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Let $a=\sqrt{2+\sqrt{5}}>0$ and $b=i/a$, so $b^2=2-\sqrt{5}$. The roots are $\pm a,\pm b$. The element $2+\sqrt{5}$ is not a square in $\mathbb{Q}(\sqrt{5})$: its field norm is $-1$, whereas the field norm of a square is a square of a rational number and hence nonnegative. Thus $[\mathbb{Q}(a):\mathbb{Q}(\sqrt{5})]=2$, and $[\mathbb{Q}(a):\mathbb{Q}]=4$. This also proves irreducibility of the quartic.

The splitting field is $K=\mathbb{Q}(a,i)$, since $b=i/a$ and $i=ab$. Because $\mathbb{Q}(a)\subset \mathbb{R}$, adjoining $i$ doubles the degree:The Galois group acts faithfully on the four roots and preserves the partition into the two opposite pairs $\{a,-a\}$ and $\{b,-b\}$, possibly interchanging the pairs. The full permutation group preserving this partition has order eight and is the dihedral group of a square. Since the Galois group has order eight, it is exactly that group. Equivalently its generators can be chosen as $r:a\mapsto b,b\mapsto -a$ and complex conjugation $s:a\mapsto a,b\mapsto -b$, with $r^4=s^2=1$ and $srs=r^{-1}$.

Write $F_n=K_n^{+}$, $F_{\infty }={\bigcup }_n{F}_n$, $G=\mathrm{Gal}(F_{\infty }/\mathbb{Q})={\Delta }^{+}\times \Gamma$, where $|{\Delta }^{+}|=(p-1)/2$ and $\Gamma \simeq {\mathbb{Z}}_p$. Set $A={\mathbb{Z}}_p[[G]]$. Choosing a topological generator $\gamma$ of $\Gamma$ identifies its Iwasawa algebra with $\Lambda ={\mathbb{Z}}_p[[S]]$, $S=\gamma -1$; this group variable is distinct from the local-unit interpolation variable $T$. The Teichmüller character decomposes $A$ into a product of copies of $\Lambda$, one for each character of ${\Delta }^{+}$, equivalently each even Teichmüller character of $\mathrm{Gal}(K_0/\mathbb{Q})$. All characteristic-ideal statements below are componentwise in this product.

Let $U_n^1$ be the local principal units at the unique prime above $p$. Let $E_n^1$ and $C_n^1$ be respectively the p-adic closures of global units and cyclotomic units inside $U_n^1$, and take their inverse limits under the field norms. Define $X_{\infty }$ as the Galois group of the maximal abelian pro-p extension of $F_{\infty }$ unramified away from $p$, and $Y_{\infty }$ as that of the maximal everywhere unramified extension. This is the real tower: its prime-ramified module is torsion, by the Leopoldt theorem for abelian number fields. It must not be confused with the positive-rank prime-ramified module of the full complex cyclotomic tower. The unramified Iwasawa torsion theorem gives torsion for $Y_{\infty }$.

A convenient formulation of the Iwasawa main conjecture uses the p-adic zeta pseudomeasure ${\zeta }_p$ on $G$. With the sign conventionits product with the augmentation ideal $I(G)$ is integral and principal. The pole on the trivial-character component is cancelled by that ideal; simply writing an integral quotient by ${\zeta }_p$ would be incorrect. The main assertion in this convention isWe explain the local calculation, the remaining global obstruction, and how the cyclotomic Euler system removes it.

The local calculation is the Iwasawa theorem on local cyclotomic units:One can see why this theorem has exactly this analytic term. For a norm-fixed Coleman power series $f$, its corrected logarithm isThe ratio lies in $1+pR$, so the expression is integral. The norm identity puts its Amice transform in the kernel of the Coleman trace operator, hence gives a measure on ${\mathbb{Z}}_p^{\times }$. Its kth moment is $(1-p^{k-1}){\delta }_k(f)$. The full local exact sequence has p-power roots of unity at its kernel and cokernel. Passing to the real, or even, part removes both, since complex conjugation acts as minus one on their Tate module and $p$ is odd. The corrected logarithm therefore identifies $U_{\infty }^1$ with $A$.

Question 2 now shows that the image of $c(a,b)$ is $([b]-[a]){\zeta }_p$: its even moments are $(1-p^{k-1})(a^k-b^k)B_k/k$. A choice of integer $e$ generating ${\mathbb{Z}}_p^{\times }$ topologically gives a norm-compatible generator $c(e,1)$, after multiplication by its constant Teichmuller lift to make it principal. That constant has zero logarithmic derivatives. Its conjugates generate the closed cyclotomic-unit module, and $[e]-[1]$ generates $I(G)$. This yields the displayed local quotient and its characteristic ideal. The generation statement is an important ingredient; the mere calculation of a few moments would not prove the theorem.

The class-field unit sequence gives the global comparisonMultiplicativity of characteristic ideals in exact sequences givesThus the local theorem proves the main conjecture precisely when one proves $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$. These terms measure the global-unit and ideal-class obstructions. Discarding them would silently impose a much stronger arithmetic hypothesis.

The first global step is the Euler-system divisibility for real cyclotomic class modules: a generator $f_Y$ of $\mathrm{char}(Y_{\infty })$ divides a generator $f_B$ of $\mathrm{char}(B)$ on every character component. Here is the mechanism. Choose auxiliary primes $q$ splitting completely in a finite layer and satisfying $q\equiv 1(\mathrm{mod}{p}^m)$. For a generator ${\sigma }_q$ of its auxiliary cyclic group, putApply products of these operators to the unit Cyclotomic Euler system of Question 3. The norm axiom and this group-ring identity make the derived elements invariant modulo $p^m$-th powers, so Kummer descent gives classes in $F_n^{\times }/(F_n^{\times }{)}^{p^m}$. The congruence axiom identifies the valuation at each newly introduced prime with a residue symbol of the preceding class; valuations away from the chosen auxiliary primes vanish modulo $p^m$. The Chebotarev density theorem supplies primes in prescribed ideal classes with the required residue-symbol behavior. Adding these primes successively forces the elementary divisors of the class group to divide the available cyclotomic-unit index. Taking the norm limit yields $f_Y\mid f_B$. Finite control defects do not survive localization at height-one primes; using two coprime annihilators of such a finite defect also shows that no hidden p-power factor is left. This is the substantive global argument furnished by the Cyclotomic Euler system, not a consequence of Iwasawa's local theorem alone.

For the reverse comparison use the cyclotomic unit index formula and norm descent, keeping invariants as well as coinvariants. At the base field put $A_0$ for its p-primary ideal class group, and let $N_{\infty }(E_0^1)$ be the universal norm of units in a Zp-extension. Class-field and unit descent, using total ramification and the principal prime over $p$, giveThese are the required norm-defect control statements. The cyclotomic unit index formula, with the p-adic closure justified by Leopoldt, gives $\#A_0=\#(E_0^1/C_0^1)$. Also $B^{\Gamma }=0$: its coinvariants are finite by the displayed control, so its invariants are finite by the Iwasawa module structure theorem, whereas $B$ is a submodule of the cyclic local quotient $H$, which has no nonzero finite submodule. One must retain $Y_{\infty }^{\Gamma }$; dropping it would incorrectly identify all global units with universal norms.

Define the finite Euler characteristic of a one-variable Iwasawa module by ${\chi }_{\Gamma }(M)=\#M_{\Gamma }/\#M^{\Gamma }$. The preceding identities now give an exact cancellation:For a torsion $\Lambda$-module with finite invariant and coinvariant groups, the structure theorem gives ${\chi }_{\Gamma }(M)=|f_M(0){|}_p^{-1}$. Write $f_{B,\chi }=f_{Y,\chi }{h}_{\chi }$ using the Euler-system divisibility. After forgetting ${\Delta }^{+}$, the characteristic series are the products of their character-component series. Equality of these Euler characteristic of a one-variable Iwasawa module values givesEvery summand is nonnegative, since $h_{\chi }\in {\mathbb{Z}}_p[[S]]$. Each is therefore zero. A formal power series with p-adic-unit constant term is a unit, so every $h_{\chi }$ is a unit. Hence $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$, and cancellation in the class-field exact sequence completes the boxed main conjecture.

The familiar reflected odd class-group formulation follows from Kummer reflection in Iwasawa theory, with the inversion of the group variable and the Tate twist both included. For example, for nontrivial even $\chi$, if $f_{\chi }((1+p{)}^s-1)=L_p(\chi ,s)$, then the p-ramified series is $g_{\chi }(S)=f_{\chi }((1+p)(1+S{)}^{-1}-1)$, and $g_{\chi }((1+p{)}^{1-s}-1)=L_p(\chi ,s)$. The proof above includes the exceptional trivial component through augmentation regularization. It assumes neither a vanishing real class group nor Vandiver's conjecture: the Euler-system divisibility and the norm-defect cancellation replace that extra hypothesis.

Let $L$ be the splitting field of the separable polynomial. Its Galois group is $G={\mathrm{Aut}}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha )\to K(\beta )$ for roots of the same irreducible factor extends to an embedding of $L$ in an algebraic closure, and normality of the splitting field makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. The action is transitive exactly when the polynomial is irreducible.

If $n=p$ is prime, a transitive action has $p\mid |G|$ by orbit-stabilizer, so Cauchy theorem for groups supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for prime degree.

The factorization theorem modulo a good prime says that if an integer polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\dots ,d_r$, its Galois group over $\mathbb{Q}$ contains a permutation with disjoint cycle lengths $d_1,\dots ,d_r$. Equivalently the finite-field Galois group, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.

For the given quintic, Eisenstein criterion at three proves irreducibility. Modulo two it factors asNeither factor has a root in ${\mathbb{F}}_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a transposition. It is also transitive of degree five.

To prove that this forces the full symmetric group here, form the graph on the five roots with an edge for each conjugate in $G$ of that transposition. The graph is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an edge, so the size cannot be one. Hence the graph is connected. Transpositions along the edges of a connected graph generate all transpositions (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving

Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A real number is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.

If current coordinates lie in a field $K\subseteq \mathbb{R}$, a line-line intersection solves linear equations over $K$, while line-circle or circle-circle intersections reduce to a quadratic equation over $K$. Thus each step requires at most adjoining a square root. Consequently every constructible number lies in a towerConversely field sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive square root by the geometric-mean construction in a semicircle. Hence every real number in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.

In particular a constructible algebraic number has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the Galois group of its normal closure is a finite two-group: a tower's normal closure remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by Galois correspondence.

The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]{2}$, whose polynomial $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos 20^{\circ }$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the unit circle requires side length $\sqrt{\pi }$, impossible because $\pi$ is transcendental whereas constructible numbers are algebraic.

For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\phi (n)$, so it is constructible exactly when this degree is a power of two. Writing the prime factorization of $n$ shows the equivalent conditionIndeed odd prime exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is prime, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic Galois group a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension:

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

An automorphism sends the primitive root of unity ${\xi }_n$ to another primitive root of unity, hence uniquely to ${\xi }_n^a$ with $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. Define $\chi (\sigma )=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore $\chi$ is an injective group homomorphism and $G$ is abelian.

The extension is finite Galois: it is the splitting field of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The Fundamental theorem of Galois theory identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives $L^H/K$ Galois, with group $G/H$.

For a non-surjective nontrivial example, take $K=\mathbb{Q}(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb{Z}/8\mathbb{Z}{)}^{\ast }$. For $K=\mathbb{Q}$ and prime $p$, ${\Phi }_p(X)=1+\cdots +X^{p-1}$ is irreducible: ${\Phi }_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the unit group's order, proving surjectivity.

For $n=7$, let $\zeta ={\xi }_7$. The cyclic Galois group of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:Here ${\sigma }_a(\zeta )={\zeta }^a$. To identify the quadratic field without guessing, put $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B{)}^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin \mathbb{Q}$, proving the quadratic identification. The real field is cubic because complex conjugation is ${\sigma }_6$, with fixed subgroup of order two.

The squared Vandermonde product defining the polynomial discriminant is symmetric in all roots. By the fundamental theorem of symmetric polynomials, it is a polynomial in the elementary symmetric functions, which are the signed coefficients of the monic polynomial. Hence it is a polynomial function of those coefficients over the base field.

For $f=X^3+pX+q$, $\Delta (f)=-\mathrm{Res}(f,f^{\prime })$. Eliminating the two leading rows in the Sylvester determinant reduces the resultant toThusFor $X^3-3X+1$, the only possible rational roots are $\pm 1$, and neither is a root. The cubic is therefore irreducible, so its Galois group is a transitive subgroup of $S_3$. Its discriminant is $81=9^2$. The Vandermonde product changes sign under odd permutations, and its square being a rational square implies that the product itself is rational; hence all Galois permutations are even. The only transitive subgroup of $A_3$ is $A_3$ itself. Consequently, writing $L$ for its splitting field,

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.

Over a finite field ${\mathbb{F}}_q$, the splitting field is a finite extension ${\mathbb{F}}_{q^m}$, where $m$ is the least common multiple of the degrees of the distinct irreducible factors. Its Galois group is cyclic of order $m$, generated by $x\mapsto x^q$.

Over ${\mathbb{F}}_2$, the polynomial has no linear factor: its values at zero and one are both one. The only irreducible quadratic is $x^2+x+1$; reduction modulo it gives $x$, not zero. A reducible quartic without linear factors would split into quadratics, so $f$ is irreducible. Thus its splitting field is ${\mathbb{F}}_{16}$ and

Irreducibility modulo two implies irreducibility over $\mathbb{Q}$ by Gauss lemma for polynomials, so the Galois group embeds transitively in $S_4$. The discriminant is $229$, hence primes two and three are unramified. The factor-degree theorem says that square-free reduction with factor degrees $d_1,\dots ,d_r$ provides an element of the Galois group with cycle lengths $d_1,\dots ,d_r$. The reductions already found therefore give a 4-cycle and a 3-cycle.

The group order divides 24 and is divisible by four and three, so is 12 or 24. An order-12 subgroup of $S_4$ has index two and must be $A_4$: in its quotient every transposition has the same nontrivial image, since transpositions are conjugate and generate $S_4$, making the quotient map the sign. But a 4-cycle is odd. Therefore the order is 24 andThe discriminant can also be obtained from $f^{\prime }=x^2(4x+3)$ and the resultant, giving $256-27=229$.

A primitive sixth root of unity satisfieswhich is irreducible over $\mathbb{Q}$ because its discriminant is $-3$. For the other polynomial choose $\alpha =i{3}^{1/6}$, so ${\alpha }^6=-3$ and ${\alpha }^3=-i\sqrt{3}$. Then ${\zeta }_6=(1-{\alpha }^3)/2\in \mathbb{Q}(\alpha )$, so this field already contains all six roots. The Eisenstein criterion at $3$ proves $x^6+3$ irreducible, giving a splitting field of degree six.

Put $r(\alpha )={\zeta }_6^2\alpha$ and $s(\alpha )={\zeta }_6\alpha$. The first automorphism fixes ${\zeta }_6$ and has order three. The second sends ${\alpha }^3$ to its negative, so sends ${\zeta }_6$ to ${\zeta }_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the Galois group. ThereforeThe presence of sixth roots of unity inside the splitting field is why this particular binomial has degree six rather than a generic twelve-degree splitting field.

If $\theta =0$, the splitting field is $K$ and its Galois group is trivial, hence cyclic. Otherwise choose $\alpha$ with ${\alpha }^n=\theta$. The roots are $\alpha {\zeta }_n^j$, so the splitting field is $K(\alpha )$ because ${\zeta }_n\in K$. The roots are distinct since the characteristic does not divide $n$, making this a Galois extension. Each automorphism satisfies $\sigma (\alpha )=c_{\sigma }\alpha$ with $c_{\sigma }\in ⟨{\zeta }_n⟩$. Because that root-of-unity group is fixed by all automorphisms, $c_{\sigma \tau }=c_{\sigma }{c}_{\tau }$. Thus $\sigma \mapsto c_{\sigma }$ is an injective homomorphism into a cyclic group. The splitting-field Galois group is cyclic, of order dividing $n$.

All the local fields here carry normalized discrete valuations. Write ${\mathcal{O}}_F$ for the valuation ring, ${\mathfrak{m}}_F$ for its maximal ideal, and $k_F$ for its residue field. We use the basic degree identity $[B:A]=e(B/A)f(B/A)$ for finite extensions of non-Archimedean local fields, with $e$ the ramification index and $f$ the residue degree.

First construct the maximal unramified subextension of a local field extension inside an arbitrary finite extension $L/K$. Choose a primitive element of a field extension $\bar{z}$ for the finite residue field extension $k_L/k_K$. Let $\bar{h}$ be its monic irreducible minimal polynomial of an algebraic element, of degree $f$, and lift its coefficients to obtain a monic $h\in {\mathcal{O}}_K[T]$ of the same degree. Finite finite fields are perfect fields, so ${\bar{h}}^{\prime }(\bar{z})\ne 0$. By Hensel lemma applied in $L$, there is $z\in {\mathcal{O}}_L$ reducing to $\bar{z}$ with $h(z)=0$. For $U=K(z)$ we haveEvery inequality is therefore an equality. In particular $U/K$ is unramified, $k_U=k_L$, and $h$ is the minimal polynomial of an algebraic element $z$.

If $L/K$ is unramified, then $[L:K]=f$, so this construction gives $U=L$. Taking $x=z$ supplies the required generator, and $h^{\prime }(x)$ is a unit. Conversely suppose $L=K(x)$ with $x$ integral and its monic minimal polynomial of an algebraic element $g$ satisfying $g^{\prime }(x)\notin {\mathfrak{m}}_L$. Its coefficients lie in ${\mathcal{O}}_K$, since all conjugates of an integral element are integral and $K$ is complete. In the constructed unramified extension $U$ the element $\bar{x}\in k_U=k_L$ is a simple root of $\bar{g}$. Hensel lemma in $U$ produces a root $x^{\prime }\in U$ with residue $\bar{x}$. In $L$ both $x$ and $x^{\prime }$ lift this same simple residue root, so Hensel uniqueness gives $x=x^{\prime }$. Thus $L\subseteq U\subseteq L$, proving

For lifting residue-field embeddings of unramified extensions, use the generator $z$ and polynomial $h$ just constructed when $L/K$ is unramified. Given a $k_K$-field homomorphism $\iota :k_L\to k_M$, its value $\iota (\bar{z})$ is a simple root of $\bar{h}$. Hensel lemma in $M$ produces the unique root $z^{\prime }\in {\mathcal{O}}_M$ with that residue. The assignment $z\mapsto z^{\prime }$ defines a $K$-field homomorphism $K[T]/(h)\to M$; it induces $\iota$ because $\bar{z}$ generates $k_L$. Conversely any such field homomorphism preserves the unique extended valuation, hence induces a residue field homomorphism, and its image of $z$ is uniquely forced by Hensel uniqueness. This proves the asserted bijection. The same proof only requires that $M$ be a non-Archimedean local field containing $K$ with its compatible valuation; its finiteness over $K$ is not essential.

Now set $M=L$. The finite residue field extension has exactly $f=[L:K]$ automorphisms, cyclically generated by $a\mapsto a^{|k_K|}$. Their unique lifts give $[L:K]$ automorphisms of $L$. Consequently $L/K$ is a Galois extension, and reduction identifies its Galois group with that cyclic residue Galois group. Its generator is the arithmetic Frobenius:

For the compositum, choose a uniformizer ${\pi }_M$ of the totally ramified extension $M/K$. The Eisenstein generator of a totally ramified extension characterization, proved in the next solution, says that $M=K({\pi }_M)$ and its minimal polynomial of an algebraic element is Eisenstein over ${\mathcal{O}}_K$. A uniformizer of $K$ remains a uniformizer in the unramified extension $L$. All its Eisenstein coefficient conditions therefore remain true over ${\mathcal{O}}_L$. Thus the same Eisenstein polynomial is irreducible over $L$ and generates $LM=L({\pi }_M)$ as a totally ramified extension. This proves $LM/L$ is totally ramified, of degree $[M:K]$.

Here is an explicit example showing that the converse decomposition of an arbitrary extension can fail. Take $K={\mathbb{Q}}_3$ and $L=K(i)$ with $i^2=-1$. Since $T^2+1$ is irreducible and separable modulo three, the degree and residue-degree argument above shows that $L/K$ is unramified quadratic. PutThe element $1+i$ is a unit in ${\mathcal{O}}_L$, so $T^2-3(1+i)$ is Eisenstein over $L$. Hence $[N:L]=2$, $e(N/K)=2$, and $f(N/K)=2$. Moreover $i={\alpha }^2/3-1$ gives $N=K(\alpha )$, withWe verify its Galois group, since that is what rules out a totally ramified complement. By Hensel lemma there is $r\in K$ with $r^2=-2$: the residue root $1$ of $T^2+2$ is simple modulo three. Set $\beta =(1+i)/r$. ThenThe assignments $\sigma (i)=-i$ and $\sigma (\alpha )=\beta \alpha$ respect the defining relation, because $(\beta \alpha {)}^2=-3i(1+i)=3(1-i)$. They therefore define a $K$-automorphism of $N$. Its square satisfies ${\sigma }^2(i)=i$ and ${\sigma }^2(\alpha )=\bar{\beta }\beta \alpha =-\alpha$, so $\sigma$ has order four. There are thus four automorphisms of the degree-four extension, andThis is the cyclic quartic local extension without a totally ramified complement. Its unique quadratic subfield is $L$, fixed by ${\sigma }^2$, and is unramified. If $N=L^{\prime }{M}^{\prime }$ with $L^{\prime }/K$ unramified and $M^{\prime }/K$ totally ramified, then $[L^{\prime }:K]\le f(N/K)=2$ and $[M^{\prime }:K]\le e(N/K)=2$. To obtain the full degree four both would have degree two, forcing $M^{\prime }$ to be a totally ramified quadratic subfield. There is no such subfield. Thus $N/{\mathbb{Q}}_3$ is the required counterexample.

Let $d=[L:K]$, with normalized discrete valuations $v_K,v_L$. Suppose first that $L/K$ is totally ramified, so $e(L/K)=d$. Choose a uniformizer $x$ of $L$. The elements $1,x,\dots ,x^{d-1}$ are linearly independent over $K$: the nonzero terms $c_j{x}^j$ in any proposed relation have valuations $d{v}_K(c_j)+j$, distinct modulo $d$, and a sum with a unique least valuation cannot vanish by the ultrametric inequality. Hence $[K(x):K]\ge d$, proving $L=K(x)$.

Extend $v_K$ to an algebraic closure with values in $\mathbb{Q}$. Every conjugate of $x$ has valuation $1/d$, by uniqueness of the extended valuation on finite extensions of the complete field $K$. In its monic minimal polynomial of an algebraic element $g(T)=T^d+a_{d-1}{T}^{d-1}+\cdots +a_0$, the coefficient $a_{d-j}$ is a sum of products of $j$ conjugates, counted with multiplicity. Thus $v_K(a_{d-j})\ge j/d>0$ for $j\ge 1$, so every nonleading coefficient is in ${\mathfrak{m}}_K$. The constant coefficient is the product of all $d$ conjugates, so has valuation exactly one. This proves that $g$ is an Eisenstein polynomial. This argument also works for inseparable finite extensions, since the conjugates can be counted with their multiplicities.

Conversely let $g$ be an Eisenstein polynomial of degree $d$, with root $x$ and $L=K(x)$. The Eisenstein criterion proves irreducibility. For completeness, monic factors would have integral coefficients; their reductions would be positive-degree powers of $T$, so both constant coefficients would be divisible by the uniformizer of $K$, contradicting $v_K(g(0))=1$. To determine the root valuation, put $t=v_K(x)$ in the extended valuation. If $t<0$, the leading term is the unique term of least valuation; if $t=0$, the leading term again has uniquely least valuation zero. Both are impossible. If $t>0$, every nonconstant lower term has valuation at least $1+jt>1$. Cancellation in $g(x)=0$ therefore requires the leading term and constant term to have equal valuation:Since $t=v_L(x)/e(L/K)$ and $v_L(x)$ is a positive integer, $e(L/K)=d{v}_L(x)\ge d$. The opposite inequality follows from $[L:K]=e(L/K)f(L/K)=d$. Hence $e=d$, $f=1$, and $v_L(x)=1$. We have proved both the equivalence and the uniformizer assertion:

Now take $q=p^a$ with $a\ge 1$ and $\zeta ={\zeta }_{p^a}$. The translated cyclotomic polynomial isIn characteristic $p$, putting $U=T^{p^{a-1}}$ reduces this to ${\sum }_{j=0}^{p-1}(1+U{)}^j=U^{p-1}=T^{p^{a-1}(p-1)}$. Its constant coefficient over the integers is $p$. It is therefore Eisenstein at $p$, of degree $d=p^{a-1}(p-1)$. Consequently ${\mathbb{Q}}_p({\zeta }_{p^a})/{\mathbb{Q}}_p$ is totally ramified of degree $d$, with uniformizer ${\zeta }_{p^a}-1$. If $q=1$, the field is simply ${\mathbb{Q}}_p$, a trivial totally ramified extension.

For a general positive integer $n=p^{a}m$ with $p\nmid m$, the field generated by the prime-to-$p$ root of unity ${\zeta }_m$ is unramified. Indeed $T^m-1$ has distinct roots modulo $p$. In a finite unramified field whose residue field contains those roots, Hensel lemma lifts all of them uniquely. Reduction preserves their orders: a prime-to-$p$ root reducing to one must itself be one, by Hensel uniqueness at the root one of $T^m-1$. The arithmetic Frobenius therefore acts as ${\zeta }_m\mapsto {\zeta }_m^p$, and the extension degree is the multiplicative order of $p$ modulo $m$, with degree one when $m=1$.

The identityfollows by taking powers of a primitive nth root of unity and using the coprimality of $p^a,m$ in the opposite direction. The prime-to-$p$ field is unramified and the $p$-power field is totally ramified; unramified base change preserves total ramification proves that their compositum has the same residue degree as the prime-to-$p$ field. Thus the total ramification criterion for local cyclotomic extensions gives the full classification:This includes trivial extensions when $a=0$. For $p=2$ only $m=1$ is possible.