion is a finite-dimensional normed real vector space over that valued real field. It is complete, hence closed, and contains the dense K, so equals the completion. V is a finite-dimensional real domain and thus a field (multiplication by any nonzero element is injective and hence surjective). By the fundamental theorem of algebra, real irreducible polynomials have degree at most two: after adjoining one nonreal element one gets $\mathbb{C}$, which has no proper finite algebraic extensions. Therefore V is $\mathbb{R}$ or $\mathbb{C}$. Uniqueness of extending values identifies its value with the ordinary modulus to the a-th power.

Each embedding into the reals or complexes gives such a place. If two embeddings give equivalent places, normalize their restriction to the same real power. The equivalence exponent must then be one on the rationals; their completions are isometrically isomorphic fixing the dense K and hence the reals. A real automorphism of the complexes sends i to i or -i, so these embeddings are equal or conjugate. Conversely conjugation preserves modulus. Finally $NP=p^f$ and ${\mathrm{ord}}_P(x)=e{\mathrm{ord}}_p(x)$ for rational x (rational units at p are local units) give the stated exponent ef.

The primitive-element theorem supplies alpha if needed. The presentation $L=K[T]\mathrm{/}(F)$ remains $K_p[T]\mathrm{/}(F)$ after scalar extension, as is seen on the power basis. Separability gives a Bezout identity for F,F' over K, hence over $K_p$, so the irreducible factors remain distinct. Polynomial CRT gives the product of factor fields.

Each factor E has the unique extending absolute value and is complete. Its element ${\alpha }_i=T<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>F_i$ generates E over $K_p$. Approximating each coefficient of a finite polynomial in ${\alpha }_i$ by elements of K shows that the image of L in E is dense. Thus E is the completion of the induced nonarchimedean place on L. Its restriction to K is the p-adic place, so [F6] classifies it by a unique prime P of ${\mathcal{O}}_L$ above p.

Conversely the inclusion K into $L_P$, using the extending power normalization, extends to $K_p$. The natural algebra map $L{\otimes }_K{K}_p\to L_P$ has finite-dimensional image over $K_p$. With its inherited norm this image is complete and therefore closed; it also contains the dense L. Hence the map is surjective onto the field $L_P$, and so factors through exactly one of the displayed factor fields. Two factors cannot induce the same place: equivalent extending values agree on K and hence have exponent one, so the completion isometry fixes K and alpha and, by density, $K_p$; the minimal polynomial of alpha over $K_p$ would then be the same factor. This establishes the bijection.

Dimensions in the finite product add to the degree of F. For x in L, scalar extension of its multiplication matrix preserves its determinant and trace; in the product it becomes block diagonal with the local multiplication matrices. The determinant of a block diagonal matrix is the product of its block determinants and its trace their sum. This proves the norm and trace formulas, including x=0 and degree one.

Factor $p{\mathcal{O}}_L$ into its finite nonempty set S of prime divisors. These primes are maximal, hence distinct ones are comaximal; G permutes S because it fixes p. If S had two orbits A,B, CRT would give $\alpha \in {\mathcal{O}}_L$ with residue zero at every prime in A and residue one at every prime in B. The remaining finitely many primes may also be assigned residue one.

The product $c={\prod }_{\sigma \in G}\sigma (\alpha )=N_{L\mathrm{/}K}(\alpha )$ belongs to K and is integral, hence belongs to ${\mathcal{O}}_K$. At a prime of A every factor is zero modulo that prime; at a prime of B every factor is one, because each inverse image prime is again in B. Thus $c\in p$ but $c\equiv 1$ at a prime above p, a contradiction. There is only one orbit.

If $\sigma P=P^{\mathrm{\prime }}$, applying $\sigma$ to the unique factorization of $p{\mathcal{O}}_L$ preserves the exponent of P. The induced map ${\mathcal{O}}_L\mathrm{/}P\to {\mathcal{O}}_L\mathrm{/}{P}^{\mathrm{\prime }}$ fixes ${\mathcal{O}}_K\mathrm{/}p$ and preserves the residue degree. Transitivity therefore makes both e and f constant.

The fundamental identity becomes ${\sum }_{P\mid p}ef=gef=[L:K]$. All three integers are positive; degree one gives e=f=g=1.

Every element of D preserves ${\mathrm{ord}}_P$, so preserves the absolute value at P and extends uniquely to the completion. It fixes $K_p$ by density of K. Extension is an injective homomorphism since L embeds in its completion.

Write $L=K(\alpha )$. The local factor description gives $L_P=K_p(\alpha )$ and a separable minimal polynomial over $K_p$ dividing the global minimal polynomial. Since L/K is normal, all global roots already lie in L. Thus the local polynomial splits in $L_P$, which proves that this finite extension is Galois.

For each prime Q above p the injection in the first step gives $\mathrm{\mid }D(Q\mathrm{/}p)\mathrm{\mid }\le [L_Q:K_p]$. Transitivity identifies all stabilizer orders with $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}g$ by orbit-stabilizer. Summing these inequalities over the g primes gives $\mathrm{\mid }G\mathrm{\mid }\le {\sum }_Q[L_Q:K_p]=[L:K]=\mathrm{\mid }G\mathrm{\mid }$. Thus every inequality is equality. In particular the injection at P accounts for every local automorphism, since the local extension is Galois. Each is consequently the continuous extension of a unique element of D; its restriction is that element.

Orbit-stabilizer on the transitive prime set gives $\mathrm{\mid }D\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}g$. Since $efg=[L:K]=\mathrm{\mid }G\mathrm{\mid }$, this order, and therefore the local Galois degree, is ef. If L=K the maps and groups are identities.

Choose a generator u of the finite cyclic group $\kappa (P{)}^{\times }$. CRT gives $\alpha \in {\mathcal{O}}_L$ reducing to u at P and to zero at every other prime above p. This also works when $\kappa (P)={\mathbb{F}}_2$, with u=1, or when P is the only prime.

The orbit polynomial $H(T)={\prod }_{\tau \in G}(T-\tau \alpha )$ has integral G-invariant coefficients, hence belongs to ${\mathcal{O}}_K[T]$. Its reduction has coefficients in $\kappa (p)$, so $H(u^q)=H(u{)}^q=0$ in $\kappa (P)$. Its displayed linear factorization implies $u^q=\overline{\sigma \alpha }$ for some $\sigma \in G$.

If ${\sigma }^{-1}P\mathrm{\ne }P$, alpha is zero at ${\sigma }^{-1}P$, giving $\overline{\sigma \alpha }=0$, contrary to $u^q\mathrm{\ne }0$. Thus $\sigma \in D$. On every nonzero residue $u^j$ it acts by $(u^q{)}^j=(u^j{)}^q$; it fixes zero as well. This is the asserted residue action.