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Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

Let $\mathcal{G}=\mathrm{Gal}(M_{\infty }/F)$. The Galois group sequence isFor $g\in \Gamma$ and $x\in X$, choose a lift $\tilde{g}\in \mathcal{G}$ and defineThis is independent of the lift because $X$ is abelian. Write $X$ additively. As an abelian pro-p group, it is a compact ${\mathbb{Z}}_p$-module, with $ax$ defined by the limit of integer powers, and the conjugation action is continuous and ${\mathbb{Z}}_p$-linear. It extends to the Iwasawa algebraIndeed, use the finite quotients of $X$ obtained by intersecting it with open normal subgroups of $\mathcal{G}$; on each quotient the action factors through one of these finite group rings. This defines the natural compact Galois module structure, even when $X$ is not finitely generated. Choosing $\gamma$ identifies $\Lambda$ with ${\mathbb{Z}}_p[[T]]$ through $T=\gamma -1$.

The subgroup $(\gamma -1)X$ is closed: it is the image of the compact group $X$ under a continuous map. Every element $(\gamma -1)x$ is a commutator of a lift of $\gamma$ with $x$. ConsequentlyConversely, in the quotient by $(\gamma -1)X$, the lift of $\gamma$ centralizes the image of $X$. Its powers are dense in the procyclic subgroup they generate. Since these powers map densely onto $\Gamma$, compactness shows that $\mathcal{G}$ is generated by $X$ and that procyclic subgroup. The quotient is therefore abelian. HenceOne can also see this by choosing a lift of $\gamma$. Since $\mathcal{G}$ is a pro-p group, the closed subgroup generated by this lift is ${\mathbb{Z}}_p$, and its map to $\Gamma$ is an isomorphism. This gives a semidirect product splitting.

The maximal abelian subextension $M/F$ is the fixed field of the closed commutator subgroup. Because $F_{\infty }/F$ is abelian, $F_{\infty }\subseteq M$. Restriction of the Galois correspondence to $X$ now gives the coinvariant formulaThe closed image above is important: the formula is a quotient of compact groups, not a quotient by an unclosed abstract subgroup. This is abelianization over a Zp-extension.

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

The cyclotomic polynomial ${\Phi }_7=1+X+\cdots +X^6$ is irreducible: replacing $X$ by $X+1$ gives an Eisenstein polynomial at seven. Hence the extension has degree six, andThere is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb{Q}(\eta )$; the whole group fixes $\mathbb{Q}$. The order-two subgroup $\{1,6\}$ is complex conjugation and fixes $\mathbb{Q}(\eta +{\eta }^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta +{\eta }^{-1})X+1$ and is not real.

The order-three subgroup $\{1,2,4\}$ fixes $u=\eta +{\eta }^2+{\eta }^4$. Its other conjugate is $v={\eta }^3+{\eta }^5+{\eta }^6$. Expanding products and using ${\sum }_{j=1}^6{\eta }^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this fixed field is $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the Galois correspondence proves they are the full fixed fields.

A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the fundamental identity for prime decomposition is ${\sum }_{\mathfrak{q}\mid \mathfrak{p}}{e}_{\mathfrak{q}}{f}_{\mathfrak{q}}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.

By part (a), the number of primes of $L$ above $\mathfrak{p}$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently whenThis condition says $gD{g}^{-1}\subseteq H$ for every $g$, orthe subgroup core.

Because $M$ is the Galois closure of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gH{g}^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The Galois correspondence therefore gives ${\mathrm{Core}}_G(H)=1$. HenceIn the Galois extension $M/K$, the number of primes above $\mathfrak{p}$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that complete splitting is preserved by Galois closure:No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.

The eleventh cyclotomic field $L=\mathbb{Q}({\xi }_{11})$ hasThe cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The Galois correspondence therefore gives exactly four subfields:The Quadratic Gaussian period in the eleventh cyclotomic field satisfies $\eta =(-1+\sqrt{-11})/2$, while $\theta$ generates the maximal real subfield of the eleventh cyclotomic field. Every subgroup of the abelian Galois group is normal, so all four subfields are Galois over $\mathbb{Q}$.

Each nontrivial proper subfield has prime degree, two or five. If $a\in M\setminus \mathbb{Q}$, then $\mathbb{Q}⊊\mathbb{Q}(a)\subseteq M$, and the tower law forces $\mathbb{Q}(a)=M$. Thus every such element is a primitive element. The displayed $\eta$ and $\theta$ are the requested examples. Finally, ${\xi }_{11}$ itself is a primitive element for $L/\mathbb{Q}$.

The seventh cyclotomic polynomial isIt is irreducible over $\mathbb{Q}$, and all its roots ${\zeta }_7^a$, $a=1,\dots ,6$, lie in $L=\mathbb{Q}({\zeta }_7)$. Hence $L/\mathbb{Q}$ is a degree-six Finite Galois extension, withThe group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ is cyclic of order six; for example,

By the Galois correspondence, the subfields are the fixed fields of the four subgroups of $C_6$:Thus these are all the subfields.

For the Quadratic Gaussian period in the seventh cyclotomic field, putThe cyclotomic relation gives $\eta +\bar{\eta }=-1$, while direct multiplication gives $\eta \bar{\eta }=2$. HenceIts discriminant is $-7$, so

For the real cubic subfield of the seventh cyclotomic field, divideby ${\zeta }_7^3$. With $\theta ={\zeta }_7+{\zeta }_7^{-1}$,soThis cubic has no rational root, and therefore

The requested primitive elements, minimal polynomials, and automorphism groups are consequentlyMore explicitly, the nontrivial automorphism of $\mathbb{Q}(\eta )$ sendsand the three automorphisms of $\mathbb{Q}(\theta )$ cyclically permute

Finally, $G$ is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that

For roots $r_1,r_2,r_3$, the discriminant isand for $f=X^3+pX+q$,If $f$ is irreducible, its Galois group is transitive in $S_3$: it is $A_3\cong C_3$ when $\Delta$ is a square in $K$, and $S_3$ otherwise. If $f$ has one root in $K$ and an irreducible quadratic factor, the group is $C_2$; if it splits, it is trivial. A repeated-root depressed cubic already has all roots in $K$, so adds no case.

For $X^3+3X+1$, the rational-root test proves irreducibility andis not a rational square. Thus $G\cong S_3$. By the fundamental theorem of Galois theory, the complete subfield list iswhere the three cubic fields fix the three order-two subgroups and the quadratic field fixes $A_3$.

By Dirichlet theorem on primes in arithmetic progressions, choose a prime $\ell$ withThe cyclotomic field $E=\mathbb{Q}({\zeta }_{\ell })$ is Galois over $\mathbb{Q}$, andwhere the last group is cyclic because the multiplicative group of a finite field is cyclic.

The cyclic group $C_{\ell -1}$ has a unique subgroup $H$ of order $(\ell -1)/r$. By the Galois correspondence, its fixed field $L=E^H$ satisfiesNormality follows because $H$ lies in an abelian group. Thus every $r>1$ occurs as the Galois group of a finite extension of $\mathbb{Q}$. This is the cyclic Galois extension of the rational numbers of every finite degree.

Krasner's lemma says that if $K$ is complete, $\alpha$ is separable over $K$, andthen $K(\alpha )\subseteq K(\beta )$.

To prove it, let $\tau$ be an automorphism of $\overline{K}$ fixing $K(\beta )$. The absolute value has a unique extension to every finite extension of the complete field $K$, so it is invariant under $\tau$. If $\tau (\alpha )\ne \alpha$, the ultrametric inequality and the displayed strict inequality giveBut $\tau (\beta )=\beta$ and invariance gives $|\tau (\alpha )-\beta |=|\alpha -\beta |$, a contradiction. Every automorphism fixing $K(\beta )$ therefore fixes $\alpha$, which proves the field inclusion by Galois correspondence.

Now let $L/{\mathbb{Q}}_p$ be finite. By the primitive element theorem, write $L={\mathbb{Q}}_p(\alpha )$ with separable minimal polynomial $f\in {\mathbb{Q}}_p[X]$. Approximate the coefficients of $f$ closely by those of a polynomial $g\in \mathbb{Q}[X]$ of the same degree. Continuity of roots over a non-Archimedean field gives a root $\beta$ of $g$ arbitrarily close to $\alpha$. Choose it close enough for Krasner's lemma; thenThe degree bound from $\mathrm{deg}g=\mathrm{deg}f$ forces equality. If $F=\mathbb{Q}(\beta )$ and $\mathfrak{p}$ is the prime selected by the embedding $F\hookrightarrow {\overline{\mathbb{Q}}}_p$, its completion isThus every finite extension of ${\mathbb{Q}}_p$ is the completion of a number field at a prime ideal.

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

For the biquadratic extension $L=K(\sqrt{a},\sqrt{b})$, the assumptions say that the square classes of $a$ and $b$ are independent. In particular $\sqrt{b}\notin K(\sqrt{a})$: writing $\sqrt{b}=u+v\sqrt{a}$ and comparing the coefficient of $\sqrt{a}$ after squaring would force either $b$ or $ab$ to be a square in $K$. Hence

The field is the splitting field of $(X^2-a)(X^2-b)$ and the characteristic is not two, so it is Galois. Its four automorphisms independently choose the signs of $\sqrt{a}$ and $\sqrt{b}$, givingBy the Galois correspondence, the intermediate-field lattice isEach of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.

Let $E=\mathbb{Q}(\alpha ,\beta )$. Sincepart (b) gives $[E:\mathbb{Q}]=8$. The roots of $X^4-6X^2+6$ are $\pm \alpha ,\pm \beta$, all in $E$, so $E$ is its splitting field and is Galois. Its action on the four roots is transitive; the order-eight entry in the list from part (a) is $D_8$. Therefore

For an explicit dihedral Galois action on four radical roots, defineThen $r^4=s^2=1$ and $srs=r^{-1}$. The subgroup lattice is determined byHere $⟨r^2⟩$ lies in all three order-four subgroups; $⟨s⟩,⟨r^{2}s⟩$ lie in $⟨r^2,s⟩$; and $⟨rs⟩,⟨r^{3}s⟩$ lie in $⟨r^2,rs⟩$.

Reversing inclusions under the Galois correspondence gives the complete field latticeThe incidence is likewise reversed: $\mathbb{Q}(\sqrt{2})$ lies in $\mathbb{Q}(\sqrt{2},\sqrt{3})$; $\mathbb{Q}(\sqrt{3})$ lies in that field, $\mathbb{Q}(\alpha )$, and $\mathbb{Q}(\beta )$; and $\mathbb{Q}(\sqrt{6})$ lies in that field, $\mathbb{Q}(\alpha +\beta )$, and $\mathbb{Q}(\alpha -\beta )$. The hint verifies the last two quartic fields through $\alpha +\beta =\sqrt{2}\gamma$ and $\alpha -\beta =\sqrt{2}\delta$.

Identify $G\simeq (\mathbb{Z}/15\mathbb{Z}{)}^{\times }\simeq C_4\times C_2$, with $\sigma (\zeta )={\zeta }^2$ and $\tau (\zeta )={\zeta }^{-1}$. Its subgroups areTheir fixed fields, in the same layers, areThe real fields are $\mathbb{Q}$, $\mathbb{Q}(\sqrt{5})$, and $L^{+}$. This is the Galois correspondence.