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Put $\omega ={\zeta }^5$, $\tau ={\zeta }^3$, $a=1+2\omega$, and $b=1+2(\tau +{\tau }^{-1})$. The primitive cubic and quintic root identities give $a^2=-3$, $b^2=5$. These elements therefore provide the following six generators:

The three quadratic fields are distinct because $-3,5,-15$ are distinct rational square classes. To justify the fifth entry, the two quadratic fields $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ are distinct, so their compositum has degree four. Moreover $a+b\ne 0$ andThus their sum alone generates the compositum.

The field $\mathbb{Q}(\tau )$ has degree four by the quintic cyclotomic polynomial. Its Galois group is $C_4$ and its unique quadratic subfield is $\mathbb{Q}(b)$, so it cannot contain $a$. It is therefore different from $\mathbb{Q}(a,b)$. The final field has degree four by part (b) and is real; the preceding two quartic fields contain nonreal $\tau$ and $a$, respectively. Thus all three quartics are distinct, and the count in part (i) proves the list exhaustive. This gives the subfields of the fifteenth cyclotomic field without relying on a choice of square-root signs.

The Galois group acts transitively on the $n$ roots because $f$ is irreducible and separable. The orbit-stabilizer theorem givesso $n$ divides $|G|$. Equivalently, the tower law gives $[K:F]=[K:F({\alpha }_1)]n$, while the finite Galois extension has group order $[K:F]$.

First assume $\mathrm{char}F\ne 2$. Set $d=\det V(\alpha )\ne 0$. If ${\Delta }_f=b^2$ with $b\in F$, then $(d-b)(d+b)=0$ in the field $K$, so $d=\pm b\in F$. For an automorphism inducing the root permutation $\pi$,Since $\sigma$ fixes $d$ and $1\ne -1$ in $F$, every induced permutation has positive sign. The root action is faithful because the roots generate $K$, givingThis is the discriminant criterion for an alternating Galois group.

As printed for arbitrary fields, the assertion is false in characteristic two. Over ${\mathbb{F}}_2$, the polynomial $T^2+T+1$ is irreducible and separable, with roots $\alpha ,\alpha +1$ in ${\mathbb{F}}_4$. Their difference is one, so the discriminant is the square $1$. Nevertheless its Galois group is $C_2$, whereas $A_2$ is trivial. In characteristic two the permutation signs become indistinguishable as field elements; square discriminants in characteristic two therefore cannot detect the alternating subgroup in this way.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

Let $H_k$ be the ordinary Hilbert class field of $k$. Its real places split completely, so $H_k$ is totally real. Since $K/k$ is quadratic and $K$ is totally imaginary, $H_k\cap K$ cannot equal $K$; hence it equals $k$. Therefore the compositum $K{H}_k$ hasand is an abelian extension of $K$, by restriction of the Galois group to $H_k$. At finite primes it is unramified because unramified extensions remain unramified under base change. The field $K$ has no real places, so there is no infinite-place ramification condition left. Thus $K{H}_k\subseteq H_K$, where $H_K$ is the Hilbert class field of $K$. Since $[H_K:K]=h_K$, the tower degree formula proves class number divisibility for a CM extension:This argument uses the ordinary class field with split real places; replacing it silently by the narrow class field would invalidate the claim that $H_k$ is totally real.

Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.

A Hilbertian field is a field over which generic irreducibility can be retained by specializing a parameter. More precisely, for polynomials $f_i(T,X)$ irreducible over $K(T)$ and separable in $X$, and a nonzero exclusion polynomial $h(T)$, the Hilbert subsetis infinite. Denominators, leading coefficients and discriminants may be incorporated into $h$. The infinite condition permits simultaneous avoidance of any finitely many bad places. Separability is part of the definition, particularly in positive characteristic.

The field-theoretic interpretation is useful in constructive Galois theory. If $L/K(T)$ is finite separable and generated by a root of $f$, specialization at a good $a$ is the residue extension at a place over $T-a$. Irreducibility of $f(a,X)$ means its residue degree is the entire degree $[L:K(T)]$. Thus a Hilbertian field has infinitely many unramified rational places which are completely inert in each prescribed finite separable extension. For a finite family, one can work in a primitive-element presentation of their separable compositum; preserving its full degree preserves the degrees of all the subextensions.

Hilbert's irreducibility theorem says that $\mathbb{Q}$ is Hilbertian, and gives the corresponding assertion for number fields, including infinitely many integral specializations in its classical form. Rational function fields $k(U)$ over arbitrary fields $k$ are also Hilbertian. In particular, global function fields are Hilbertian just as number fields are. These standard field classes and the specialization viewpoint are discussed in Haran and Jarden's introduction to Hilbertian fields. The theorem does not say every parameter is good: $X^2-T$ is generically irreducible, but a square parameter makes it split. For simultaneous irreducibility, one excludes more exceptional parameters, not the entire field.

Hilbertianity survives finite extensions. For the separable case, let $K$ be Hilbertian, $M/K$ finite separable and $f\in M(T)[X]$ irreducible and separable. A primitive element of $M(T)(\alpha )$ over $K(T)$ has degree $[M:K]\mathrm{deg}f$. Apply Hilbertianity to its minimal polynomial and exclude the presentation's denominators. The resulting residue field contains the constant copy of $M$ and has the same total degree over $K$, so its degree over $M$ is $\mathrm{deg}f$. Therefore $f(a,X)$ remains irreducible over $M$, even for infinitely many $a$ chosen in $K$. This argument also handles finitely many polynomials by a compositum. Purely inseparable finite extensions are treated by separating the separable part and transporting its rational places through the unique purely inseparable extension; the finite-extension stability statement remains valid.

The property is not inherited by arbitrary algebraic extensions. An algebraic closure is not Hilbertian: in characteristic zero, $X^2-T$ is irreducible over $K(T)$ but every specialization has a root in $K$. In characteristic $p$, use $X^p-X-T$ instead; a hypothetical rational-function root would give its right side a pole of order divisible by $p$, contradicting its simple pole at infinity. All constant specializations split over the algebraically closed field.

Finite fields also fail. For $K={\mathbb{F}}_q$ of characteristic $p$, considerIts right-hand additive datum has a pole of order $q+1$, not divisible by $p$, and cannot be $b^p-b$ for a rational function $b$. The polynomial is therefore irreducible over $K(T)$: its roots are $\alpha +c$ for $c\in {\mathbb{F}}_p$, so its splitting-field group has prime order unless a root already belongs to the base field. At every $a\in K$, however, $a^q-a=0$ and the specialization is $X^p-X$, which splits. This gives a concrete failure of the required specialization property.

The main constructive application is Galois-preserving Hilbert specialization. Suppose $E/K(T)$ is finite Galois with group $G$, and use a primitive element of the entire splitting field, not merely a generator of one root field. A good specialization retaining degree $|G|$ has residue group embedded in $G$ by unramified reduction. Since its order is still $|G|$, it is exactly $G$. Preserving only the irreducibility of the original degree-$n$ polynomial would prove transitivity, which is weaker than preserving its full group. Resolvents or the primitive generator eliminate that ambiguity.

Consequently a regular realization of $G$ over $K(T)$, with constant field $K$, yields realizations over the Hilbertian field $K$. This is the bridge from branch-cycle constructions to the Inverse Galois problem over number fields. A realization over $\mathbb{C}(T)$ alone cannot be specialized to $\mathbb{Q}$; descent of its coefficients and constants is an additional step. Rational rigidity can supply that descent under its appropriate hypotheses, while ordinary complex rigidity answers only the geometric uniqueness question.

For example, the family $X^n-X-T$ has full symmetric geometric monodromy in characteristic zero. It is irreducible because it is linear in $T$. Its derivative has $n-1$ distinct critical roots satisfying $n{X}^{n-1}=1$, and their distinct critical values are $-(n-1)X/n$. Each simple critical point gives a transposition. These finite branch cycles generate the monodromy, since the infinity loop is the inverse product of the finite loops. Their transitive connected-edge action gives $S_n$. The arithmetic group is also $S_n$, so the constant field is $\mathbb{Q}$. Hilbert specialization then produces symmetric Galois groups over $\mathbb{Q}$. For $n=2$ the same conclusion follows directly from the square-root family.

Finally, the good parameters form a large arithmetic set without being every point or generally a Zariski-open set. Reducibility can occur at infinitely many exceptional points, as the square example shows. The language of thin sets packages those exceptions as images of generically finite separable covers of degree greater than one, together with finitely many points. Hilbertianity asserts that such finitely many exceptional constructions cannot exhaust the rational line. It is an existence principle with powerful constructive consequences, not by itself an algorithm that certifies a particular chosen parameter.

The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic field. We prove its quadratic case by identifying the ramification that a quadratic number field must have.

For an odd prime $\ell$, the cyclic Galois group of $\mathbb{Q}({\zeta }_{\ell })/\mathbb{Q}$ has exactly one subgroup of index two. Its fixed field $F_{\ell }$ is quadratic and is unramified at every finite prime other than $\ell$. For a squarefree integer $d$, the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d\equiv 1(\mathrm{mod}4)$, and $4d$ otherwise. A nontrivial quadratic discriminant supported on the odd prime $\ell$ must therefore beIndeed $d=\pm 1$ cannot give a nontrivial extension unramified at two, whereas exactly one of $\ell ,-\ell$ is congruent to one modulo four. ThusThis identifies the quadratic subfield of a prime cyclotomic field using ramification, without a Gaussian-sum computation.

Now let $F=\mathbb{Q}(\sqrt{d})$ with $d$ squarefree. Multiply $d$ by ${\prod }_{\ell \mid d,\ell \text{ odd}}{\ell }^{\ast }$. Every odd prime then occurs with even exponent, so its square class is represented by one of $1,-1,2,-2$. Thus adjoining the square roots of the ${\ell }^{\ast }$ removes all possible odd-prime ramification; the remaining quadratic factor is contained inThe original $\sqrt{d}$ is a rational multiple of a product of these square roots. It belongs to the compositum of $\mathbb{Q}({\zeta }_8)$ and the prime cyclotomic fields just used. This compositum is itself contained in a cyclotomic field, proving the theorem for $F$.

The sharper conductor statement follows from the same ramification calculation. For the fundamental discriminant $D$ of $F$, putThe possibilities for $D_2$ follow immediately from the quadratic discriminant formula. The corresponding fields are trivial, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$, lying in the cyclotomic fields of conductors $1,4,8,8$. Their conductors are coprime to the odd primes in the product. Hence quadratic Kronecker–Weber theorem by ramification gives the explicit conclusion

Write $\mathcal{O}$ for the valuation ring of $K$, $\pi$ for a uniformizer, $v(\pi )=1$, and $k={\mathbb{F}}_q$ for the residue field. A useful strong form of Hensel lemma is the following: if $f\in \mathcal{O}[X]$ and $a_0\in \mathcal{O}$ satisfythen there is a root $a\in \mathcal{O}$ with $v(a-a_0)>v(f^{\prime }(a_0))$, unique among roots in that ball. If $f(a_0)\ne 0$, its distance satisfies $v(a-a_0)=v(f(a_0))-v(f^{\prime }(a_0))$.

Here is a proof by successive correction. Put $s=v(f^{\prime }(a_0))$, $d_j=v(f(a_j))$, andStop if a root has already been reached. Otherwise the correction has valuation $d_j-s>s$. Integral Taylor expansions show that $v(f^{\prime }(a_{j+1}))=s$ andThus $d_j-2s$ grows at least geometrically, the corrections tend to zero, and completeness gives a limit $a$ with $f(a)=0$. The first correction has strictly smaller valuation than every later nonzero correction, proving the asserted distance. If $x,y$ are in the ball $v(x-a_0),v(y-a_0)>s$, thenThe second factor cannot vanish. Two roots in the ball must therefore coincide. This proves Hensel lemma; in particular a simple root modulo $\pi$ has a unique lift with that residue.

We use this to establish existence and uniqueness of unramified local extensions. For each integer $d\ge 1$, choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $d$ and lift it to a monic $f\in \mathcal{O}[X]$. Reduction and Gauss lemma show that $f$ remains irreducible over $K$. LetThe ring $A$ is finite free and complete over $\mathcal{O}$, and $A/\pi A\cong {\mathbb{F}}_{q^d}$. An element outside $\pi A$ is invertible: lift an inverse from the residue field and invert the remaining $1+\pi z$ by a convergent geometric series. Every nonzero element of $A$ is ${\pi }^t$ times such a unit, since ${\bigcap }_t{\pi }^{t}A=0$ in a finite free module. Thus $A$ is a discrete valuation ring, its fraction field is $L$, andThis constructs an unramified extension of each degree.

Conversely, let $M/K$ be an unramified extension of degree $d$. Its residue field is the unique degree-$d$ finite field extension ${\mathbb{F}}_{q^d}/{\mathbb{F}}_q$. Choose a residue root of $\bar{f}$ in it. The root is simple, so Hensel lemma lifts it to a root of $f$ in $M$. This embeds $L$ into $M$, and equality of degrees makes it an isomorphism. All $d$ residue conjugates lift in $L$ itself; hence $L/K$ is Galois, and reduction identifies its Galois group with the cyclic Galois group of the residue field extension. Its distinguished Frobenius automorphism reduces to $x\mapsto x^q$.

There is exactly one unramified extension $K_d$ of each degree inside a fixed algebraic closure, andFor the middle equality, every nonzero element of ${\mathbb{F}}_{q^d}$ lifts uniquely to a $(q^d-1)$st root of unity, because that exponent is prime to $p$. A primitive such residue generates the finite field extension, so its lift generates $K_d$. Taking the union gives the maximal unramified extension, with Galois group $\hat{\mathbb{Z}}$.

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

Put $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.

It is not one. The algebraic integerso it is a unit. If $P$ were principal, its generator $a=x+y\omega$ would have norm $\pm 3$. Multiply $a$ by a suitable integer power of $ϵ$ so thatIts conjugate then also has absolute value at most $\sqrt{3ϵ}$ because $|a{a}^{\prime }|=3$. ThereforeThe final inequality is $12ϵ<229$, immediate from the displayed value of $ϵ$. Thus the integer $y$ is zero, which makes $|N(a)|=x^2=3$ impossible. ConsequentlyThis proves the ideal class group of Q of square root 229. No claim that $ϵ$ is a fundamental unit is needed. Its norm minus one also realizes all real sign patterns after multiplication by $-1$, so the ordinary and narrow class numbers agree.

Now let $f(X)=X^3-4X-1$ and let $H$ be its splitting field. The possible rational roots $\pm 1$ are not roots, so $f$ is irreducible. Its discriminant isThus its Galois group is $S_3$, since an irreducible cubic has group $A_3$ precisely when its discriminant is a square. Its quadratic subfield is $\mathbb{Q}(\sqrt{229})=K$, and $H/K$ is cyclic of degree three. If $\alpha$ is one root, then $H=K(\alpha )$.

For every rational prime $\ell \ne 229$, the reduction of $f$ is separable. Its roots lift in an unramified extension containing its residue splitting field, by Hensel's lemma. Hence its splitting field is unramified over $\mathbb{Q}$ at all these primes, and consequently over $K$ there.

At 229 there is exactly one double residue root and one simple root:The simple root lifts to $r\in {\mathbb{Q}}_{229}$, and division gives $f=(X-r)q$ with $q$ monic quadratic. The discriminant-product identity isHere $q(r)=f^{\prime }(r)$ is a unit, since $r$ has the simple residue root. Therefore the quadratic discriminant has valuation one, and its root field is the ramified quadratic field ${\mathbb{Q}}_{229}(\sqrt{229})$. This is already the completion of $K$. The local splitting field has no further extension over it: the simple root lies in ${\mathbb{Q}}_{229}$ and the other two lie in that quadratic field. Thus $H/K$ is unramified even at 229. Equivalently its rational inertia is a transposition and has trivial intersection with $A_3$.

The positive cubic discriminant gives three distinct real roots, so $H$ is totally real and there is no infinite ramification over $K$. We have exhibited an everywhere-unramified cyclic extension of degree $h_K=3$. The Hilbert-class-field characterization in part (i) now proves

Finally let $p\ne 229$ be rational prime. A prime ideal of norm $p$ is principal if and only if its Artin symbol in $H_K/K$ is trivial. If it is principal, a generator has norm $\pm p$; multiplying by the norm-minus-one unit if necessary gives norm $+p$. Conversely an element of norm $p$ generates a prime ideal of norm $p$. Hence representation by $x^2+xy-57y^2$ is equivalent to $p$ splitting in $K$ with trivial Artin symbol in $H_K/K$, or to complete splitting of $p$ in $H_K/\mathbb{Q}$.

Let ${\tau }_p\in S_3$ be its rational Frobenius, defined up to conjugacy. Its restriction to the discriminant quadratic field is its sign. The splitting criterion for that quadratic field isFor odd $p$ this follows by quadratic reciprocity from the character of discriminant 229; for $p=2$ both criteria are negative since $229\equiv 5(\mathrm{mod}8)$. The polynomial is separable modulo $p$, and the degrees of its factors are the cycle lengths of ${\tau }_p$. Thus a root modulo $p$ means that ${\tau }_p$ fixes a root. An even permutation of three letters with a fixed point is the identity: a nonidentity three-cycle has none. Therefore the two stated conditions together are exactly complete splitting in $H_K$, andThis is prime representation by the norm form of Q of square root 229, with the norm sign and the prime two explicitly covered.

The existence theorem of global class field theory, together with reciprocity, gives an inclusion-reversing bijection between finite abelian extensions $L/K$ inside a fixed algebraic closure and open finite-index subgroups of the idèle class group $C_K$. The subgroup for $L$ is $N_{L/K}{C}_L$, andEquivalently, for each modulus $\mathfrak{m}$, subgroups between the ray-principal group $P_K(\mathfrak{m})$ and the ideal group $I_K(\mathfrak{m})$ classify abelian extensions whose conductor divides $\mathfrak{m}$. Their quotients are identified with the corresponding Galois groups by the Artin map. Changing the modulus gives the same extension when the induced congruence subgroup is the same.

The conductor criterion says that a finite place has conductor exponent zero exactly when the extension is unramified there; a real place is absent from the conductor exactly when it stays real. Thus conductor one means unramified at every finite place and no real-to-complex ramification.

Apply the existence theorem to the open subgroup of $C_K$ which is the image ofThe idele-to-ideal map identifies the quotient by this subgroup with the ordinary ideal class group, which is finite. Therefore its class field $H$ exists and satisfiesIts conductor is one, so it is everywhere unramified. Conversely any everywhere-unramified abelian extension has conductor one; its Artin map kills $U$, hence factors through this quotient and corresponds to an extension contained in $H$. This proves that $H$ is exactly the Hilbert class field, the maximal abelian everywhere-unramified extension. Here the ordinary ideal class group is appropriate because the infinite factors include all $K_v^{\ast }$, rather than only the positive real factors used in a narrow ray class group.

A cyclotomic Euler system must relate units at many auxiliary conductors, not just the successive layers of one p-power tower. A useful axiomatization is the root-of-unity Euler system. Fix a finite set of excluded primes $S$ containing $2$, and let $W_S$ be the roots of unity whose orders are prime to every prime in $S$. A system is a nonzero-valued function $\Phi :W_S\to {\overline{\mathbb{Q}}}^{\times }$ with three properties. It is even and Galois equivariant:It satisfies the distribution relation at every prime $\ell \notin S$:Finally, if $\ell \nmid \mathrm{ord}(\xi )$, it satisfies the local congruenceThe values in this congruence are required to be integral at the prime; in the unit example below they are local units. Evenness puts the values in the maximal real cyclotomic subfields. Equivariance identifies conjugates, distribution gives exact norm identities, and the congruence supplies information at newly introduced primes. These are the three distinct inputs used in the descent, rather than a single norm-coherence assumption.

Here is a concrete system extending the units in Q2. Fix a nonzero integer $a$ prime to $p$ and let $S$ contain $2$ and the prime divisors of $a$, but not $p$. PutEvery value is defined and nonzero: the excluded primes make $a$ invertible modulo the order of $\xi$, and that order is odd. The quotient is the same Laurent polynomial $b_a$ used in Q2. Replacing $\xi$ by ${\xi }^{-1}$ changes both numerator and denominator by a minus sign, so it is even; its rational coefficients give Galois equivariance.

For the distribution axiom, $\ell \notin S$ makes $\ell$ odd and prime to $a$. Both exponents $2$ and $2a$ permute ${\mu }_{\ell }$, and ${\prod }_{\rho \in {\mu }_{\ell }}\rho =1$. The calculation in Q2 therefore holds with $p$ replaced by $\ell$:Raising to $p-1$ proves the axiom. At removable points it holds by evaluating the regular Laurent polynomials, so it includes $\xi =1$ and ${\xi }^{\ell }=1$.

For the congruence axiom, a root $\rho$ of order $\ell$ reduces to one at every prime above $\ell$. The integral Laurent polynomial $b_a$ thus gives $b_a(\rho \xi )\equiv b_a(\xi )$. These values are local units when $\xi$ has order prime to $\ell$: its reduction retains that order, and, unless $\xi =1$, both $1-{\xi }^2$ and $1-{\xi }^{2a}$ remain nonzero; at $\xi =1$ the value is the unit $a$. The stated power preserves the congruence. Thus ${\Phi }_a$ satisfies all three Euler-system axioms, and its values at ${\zeta }_n$ are precisely $c_n(a)$. Products, Galois translates, suitable field norms and integral group-ring powers produce the other cyclotomic-unit systems needed in the theory.

The distribution relation explains the Euler factors. If $\ell \nmid m$, $\xi$ has order $m$ and $\rho$ is a primitive $\ell$th root, the conjugates over $\mathbb{Q}({\mu }_m)$ run through the nontrivial $\ell$th roots. Removing the single factor with $\rho =1$ from the distribution product givesWe use arithmetic Frobenius, so ${\mathrm{Fr}}_{\ell }(\xi )={\xi }^{\ell }$. This is the cyclotomic Euler norm factor at a new prime. At a higher power of a prime already dividing the conductor, all $\ell$ translates occur, and the ordinary successive-layer norm relation has no missing-factor quotient. A mere p-power compatible sequence therefore contains only part of the Euler-system information.

To explain the axiomatic method, fix a p-power modulus $P=p^r$ and auxiliary primes $\ell$ with $P\mid \ell -1$, split in the chosen real base field. For the cyclic auxiliary Galois group of order $\ell -1$, choose a generator ${\sigma }_{\ell }$ and defineSubtracting consecutive coefficients gives the concrete identityThis is the Kolyvagin derivative operator for a cyclic group. Apply products of these operators to Euler-system values at squarefree auxiliary conductors. The norm relations control the apparent failure of invariance modulo $P$; Kummer descent, using Hilbert 90 and retaining roots-of-unity corrections where present, turns the derivative data into classes over the base field. They have controlled valuations and are unramified away from the selected auxiliary primes and the excluded set. The local congruence relates the valuation component at a newly added prime to the finite residue-symbol component of the preceding class, with the sign and normalization determined by the chosen generator. This finite-to-singular relation is precisely the additional information not obtainable from norm compatibility alone.

The Chebotarev density theorem supplies auxiliary prime ideals in specified ideal classes while satisfying the splitting and congruence conditions. Reading the controlled valuations as principal-ideal relations successively bounds the p-primary elementary divisors of the ideal class group by the divisibility of the initial cyclotomic-unit class. On suitable nontrivial character components this gives the finite bound $\#A^{\chi }\le \#(E/C{)}^{\chi }$; passing through the norm tower yieldswhere divisibility is between characteristic generators, so the associated ideal containment is reversed. The Euler-system divisibility for real cyclotomic class modules requires the usual local conditions and control of finite errors; it is not a claim that arbitrary compatible units automatically annihilate every class group. The cyclotomic unit index formula, together with the global norm-defect comparison, gives the reverse equality. Q3's exact sequence then proves the main conjecture.