Kronecker-Weber Theorem

The Kronecker–Weber theorem says that every finite abelian extension of the rational numbers is contained in a field obtained by adjoining roots of unity. The statement is surprisingly strong. We begin with an arbitrary polynomial, whose roots may have no obvious connection with roots of unity. We assume only that its Galois group is abelian. Somehow, by adjoining sufficiently many roots of unity, we should recover every root of the original polynomial. We want to understand why this should be true by working with an actual element of the extension. Rather than begin with ramification groups, we will start with its conjugates, write a Fourier expansion, and try to replace the terms in that expansion by expressions involving roots of unity.

For cubic extensions, this works remarkably well. We can construct a Fourier expression, calculate its cube, and remove the unwanted prime factors by dividing by suitable Gauss sums. Eventually, the only cube equations left involve ninth roots of unity. This gives a direct proof for every cyclic cubic extension. For higher degrees, the same cancellation of ordinary ramified primes still works. The remaining calculation is more difficult because powers of a prime may occur as the extension degree. We will return to that difficulty only after understanding exactly what the Fourier calculations accomplish.

Cubic Case

Suppose $\displaystyle K/\mathbb Q$ is a cyclic cubic extension. Choose an element $\displaystyle \alpha\in K$ that generates the field. Write its three conjugates as $\displaystyle x_0,x_1,x_2$ , ordered so that the Galois symmetry $\displaystyle \sigma$ sends $\displaystyle x_0$ to $\displaystyle x_1$ , then to $\displaystyle x_2$ , and finally back to $\displaystyle x_0$ . Our goal is to find an expression for $\displaystyle \alpha=x_0$ in terms of roots of unity. Let $\displaystyle \omega$ be a primitive cube root of unity, so $\displaystyle 1+\omega+\omega^2=0$ . The natural thing to try is to combine the three conjugates with the coefficients $\displaystyle 1,\omega,\omega^2$ . Define

$\displaystyle \begin{aligned} T&=x_0+x_1+x_2,\ R&=x_0+\omega^2x_1+\omega x_2,\ S&=x_0+\omega x_1+\omega^2x_2. \end{aligned}$

The first expression $\displaystyle T$ is rational because it is unchanged by every Galois symmetry. The other two are more interesting. When we cyclically permute the roots, we find $\displaystyle \sigma(R)=\omega R$ and $\displaystyle \sigma(S)=\omega^2S$ . Instead of permuting three different terms, the Galois symmetry now simply multiplies an expression by a fixed cube root of unity. This is the whole reason for using the Fourier coefficients. We have not lost the original number. Adding the three expressions gives

$\displaystyle {\alpha=\frac{T+R+S}{3}.}$

Because a cyclic cubic extension of $\displaystyle \mathbb Q$ is totally real, complex conjugation fixes the $\displaystyle x_j$ and exchanges $\displaystyle \omega$ with $\displaystyle \omega^2$ . Consequently $\displaystyle S=\overline R$ . Our problem is therefore to find an expression for $\displaystyle R$ . Once we know $\displaystyle R$ , we recover the original element by adding its complex conjugate and the rational number $\displaystyle T$ .

Now the first simplification appears. Since $\displaystyle \sigma(R)=\omega R$ , we have $\displaystyle \sigma(R^3)=R^3$ . The field $\displaystyle K$ has degree three, while $\displaystyle \mathbb Q(\omega)$ has degree two, so the two fields have no nontrivial intersection. Therefore

$\displaystyle R^3=b,\qquad b\in\mathbb Q(\omega).$

This is already a remarkable reduction. We began with an arbitrary cubic polynomial, but the entire extension is now described by a cube root of an element of the simple quadratic field $\displaystyle \mathbb Q(\omega)$ . There is another relation worth recording. Since $\displaystyle \sigma(RS)=RS$ , the product $\displaystyle RS$ belongs to $\displaystyle \mathbb Q(\omega)$ . But complex conjugation also fixes it. Hence $\displaystyle RS\in\mathbb Q$ , and therefore

$\displaystyle b\overline b=(RS)^3\in\mathbb Q^3.$

The norm of $\displaystyle b$ from $\displaystyle \mathbb Q(\omega)$ to $\displaystyle \mathbb Q$ must be a rational cube. This will severely restrict the prime factors of $\displaystyle b$ .

The ring of integers of $\displaystyle \mathbb Q(\omega)$ is $\displaystyle \mathbb Z[\omega]$ . Every element is $\displaystyle a+c\omega$ , and its norm is $\displaystyle a^2-ac+c^2$ . This ring has unique factorization, much like the ordinary integers. One can see why from its triangular lattice: given any complex number $\displaystyle x+y\omega$ , choose integers nearest to $\displaystyle x,y$ . The difference has norm at most $\displaystyle 3/4$ . Applying this approximation to a quotient of two elements gives division with a remainder of smaller norm, and Euclid’s algorithm follows. We can therefore factor $\displaystyle b$ into prime elements of $\displaystyle \mathbb Z[\omega]$ , allowing negative powers if $\displaystyle b$ is not integral.

A prime factor appearing to a power divisible by three is harmless. If $\displaystyle \pi^3$ divides $\displaystyle b$ , we can replace $\displaystyle R$ by $\displaystyle R/\pi$ . This changes the cube from $\displaystyle b$ to $\displaystyle b/\pi^3$ , without changing the field obtained by adjoining $\displaystyle R$ over $\displaystyle \mathbb Q(\omega)$ . Thus only prime exponents modulo three matter. Suppose an ordinary rational prime $\displaystyle q\neq3$ occurs in the factorization. There are two possibilities.

If $\displaystyle q\equiv2\pmod3$ , then $\displaystyle q$ remains prime in $\displaystyle \mathbb Z[\omega]$ . Its norm is $\displaystyle q^2$ . If it appears in $\displaystyle b$ to exponent $\displaystyle e$ , it appears in $\displaystyle b\overline b$ to exponent $\displaystyle 2e$ . But we proved that $\displaystyle b\overline b$ is a rational cube. Hence $\displaystyle 3\mid2e$ , so $\displaystyle 3\mid e$ . We can remove this factor by dividing $\displaystyle R$ by a power of $\displaystyle q$ . The other case is more interesting. If $\displaystyle q\equiv1\pmod3$ , then $\displaystyle q$ splits:

$\displaystyle q=\pi\overline\pi,\qquad N(\pi)=q.$

Suppose $\displaystyle \pi$ and $\displaystyle \overline\pi$ occur in $\displaystyle b$ to exponents $\displaystyle e,f$ . Since $\displaystyle b\overline b$ is a cube, their exponents satisfy $\displaystyle e+f\equiv0\pmod3$ . After removing ordinary cubes, the only possible nontrivial patterns are therefore

$\displaystyle \pi^2\overline\pi\qquad\text{or}\qquad\pi\overline\pi^2.$

These are exactly the factors we need to remove. Multiplication by a number of $\displaystyle \mathbb Q(\omega)$ cannot remove them, because that only changes exponents by multiples of three. We need an algebraic expression whose cube contains $\displaystyle \pi^2\overline\pi$ . Where can we find one? We will construct it explicitly from Gauss sums, $\displaystyle q$ th roots of unity. Take $\displaystyle q\equiv1\pmod3$ . Choose a multiplicative character $\displaystyle \chi$ of order three modulo $\displaystyle q$ . Its values are $\displaystyle 1,\omega,\omega^2$ . Let $\displaystyle \zeta_q$ be a primitive $\displaystyle q$ th root of unity, and form the weighted sum

$\displaystyle G_q=\sum_{a=1}^{q-1}\chi(a)\zeta_q^a.$

This is another Fourier expression, but now it is built entirely from roots of unity. If we replace $\displaystyle \zeta_q$ by $\displaystyle \zeta_q^u$ , and change the summation variable from $\displaystyle a$ to $\displaystyle ua$ , we find that $\displaystyle G_q$ is multiplied by $\displaystyle \chi(u)^{-1}$ . So it behaves exactly like the Fourier expression $\displaystyle R$ from our unknown cubic field: under the relevant symmetry, both are multiplied by cube roots of unity. Now calculate its cube. Put

$\displaystyle G'_q=\sum_a\chi(a)^2\zeta_q^a$ .

Multiplying $\displaystyle G_qG'_q$ and writing $\displaystyle x=ty$ in the double sum gives $\displaystyle G_qG'_q=q$ . Similarly, grouping the terms of $\displaystyle G_q^2$ according to $\displaystyle x+y=z$ gives $\displaystyle G_q^2=J_qG'_q$ , where

$\displaystyle J_q=\sum_{a\bmod q}\chi(a)\chi(1-a).$

Consequently,

$\displaystyle G_q^3=qJ_q.$

The advantage is that $\displaystyle J_q$ is a sum of cube roots of unity, so it belongs to $\displaystyle \mathbb Z[\omega]$ . We can determine its prime factorization directly. The Gauss-sum identities give $\displaystyle J_q\overline{J_q}=q$ . Thus $\displaystyle J_q$ is a prime element of norm $\displaystyle q$ , associated to either $\displaystyle \pi$ or $\displaystyle \overline\pi$ . To identify the correct factor, choose the cubic character by $\displaystyle \chi(a)\equiv a^{(q-1)/3}\pmod\pi$ . Then reduce the Jacobi sum modulo $\displaystyle \pi$ . Writing $\displaystyle m=(q-1)/3$ , we obtain $\displaystyle J_q\equiv\sum_a a^m(1-a)^m\pmod\pi$ . When the binomial is expanded, every power sum has exponent between $\displaystyle m$ and $\displaystyle 2m$ , strictly between zero and $\displaystyle q-1$ . Each power sum vanishes modulo $\displaystyle q$ . Hence $\displaystyle \pi\mid J_q$ . Finally, the congruence modulo three fixes the unit. Because $\displaystyle (x+y+z)^3\equiv x^3+y^3+z^3\pmod3$ , we have $\displaystyle G_q^3\equiv\sum_{a=1}^{q-1}\zeta_q^{3a}=-1\pmod3$ . Since $\displaystyle q\equiv1\pmod3$ , this gives $\displaystyle J_q\equiv-1\pmod3$ . Choosing the associate $\displaystyle \pi\equiv-1\pmod3$ , we obtain $\displaystyle J_q=\pi$ . Our calculation has therefore produced exactly what we wanted:

$\displaystyle G_q^3=q\pi=\pi^2\overline\pi,\qquad G_q\overline{G_q}=q.$

This is the central identity of the cubic proof. It constructs from roots of unity a Fourier expression with precisely the prime exponents we were unable to remove by multiplying by an element of $\displaystyle \mathbb Q(\omega)$ .

Return to the Fourier expression $\displaystyle R$ of our unknown cubic field. Suppose the factors above $\displaystyle q$ occur in $\displaystyle R^3$ with exponents $\displaystyle (2,1)$ modulo three. Divide by the Gauss sum:

$\displaystyle U=\frac{R}{G_q}.$

Since $\displaystyle G_q^3=\pi^2\overline\pi$ , the new cube $\displaystyle U^3$ has exponents divisible by three at both primes above $\displaystyle q$ . We can remove those remaining multiples of three by multiplying $\displaystyle U$ by an appropriate element of $\displaystyle \mathbb Q(\omega)$ . If the original exponents are $\displaystyle (1,2)$ , divide by $\displaystyle G_q^2$ instead. Its cube has exponent pattern $\displaystyle (4,2)$ , which is congruent to $\displaystyle (1,2)$ modulo three. Thus either possible pattern can be removed.

Why is this really removing ramification, rather than merely changing an equation? Consider a prime $\displaystyle \pi$ above $\displaystyle q$ . After dividing by the Gauss sum and removing cubes of local prime elements, the remaining equation near $\displaystyle \pi$ has the form $\displaystyle U^3=u$ , where $\displaystyle u$ is a local unit. Since $\displaystyle q\neq3$ , the derivative of $\displaystyle X^3-u$ is $\displaystyle 3X^2$ , which does not vanish at a nonzero root modulo $\displaystyle \pi$ . The reduced polynomial has no repeated roots. Its factors lift to the local integer ring without requiring a ramified extension. So the new cubic equation is unramified at $\displaystyle q$ .

This is the concrete prime-removal calculation: divide a resolvent expression by a Gauss sum, and the troublesome prime exponents disappear from the cube equation.

There is one small issue. The new expression $\displaystyle U$ need not belong to the original cubic field, because $\displaystyle G_q$ belongs to a cyclotomic field. But that is not a problem. We are allowed to enlarge the field temporarily, since our goal is to show that the original number can be expressed using roots of unity. After removing the prime, we repeat the operation for every remaining rational prime different from three. Let’s construct an example where we can see the entire cancellation and the new polynomial.

For $\displaystyle q=7$ , choose the cubic character with $\displaystyle \chi(2)=\omega$ . The Jacobi sum calculation gives $\displaystyle J_7=2+3\omega$ . Therefore the corresponding Gauss sum satisfies

$\displaystyle G_7^3=7(2+3\omega),\qquad G_7\overline{G_7}=7.$

For $\displaystyle q=13$ , choose the cubic character with $\displaystyle \chi(2)=\omega^2$ . A direct evaluation of $\displaystyle \sum_a\chi(a)\chi(1-a)$ gives two terms equal to $\displaystyle 1$ , six equal to $\displaystyle \omega$ , and three equal to $\displaystyle \omega^2$ . Thus its Jacobi sum is $\displaystyle 2+6\omega+3\omega^2=-1+3\omega$ , and

$\displaystyle G_{13}^3=13(-1+3\omega),\qquad G_{13}\overline{G_{13}}=13.$

Now multiply the two Fourier sums:

$\displaystyle R=G_7G_{13}.$

Its complex conjugate is $\displaystyle \overline R=\overline{G_7}\overline{G_{13}}$ , and their product is $\displaystyle R\overline R=7\cdot13=91$ .

Consider the real algebraic number

$\displaystyle x=R+\overline R.$

Because each Gauss sum transforms by a cubic character, $\displaystyle R$ has three conjugates $\displaystyle R,\omega R,\omega^2R$ . The corresponding conjugates of $\displaystyle x$ are therefore $\displaystyle R+\overline R$ , $\displaystyle \omega R+\omega^2\overline R$ , and $\displaystyle \omega^2R+\omega\overline R$ . This produces a cyclic cubic field. We can find its polynomial without guessing. Expand $\displaystyle x^3$ :

$\displaystyle x^3=R^3+\overline R^3+3R\overline R(R+\overline R).$

We know $\displaystyle R\overline R=91$ . We also have $\displaystyle R^3=91(2+3\omega)(-1+3\omega)=91(-11-6\omega)$ , and hence $\displaystyle R^3+\overline R^3=-1456$ . Therefore

$\displaystyle x^3-273x+1456=0.$

This is a concrete cubic polynomial obtained by multiplying Fourier sums from seventh and thirteenth roots of unity. It is irreducible over $\displaystyle \mathbb Q$ ; for example, modulo five it becomes $\displaystyle X^3+2X+1$ , which has no root. Now suppose we want to remove the prime seven. Instead of manipulating the cubic polynomial directly, return to the Fourier expression $\displaystyle R$ and divide by the known seventh-root Gauss sum:

$\displaystyle \frac{R}{G_7}=G_{13}.$

The unwanted factor has disappeared exactly. The new real number is

$\displaystyle y=G_{13}+\overline{G_{13}}.$

Using $\displaystyle G_{13}\overline{G_{13}}=13$ and $\displaystyle G_{13}^3+\overline{G_{13}}^3=13((-1+3\omega)+(-1+3\omega^2))=-65$ , we obtain

$\displaystyle y^3-39y+65=0.$

We have gone from a cubic equation involving seventh and thirteenth roots of unity to one involving only thirteenth roots of unity. And we can reconstruct the original number. We know $\displaystyle R=G_7G_{13}$ , so

$\displaystyle x=G_7G_{13}+\overline{G_7G_{13}}.$

The field generated by $\displaystyle x$ is contained in the field generated by $\displaystyle G_7$ and $\displaystyle y$ . Thus removing seven did not destroy the original information. We moved it into a known cyclotomic factor. This is the calculation to keep in mind throughout the general proof. We are not trying to transform a complicated polynomial into a cyclotomic polynomial by factoring it directly. We are finding an expression for one root, separating its Fourier components, and replacing those components by known cyclotomic expressions.

Now return to an arbitrary cyclic cubic field, with Fourier expression $\displaystyle R$ . For each prime $\displaystyle q\equiv1\pmod3$ whose factors occur in $\displaystyle R^3$ with nonzero exponents modulo three, divide by $\displaystyle G_q$ or $\displaystyle G_q^2$ , whichever matches those exponents. At primes $\displaystyle q\equiv2\pmod3$ , the exponents were already divisible by three. After doing this for finitely many primes, we have replaced $\displaystyle R$ by another expression $\displaystyle U$ whose cube has no prime exponents outside three, except multiples of three. Remove those remaining cubes using elements of $\displaystyle \mathbb Q(\omega)$ . The new equation must therefore have the form

$\displaystyle U^3=\varepsilon(1-\omega)^k,$

after multiplying $\displaystyle U$ by an element of $\displaystyle \mathbb Q(\omega)$ , where $\displaystyle \varepsilon$ is a unit of $\displaystyle \mathbb Z[\omega]$ . Why is this so restrictive? Because the units of $\displaystyle \mathbb Z[\omega]$ are only $\displaystyle \pm1,\pm\omega,\pm\omega^2$ , and $\displaystyle 3$ has, up to a unit, just one prime factor in this ring: $\displaystyle 1-\omega$ . We still have the important relation that the norm of the cube’s right-hand side is a rational cube. This relation survives all our divisions, because $\displaystyle R\overline R$ is rational and $\displaystyle G_q\overline{G_q}=q$ . But $\displaystyle N(1-\omega)=3$ . Therefore the norm of $\displaystyle \varepsilon(1-\omega)^k$ contains $\displaystyle 3^k$ . Since the norm is a rational cube, $\displaystyle k$ must be divisible by three. We can absorb that factor into $\displaystyle U$ as well. We are left with

$\displaystyle U^3=\varepsilon.$

Every unit is, modulo cubes of units, one of $\displaystyle 1,\omega,\omega^2$ . Indeed, the cubes of the six units are just $\displaystyle \pm1$ . After one last multiplication by a unit, our equation is simply

$\displaystyle U^3=1,\quad U^3=\omega,\quad\text{or}\quad U^3=\omega^2.$

These equations are completely solved by ninth roots of unity. If $\displaystyle U^3=\omega$ , we can take $\displaystyle U=\zeta_9$ . If $\displaystyle U^3=\omega^2$ , we can take $\displaystyle U=\zeta_9^2$ , up to multiplication by a cube root of unity. And if $\displaystyle U^3=1$ , then $\displaystyle U$ already belongs to $\displaystyle \mathbb Q(\omega)$ .

So the unknown Fourier expression $\displaystyle R$ can be reconstructed from ninth roots of unity and the Gauss sums we used to remove its other prime factors. More precisely, there are finitely many primes $\displaystyle q_1,\ldots,q_s\equiv1\pmod3$ , exponents $\displaystyle e_j\in{1,2}$ , an integer $\displaystyle a\in{0,1,2}$ , and a nonzero number $\displaystyle c\in\mathbb Q(\omega)$ such that

$\displaystyle R=c\zeta_9^a\prod_{j=1}^sG_{q_j}^{e_j}.$

There may be no need for the ninth-root factor, in which case $\displaystyle a=0$ . This is the explicit expression we were looking for. Substitute it back into the Fourier inversion formula:

$\displaystyle \alpha=\frac13\left(T+c\zeta_9^a\prod_jG_{q_j}^{,e_j}+\overline{c\zeta_9^a\prod_jG_{q_j}^{e_j}}\right).$

Every $\displaystyle G_q$ is a finite sum of $\displaystyle q$ th roots of unity, with coefficients among $\displaystyle 1,\omega,\omega^2$ . The number $\displaystyle c$ belongs to $\displaystyle \mathbb Q(\omega)$ , and $\displaystyle T$ is rational. Thus the formula expresses $\displaystyle \alpha$ as an element of a cyclotomic field. We have proved Kronecker–Weber for every cyclic cubic extension by explicitly reconstructing its Fourier expression from Gauss sums and ninth roots of unity. No general class field theory was needed. The essential calculations were unique factorization in $\displaystyle \mathbb Z[\omega]$ , the Jacobi-sum identity, and the fact that the norm of the original cube is itself a cube. This is the most concrete form of the proof.

Prime Order

We have finished the cubic case. We began with a Fourier expression $\displaystyle R$ , found that its cube belonged to $\displaystyle \mathbb Q(\omega)$ , and examined its prime factors. At a prime $\displaystyle q\neq3$ , the only nontrivial patterns were $\displaystyle (2,1)$ and $\displaystyle (1,2)$ . The Gauss sum gave exactly the pattern $\displaystyle (2,1)$ , and its square gave $\displaystyle (1,2)$ modulo three. So we could remove either pattern. We want to repeat this for a cyclic extension of degree $\displaystyle p$ , where $\displaystyle p$ is an odd prime. Afterwards, we will see what changes for degree $\displaystyle p^r$ . Let $\displaystyle K/\mathbb Q$ be cyclic of degree $\displaystyle p$ , with Galois group generated by $\displaystyle \sigma$ . Choose an element $\displaystyle \alpha$ generating the field, and write its conjugates as $\displaystyle x_j=\sigma^j(\alpha)$ . Let $\displaystyle \eta$ be a primitive $\displaystyle p$ th root of unity. As in the cubic case, define

$\displaystyle R_k=\sum_{j=0}^{p-1}\eta^{-kj}x_j$ ,

so that $\displaystyle \sigma(R_k)=\eta^kR_k$ . Let’s concentrate on $\displaystyle R=R_1$ . Since $\displaystyle \sigma(R)=\eta R$ , its $\displaystyle p$ th power is fixed by $\displaystyle \sigma$ . Also, $\displaystyle K$ and $\displaystyle \mathbb Q(\eta)$ have coprime degrees $\displaystyle p$ and $\displaystyle p-1$ . Thus, exactly as before,

$\displaystyle R^p=B,\qquad B\in\mathbb Q(\eta).$

We also get a constraint analogous to the relation $\displaystyle B\bar B\in\mathbb Q^3$ in the cubic case. Multiply all the conjugates of $\displaystyle R$ obtained by replacing $\displaystyle \eta$ with $\displaystyle \eta^a$ , for $\displaystyle a=1,\ldots,p-1$ . Their product is fixed by $\displaystyle \sigma$ , because the multipliers multiply to $\displaystyle \eta^{1+2+\cdots+(p-1)}=1$ . It is also fixed by every symmetry of $\displaystyle \mathbb Q(\eta)$ , since these merely permute the factors. Therefore the product is rational, and its $\displaystyle p$ th power is the norm of $\displaystyle B$ . Hence $\displaystyle N_{\mathbb Q(\eta)/\mathbb Q}(B)$ is a rational $\displaystyle p$ th power.

Now suppose $\displaystyle q\neq p$ ramifies in $\displaystyle K$ .Choose a prime factor above it in the integer ring of $\displaystyle K$ , and choose an algebraic integer $\displaystyle \pi$ that has exactly one factor of this prime locally. Take a Galois symmetry $\displaystyle \tau$ that does not change the residues modulo this prime. Then $\displaystyle \tau(\pi)/\pi$ is a local unit. Reducing it modulo the prime gives a nonzero element of a finite field. Moreover, the multipliers for two such symmetries multiply. Thus we have another multiplicative character, coming directly from how the roots move near $\displaystyle q$ . Now look at a symmetry whose multiplier is one modulo the prime. Its movement begins at the next order: $\displaystyle \tau(\pi)=\pi+c\pi^2+\cdots$ . The first nonzero coefficients at this order add when symmetries are composed. The possible coefficients form an additive group of characteristic $\displaystyle q$ . But our Galois group has order a power of $\displaystyle p$ , and $\displaystyle q\neq p$ . So this additive part must be trivial. In ordinary language, every symmetry responsible for ramification at $\displaystyle q$ is already visible through the first multiplicative coefficient. There is no hidden higher-order movement. Furthermore, because the whole Galois group is abelian, applying the residue-field Frobenius does not alter the multiplier. Thus the multiplier lies in $\displaystyle \mathbb F_q^\times$ , rather than a larger residue field. If the ramification group has order $\displaystyle e=p^s$ , this proves $\displaystyle p^s\mid q-1.$ That is just what we need. The group $\displaystyle (\mathbb Z/q\mathbb Z)^\times$ is cyclic of order $\displaystyle q-1$ , so it contains a character of order $\displaystyle p^s$ .

Consequently in the prime order case, $\displaystyle q$ splits into $\displaystyle p-1$ distinct prime ideals in $\displaystyle \mathbb Q(\eta)$ . Let us call them $\displaystyle \mathfrak q_1,\ldots,\mathfrak q_{p-1}$ . We now want to know the exponents with which these primes occur in $\displaystyle B=R^p$ . The norm relation says that the sum of their exponents is divisible by $\displaystyle p$ . But that is not enough. For example, when $\displaystyle p=5$ , there are many possible vectors whose entries add to a multiple of five. We need a stronger restriction. That restriction comes from the equation $\displaystyle \sigma(R)=\eta R$ . It turns out that the exponent vector is not arbitrary at all: modulo $\displaystyle p$ , it must be one of the patterns obtained from a single Gauss sum.

Since $\displaystyle p\mid q-1$ , choose a character $\displaystyle \chi$ of order $\displaystyle p$ modulo $\displaystyle q$ . Form

$\displaystyle G_q=\sum_{a=1}^{q-1}\chi(a)\zeta_q^a.$

We know that replacing $\displaystyle \zeta_q$ by $\displaystyle \zeta_q^u$ multiplies $\displaystyle G_q$ by $\displaystyle \chi(u)^{-1}$ . Thus its $\displaystyle p$ th power is fixed by every symmetry acting on $\displaystyle \zeta_q$ , and so $\displaystyle G_q^p\in\mathbb Q(\eta)$ . We can calculate the prime factors of $\displaystyle G_q^p$ almost as directly as in the cubic case. Label the primes $\displaystyle \mathfrak q_a$ so that, modulo $\displaystyle \mathfrak q_a$ , the character satisfies $\displaystyle \chi(x)\equiv x^{a(q-1)/p}$ . Write $\displaystyle \lambda=\zeta_q-1$ and expand $\displaystyle \zeta_q^x=(1+\lambda)^x$ . Then $\displaystyle G_q=\sum_j\lambda^j\sum_x\chi(x)\binom{x}{j}$ . Modulo $\displaystyle \mathfrak q_a$ , the coefficient of $\displaystyle \lambda^j$ becomes $\displaystyle \sum_x x^{a(q-1)/p}\binom{x}{j}$ . All these coefficients vanish until $j=(q-1)(p-a)/p$ . At that value, the highest power of $\displaystyle x$ is $\displaystyle x^{q-1}$ , whose sum over the nonzero residues is $\displaystyle -1$ . Since $\binom{x}{j}$ has leading coefficient $\displaystyle 1/j!$ , the coefficient is nonzero. The earlier coefficients are divisible by $\displaystyle q$ locally, so they cannot change the first nonzero power of $\displaystyle \lambda$ . Since $\displaystyle q$ is, up to a unit, $\displaystyle \lambda^{q-1}$ , this gives the exponent of each prime in $\displaystyle G_q^p$ :

$\displaystyle v_{\mathfrak q_a}(G_q^p)=p-a,\qquad 1\le a<p.$

So the Gauss sum gives the pattern $\displaystyle (p-1,p-2,\ldots,1)$ , with the primes labelled as above. Its square gives twice that pattern modulo $\displaystyle p$ , its cube gives three times the pattern, and so on. But how do we know that the unknown $\displaystyle B=R^p$ has one of precisely these patterns?

Temporarily adjoin the degree- $\displaystyle p$ subfield of $\displaystyle \mathbb Q(\zeta_q)$ to $\displaystyle K$ . At $\displaystyle q$ , the ramification group of the combined extension is still cyclic of order $\displaystyle p$ . We can choose its generator $\displaystyle \tau$ to act as $\displaystyle \sigma$ on $\displaystyle K$ , and choose $\displaystyle \chi$ so that it acts on the Gauss sum by $\displaystyle \tau(G_q)=\eta G_q$ . Our unknown resolvent satisfies the identical equation $\displaystyle \tau(R)=\eta R$ . Take a prime element $\displaystyle \pi$ in the local combined field. Because the ramification is tame, $\displaystyle \tau(\pi)/\pi$ reduces to an element $\displaystyle c$ of order $\displaystyle p$ . Write $\displaystyle R=\pi^t u$ and $\displaystyle G_q=\pi^h v$ , with $\displaystyle u,v$ local units. The symmetry $\displaystyle \tau$ does not change the residues of these units. Therefore reducing the two transformation equations gives $\displaystyle c^t=\eta$ and $\displaystyle c^h=\eta$ . Hence $\displaystyle t\equiv h\pmod p$ . Taking $\displaystyle p$ th powers and returning to the field $\displaystyle \mathbb Q(\eta)$ , we obtain the desired constraint:

$\displaystyle v_{\mathfrak q_a}(B)\equiv v_{\mathfrak q_a}(G_q^p)\equiv p-a\pmod p.$

The equality is understood with the primes labelled consistently by the chosen character. Thus the unknown resolvent can only have the same exponent pattern as the Gauss sum, modulo $\displaystyle p$ . Dividing $\displaystyle R$ by $\displaystyle G_q$ removes it. For a Fourier component $\displaystyle R_k$ , whose multiplier is $\displaystyle \eta^k$ , the same argument gives $\displaystyle v_{\mathfrak q_a}(R_k^p)\equiv k(p-a)\pmod p$ , and we divide by $\displaystyle G_q^k$ . There are no other possible patterns because there are only $\displaystyle p$ possible multipliers $\displaystyle 1,\eta,\ldots,\eta^{p-1}$ . Each multiplier forces one of the $\displaystyle p$ exponent patterns.

Take $\displaystyle p=5$ and $\displaystyle q=11$ . Since $\displaystyle 5\mid10$ , we have characters of order five modulo eleven. In $\displaystyle \mathbb Q(\zeta_5)$ , the prime eleven splits into four primes $\displaystyle \mathfrak q_1,\ldots,\mathfrak q_4$ . Our Gauss sum satisfies $\displaystyle G_{11}^5\in\mathbb Q(\zeta_5)$ . Its prime exponents are $\displaystyle (4,3,2,1)$ . For example, at $\displaystyle \mathfrak q_1$ , the character reduces to $\displaystyle \chi(x)\equiv x^2\pmod{\mathfrak q_1}$ . In the expansion using $\displaystyle \lambda=\zeta_{11}-1$ , the first nonzero coefficient occurs at $\displaystyle j=8$ , because $\displaystyle x^2\binom{x}{j}$ first contains $\displaystyle x^{10}$ when $\displaystyle j=8$ . Since eleven corresponds to ten powers of $\displaystyle \lambda$ , the exponent in $\displaystyle G_{11}^5$ is $\displaystyle 5\cdot8/10=4$ . Repeating this at the other three primes gives exponents $\displaystyle 3,2,1$ .

Now take the powers of this pattern modulo five:

Fourier character	Prime exponents modulo 5	Divide by
$\displaystyle k=0$	$\displaystyle (0,0,0,0)$	$\displaystyle 1$
$\displaystyle k=1$	$\displaystyle (4,3,2,1)$	$\displaystyle G_{11}$
$\displaystyle k=2$	$\displaystyle (3,1,4,2)$	$\displaystyle G_{11}^2$
$\displaystyle k=3$	$\displaystyle (2,4,1,3)$	$\displaystyle G_{11}^3$
$\displaystyle k=4$	$\displaystyle (1,2,3,4)$	$\displaystyle G_{11}^4$

These are all five possible patterns for a Fourier component in a cyclic degree-five extension ramified at eleven. We have not assumed this from the norm constraint. We derived it from the transformation law of the resolvent.

Prime Power Case

We have seen how to remove a prime from a cyclic extension of degree $\displaystyle p$ . The Fourier expression satisfies an equation $\displaystyle R^p=B$ , and the prime factors of $\displaystyle B$ can be matched with those of a Gauss sum. We want to do the same thing when the extension has degree $\displaystyle p^r$ . The only new difficulty is that a prime may affect just part of the Galois group. Suppose $\displaystyle K/\mathbb Q$ is cyclic of degree nine, with Galois group generated by $\displaystyle \sigma$ . Choose $\displaystyle \alpha\in K$ and write its conjugates as $\displaystyle x_j=\sigma^j(\alpha)$ . Let $\displaystyle \zeta=\zeta_9$ . We take the usual Fourier expressions

$\displaystyle R_k=\sum_{j=0}^8\zeta^{-kj}x_j,\qquad \alpha=\frac19\sum_{k=0}^8R_k.$

Now suppose seven ramifies, but its ramification group has order only three. Since the full Galois group has order nine, this subgroup is generated by $\displaystyle \tau=\sigma^3$ . Let $\displaystyle \omega=\zeta^3$ . Applying $\displaystyle \tau$ to a Fourier expression shifts the conjugates by three places. Since seven is different from three, we can let this ramification symmetry fix $\displaystyle \zeta$ . The Fourier calculation then gives $\displaystyle \tau(R_k)=\omega^kR_k$ . Look at what this means. The nine Fourier expressions fall into three groups. The expressions $\displaystyle R_0,R_3,R_6$ are unchanged by $\displaystyle \tau$ . The expressions $\displaystyle R_1,R_4,R_7$ are multiplied by $\displaystyle \omega$ . Finally, $\displaystyle R_2,R_5,R_8$ are multiplied by $\displaystyle \omega^2$ . Therefore, although the extension has degree nine, we have only three possible types of movement at seven. What equations do these expressions satisfy? Since $\displaystyle \tau^3=1$ , taking cubes removes the movement:

$\displaystyle R_k^3=B_k,\qquad B_k\in K^{\langle\tau\rangle}(\zeta_9).$

Here $\displaystyle K^{\langle\tau\rangle}$ means the elements of $\displaystyle K$ unchanged by $\displaystyle \tau$ . It is a cubic field, and seven is unramified in it. The field $\displaystyle \mathbb Q(\zeta_9)$ is also unramified at seven. Thus the coefficients $\displaystyle B_k$ belong to a field with no ramification at seven. We can now examine the prime factors of these cube equations, just as we did in the original cubic proof. In the cubic case we calculated a Gauss sum from seventh roots of unity satisfying $\displaystyle G_7^3=\pi^2\bar\pi$ , where $\displaystyle \pi=2+3\omega$ and $\displaystyle \pi\bar\pi=7$ . We will use exactly the same Gauss sum here.

Temporarily combine $\displaystyle K$ with the cyclic cubic subfield of $\displaystyle \mathbb Q(\zeta_7)$ . The ramification group in the combined field still has order three. We can choose the cubic character defining $\displaystyle G_7$ so that our generator $\displaystyle \tau$ acts by $\displaystyle \tau(G_7)=\omega G_7$ . The expressions $\displaystyle R_1,R_4,R_7$ have this same movement. Therefore their cubes must have the same prime-exponent pattern as $\displaystyle G_7^3$ , modulo three. Similarly, $\displaystyle R_2,R_5,R_8$ have the movement of $\displaystyle G_7^2$ , so their cubes have the pattern of $\displaystyle G_7^6$ .

Thus all nine Fourier expressions are covered by just three patterns:

Fourier expressions	Exponents at $\displaystyle \pi,\bar\pi$ modulo 3	Correction
$\displaystyle R_0,R_3,R_6$	$\displaystyle (0,0)$	$\displaystyle 1$
$\displaystyle R_1,R_4,R_7$	$\displaystyle (2,1)$	$\displaystyle G_7$
$\displaystyle R_2,R_5,R_8$	$\displaystyle (1,2)$	$\displaystyle G_7^2$

So we divide each Fourier expression by the indicated Gauss sum. The first group needs no correction. For the second group, put $\displaystyle U_k=R_k/G_7$ , and for the third group put $\displaystyle U_k=R_k/G_7^2$ . In every case, $\displaystyle \tau(U_k)=U_k$ . Moreover, the cube equation for each $\displaystyle U_k$ has prime exponents divisible by three at all primes above seven. We have removed seven from the unknown coefficients. The Fourier expansion still reconstructs $\displaystyle \alpha$ , but now every occurrence of the movement at seven is carried by the explicit Gauss sum $\displaystyle G_7$ . This is the entire idea. We did not need nine different Gauss sums. We needed only three corrections because the ramification group at seven had order three. Now do the same calculation for degree $\displaystyle p^r$

Suppose $\displaystyle K/\mathbb Q$ is cyclic of degree $\displaystyle n=p^r$ , with generator $\displaystyle \sigma$ . Choose $\displaystyle \alpha\in K$ and form the Fourier expressions $\displaystyle R_k=\sum_{j=0}^{n-1}\zeta_n^{-kj}\sigma^j(\alpha)$ , so $\displaystyle \alpha=\frac1n\sum_kR_k$ . Let $\displaystyle q\neq p$ be a ramified prime, and suppose its ramification group has order $\displaystyle e=p^s$ . This group is generated by $\displaystyle \tau=\sigma^{n/e}$ . Put $\displaystyle \eta=\zeta_n^{n/e}$ , a primitive $\displaystyle e$ th root of unity. Exactly as in the degree-nine example, applying $\displaystyle \tau$ shifts the Fourier sum by $\displaystyle n/e$ places. Therefore $\displaystyle \tau(R_k)=\eta^kR_k$ . Taking the $\displaystyle e$ th power removes this movement:

$\displaystyle R_k^e=B_k,\qquad B_k\in K^{\langle\tau\rangle}(\zeta_n).$

This is the correct analogue of the cubic equation. Although $\displaystyle R_k$ may have degree much larger than $\displaystyle e$ , its $\displaystyle e$ th power belongs to a field unramified at $\displaystyle q$ . We now want to remove the factors above $\displaystyle q$ from these equations. Since $\displaystyle e\mid q-1$ , choose a character $\displaystyle \chi$ of order $\displaystyle e$ modulo $\displaystyle q$ , and form the Gauss sum $\displaystyle G_q=\sum_{a=1}^{q-1}\chi(a)\zeta_q^a$ . We choose the character so that the same ramification symmetry satisfies $\displaystyle \tau(G_q)=\eta G_q$ . Now recall the prime-factor calculation for Gauss sums. The prime $\displaystyle q$ splits into $\displaystyle \varphi(e)$ primes in $\displaystyle \mathbb Q(\eta)$ . If we label these primes $\displaystyle \mathfrak q_a$ consistently with the character, then

$\displaystyle v_{\mathfrak q_a}(G_q^e)=e-a,\qquad (a,e)=1,\quad 1\le a<e.$

So the Gauss sum supplies one definite vector of prime exponents. Its $\displaystyle k$ th power supplies $\displaystyle k$ times that vector, modulo $\displaystyle e$ . Why must the unknown equation $\displaystyle R_k^e=B_k$ have that same pattern? Choose a prime element $\displaystyle \pi$ in the combined local field. The ramification symmetry changes it by $\displaystyle \tau(\pi)/\pi\equiv c\pmod\pi$ , where $\displaystyle c$ has order $\displaystyle e$ . Write $\displaystyle R_k=\pi^t u$ and $\displaystyle G_q=\pi^h v$ , with $\displaystyle u,v$ local units. Because $\displaystyle \tau$ leaves the residues of units unchanged, reducing the equations $\displaystyle \tau(R_k)=\eta^kR_k$ and $\displaystyle \tau(G_q)=\eta G_q$ gives $\displaystyle c^t=\eta^k$ and $\displaystyle c^h=\eta$ in the residue field. Hence $\displaystyle c^{t-kh}=1$ . Since $\displaystyle c$ has order $\displaystyle e$ , we obtain $\displaystyle t\equiv kh\pmod e$ .

The equations $\displaystyle B_k=R_k^e$ and $\displaystyle G_q^e\in\mathbb Q(\eta)$ belong to fields unramified at $\displaystyle q$ . Passing from the larger local field to these unramified fields converts $\displaystyle t,h$ into the corresponding prime exponents of $\displaystyle B_k,G_q^e$ . Thus, at every prime above $\displaystyle q$ ,

$\displaystyle v(B_k)\equiv k v(G_q^e)\pmod e.$

This is exactly the constraint we need. The prime exponents of the unknown resolvent equation are forced by its Fourier character. They cannot have arbitrary patterns. There are only $\displaystyle e$ possibilities, corresponding to $\displaystyle k=0,\ldots,e-1$ , and the powers $\displaystyle 1,G_q,\ldots,G_q^{e-1}$ match every one of them. Therefore, if $\displaystyle a_k$ is the remainder of $\displaystyle k$ modulo $\displaystyle e$ , we put $\displaystyle U_k=R_k/G_q^{a_k}$ . The corrected expression satisfies $\displaystyle \tau(U_k)=U_k$ , and its $\displaystyle e$ th power has exponents divisible by $\displaystyle e$ at every prime above $\displaystyle q$ . We have rewritten our original element as

$\displaystyle \alpha=\frac1n\sum_{k=0}^{n-1}U_kG_q^{a_k}.$

Every Gauss sum is an explicit expression in roots of unity. The new coefficients $\displaystyle U_k$ no longer move under the ramification at $\displaystyle q$ . To describe where these coefficients live, let $\displaystyle C_q$ be the cyclic subfield of $\displaystyle \mathbb Q(\zeta_q)$ of degree $\displaystyle e$ , and form $\displaystyle M=KC_q$ . In this combined field, the ramification group at $\displaystyle q$ still has order $\displaystyle e$ . Let $\displaystyle K'$ be the elements of $\displaystyle M$ unchanged by this group. Then $\displaystyle K'$ has no ramification at $\displaystyle q$ , and the corrected coefficients $\displaystyle U_k$ belong to $\displaystyle K'(\zeta_n)$ . Also, $\displaystyle KC_q=K'C_q$ , since the common ramification group maps isomorphically onto the Galois group of $\displaystyle C_q/\mathbb Q$ .

Thus we have removed one ramified prime while keeping an explicit expression for our original element. We can repeat the calculation for each remaining prime different from $\displaystyle p$ . Eventually, only the prime $\displaystyle p$ remains to be handled. The remaining case, where the prime equals $\displaystyle p$ , needs a separate argument.

Wild Ramification

The previous argument depended on the fact that the ramification group at $\displaystyle q$ was detected by a character of $\displaystyle \mathbb F_q^\times$ . But at the prime $\displaystyle p$ , a group of order $\displaystyle p^r$ can move an algebraic number by extremely small amounts without changing its first multiplier. For example, in $\displaystyle \mathbb Q(\zeta_9)$ , take $\displaystyle \pi=\zeta_9-1$ . The symmetry $\displaystyle \zeta_9\mapsto\zeta_9^4$ changes $\displaystyle \pi$ by $\displaystyle \zeta_9^4-\zeta_9=\zeta_9(\zeta_9^3-1)$ . This is divisible by three powers of $\displaystyle \pi$ , not just one.

The first-order character cannot see this movement. It only becomes visible after examining higher powers of the local prime element. For cubic fields, we avoided this difficulty by unique factorization in $\displaystyle \mathbb Z[\omega]$ . After cancelling the other factors, the remaining equation reduced to a cube root of $\displaystyle \omega$ . For arbitrary $\displaystyle p^r$ , such a simple reduction is not available. The larger cyclotomic ring need not even have unique factorization of elements. Instead, we study how closely the conjugate roots of an unknown polynomial can approach one another near $\displaystyle p$ .

Suppose $\displaystyle p$ is odd, and we have a cyclic extension of degree $\displaystyle p$ in which only $\displaystyle p$ ramifies. First, $\displaystyle p$ must be totally ramified. Otherwise there would be a nontrivial intermediate field in which no finite rational prime ramifies. This is impossible: the lattice discriminant bound from the geometry of numbers shows that every number field other than $\displaystyle \mathbb Q$ has discriminant of absolute value greater than one, hence at least one ramified finite prime.

Choose a local prime element $\displaystyle \varpi$ in our extension. Its minimal polynomial over $\displaystyle \mathbb Q_p$ is Eisenstein:

$\displaystyle f(X)=X^p+a_{p-1}X^{p-1}+\cdots+a_1X+a_0.$

All coefficients except the leading one are divisible by $\displaystyle p$ , while the constant term is not divisible by $\displaystyle p^2$ . Normalize the valuation by $\displaystyle v(\varpi)=1$ . Since the extension is totally ramified of degree $\displaystyle p$ , we have $\displaystyle v(p)=p$ . Now write the polynomial using all its conjugate roots:

$\displaystyle f(X)=\prod_{\sigma}(X-\sigma(\varpi)).$

Differentiate and substitute $\displaystyle X=\varpi$ . Every term vanishes except the one obtained by differentiating the factor $\displaystyle X-\varpi$ . Thus

$\displaystyle f'(\varpi)=\prod_{\sigma\neq1}(\varpi-\sigma(\varpi)).$

This is the important expression. Its valuation measures the total number of factors of $\displaystyle \varpi$ occurring in the differences between all the conjugate roots. Because the Galois group has prime order $\displaystyle p$ , all nonidentity symmetries remain together until the first level at which the group becomes trivial. Suppose each difference $\displaystyle \sigma(\varpi)-\varpi$ is divisible by exactly $\displaystyle b+1$ powers of $\displaystyle \varpi$ . The product formula gives $\displaystyle v(f'(\varpi))=(p-1)(b+1)$ .

But we can also differentiate the polynomial in the usual way. The leading term of the derivative is $\displaystyle p\varpi^{p-1}$ , which has valuation $\displaystyle 2p-1$ . Every other nonzero term has valuation congruent to a different number modulo $\displaystyle p$ , because it is of the form $\displaystyle ja_j\varpi^{j-1}$ , with $\displaystyle 1\le j<p$ . The smallest-valuation terms therefore cannot cancel one another. We conclude that

$\displaystyle (p-1)(b+1)=v(f'(\varpi))\le2p-1.$

Since $\displaystyle p$ is odd, this forces $\displaystyle b\le1$ . On the other hand, the first-order multiplicative character is trivial for a group of order $\displaystyle p$ , so every nonidentity symmetry moves $\displaystyle \varpi$ by at least $\displaystyle \varpi^2$ . Therefore $\displaystyle b\ge1$ . Hence $\displaystyle b=1$ .

We have proved a concrete restriction: in a cyclic degree- $\displaystyle p$ extension ramified only at the odd prime $\displaystyle p$ , every nonidentity symmetry moves a local prime element by exactly two powers of that element. The total exponent of the derivative is therefore $\displaystyle 2(p-1)$ . We obtained this without identifying the unknown polynomial. We used only its degree, its Eisenstein coefficients, and the differences of its conjugate roots.

Suppose now that $\displaystyle M/\mathbb Q$ is any abelian extension of $\displaystyle p$ -power degree, with $\displaystyle p$ odd and no other finite prime ramified. We first show that its Galois group must be cyclic. At the prime above $\displaystyle p$ , look at the groups of symmetries whose differences $\displaystyle \sigma(\varpi)-\varpi$ are divisible by increasingly high powers of $\displaystyle \varpi$ . The first two groups are the whole Galois group. Every time the group becomes smaller, the quotient embeds into the additive group of $\displaystyle \mathbb F_p$ , so a nontrivial step removes exactly a factor of $\displaystyle p$ . Let $\displaystyle J$ be the group remaining after the first strict drop. Then $\displaystyle J$ has index $\displaystyle p$ . Suppose another subgroup $\displaystyle H\neq J$ also had index $\displaystyle p$ . The two fixed fields $\displaystyle M^J$ and $\displaystyle M^H$ would both be degree- $\displaystyle p$ extensions ramified only at $\displaystyle p$ . By the polynomial calculation above, the exponent measuring the differences of conjugate roots in each of them is $\displaystyle 2(p-1)$ . Now compare the extension $\displaystyle M/M^J$ with $\displaystyle M/M^H$ . At the initial levels, the relevant groups have the same size. At the first drop, $\displaystyle J$ contains the entire remaining group, whereas $\displaystyle H$ does not. At every later level, the intersection with $\displaystyle J$ is at least as large as the intersection with $\displaystyle H$ . Therefore the total exponent obtained from the product of root differences is strictly larger for $\displaystyle M/M^J$ than for $\displaystyle M/M^H$ . But computing those exponents through the two towers $\displaystyle M/M^J/\mathbb Q$ and $\displaystyle M/M^H/\mathbb Q$ gives the same answer. Indeed, the different of a tower satisfies $\displaystyle d(M/\mathbb Q)=d(M/E)+[M]d(E/\mathbb Q)$ in the totally ramified situation. Both intermediate fields have degree $\displaystyle p$ and the same different exponent $\displaystyle 2(p-1)$ . This contradicts the strict inequality. Hence $\displaystyle J$ is the only subgroup of index $\displaystyle p$ . A finite abelian $\displaystyle p$ -group with only one index- $\displaystyle p$ subgroup is cyclic. If it had two independent cyclic factors, reducing either factor modulo $\displaystyle p$ would give two different homomorphisms onto a group of order $\displaystyle p$ , and hence two different index- $\displaystyle p$ subgroups. So $\displaystyle M/\mathbb Q$ must be cyclic.

Now suppose our remaining field $\displaystyle K/\mathbb Q$ has degree $\displaystyle p^r$ . Inside $\displaystyle \mathbb Q(\zeta_{p^{r+1}})$ , there is a cyclic subfield $\displaystyle C$ of degree $\displaystyle p^r$ , because the full Galois group is cyclic of order $\displaystyle (p-1)p^r$ . Form $\displaystyle M=KC$ . This extension is abelian of $\displaystyle p$ -power degree, and only $\displaystyle p$ ramifies. We just proved that its Galois group must be cyclic. But this cyclic group embeds into the product of the two cyclic groups $\displaystyle \text{Gal}(K/\mathbb Q)$ and $\displaystyle {\text{Gal}}(C/\mathbb Q)$ , each of order $\displaystyle p^r$ . No element of this product has order larger than $\displaystyle p^r$ . Therefore a cyclic subgroup of the product has order at most $\displaystyle p^r$ . Since $\displaystyle M$ contains both $\displaystyle K$ and $\displaystyle C$ , each of degree $\displaystyle p^r$ , this forces $\displaystyle K=C$ . Thus

$\displaystyle K\subseteq\mathbb Q(\zeta_{p^{r+1}})\qquad(p\text{ odd}).$

The last unknown extension has been identified by the restrictions on how its conjugate roots can move near $\displaystyle p$ .

Prime Two: The prime two needs a separate calculation because the degree-two version of the derivative inequality does not restrict the movement as strongly. The first observation is very elementary. A quadratic field $\displaystyle \mathbb Q(\sqrt d)$ , with $\displaystyle d$ squarefree, has discriminant $\displaystyle d$ when $\displaystyle d\equiv1\pmod4$ , and $\displaystyle 4d$ otherwise. If only two ramifies, the only possibilities are $\displaystyle d=-1,2,-2$ . Thus the three quadratic fields are $\displaystyle \mathbb Q(i),\mathbb Q(\sqrt2),\mathbb Q(\sqrt{-2})$ , all contained in $\displaystyle \mathbb Q(\zeta_8)$ . In particular, there is only one real quadratic field ramified only at two, namely $\displaystyle \mathbb Q(\sqrt2)$ .

Now let $\displaystyle K/\mathbb Q$ be a real cyclic extension of degree $\displaystyle 2^r$ , ramified only at two. Let $\displaystyle C$ be the real subfield of $\displaystyle \mathbb Q(\zeta_{2^{r+2}})$ . It is also cyclic of degree $\displaystyle 2^r$ . Form $\displaystyle M=KC$ . The field $\displaystyle M$ is real and abelian, and only two ramifies in it. Hence it can have only one quadratic subfield, namely $\displaystyle \mathbb Q(\sqrt2)$ . But if a finite abelian group of power-of-two order is not cyclic, it has at least three subgroups of index two, and the corresponding fixed fields give at least three quadratic subfields.

Therefore $\displaystyle M/\mathbb Q$ is cyclic. Its Galois group embeds into $\displaystyle C_{2^r}\times C_{2^r}$ , so its degree cannot exceed $\displaystyle 2^r$ . Since it contains $\displaystyle K$ and $\displaystyle C$ , both of that degree, they coincide. Thus every real cyclic $\displaystyle 2$ -power extension ramified only at two is contained in a power-of-two cyclotomic field.

For a nonreal cyclic extension $\displaystyle K/\mathbb Q$ of degree at least four, its unique quadratic subfield is real. Indeed, complex conjugation is the unique element of order two in its cyclic Galois group, and it fixes that quadratic subfield. Therefore $\displaystyle i\notin K$ . Form $\displaystyle M=K(i)$ . Its Galois group is $\displaystyle C_{2^r}\times C_2$ . Complex conjugation is the element whose first component is the unique involution in $\displaystyle C_{2^r}$ and whose second component is the nonidentity element of $\displaystyle C_2$ . Quotienting by that diagonal involution gives a cyclic group of order $\displaystyle 2^r$ .

Therefore the real subfield $\displaystyle M^+$ is a real cyclic extension of degree $\displaystyle 2^r$ , ramified only at two. The real case shows that $\displaystyle M^+$ is cyclotomic. Since $\displaystyle M=M^+(i)$ , the original field $\displaystyle K$ is cyclotomic as well. The prime-two case is complete.

Final Proof

We can now finish the proof. Suppose $\displaystyle K/\mathbb Q$ has an arbitrary finite abelian Galois group. Such a group is a product of cyclic groups of prime-power order. For each cyclic factor, take the field fixed by all the other factors. The original field is the compositum of these smaller cyclic fields. For each cyclic prime-power extension, our Fourier and prime-removal argument eliminates all ramification away from its degree prime, at the cost of adjoining suitable roots of unity. The final calculation at that prime shows that the remaining extension also lies in a cyclotomic field. Thus every cyclic factor belongs to a cyclotomic field. Finally, adjoining roots of unity of orders $\displaystyle m$ and $\displaystyle n$ is the same as adjoining roots of unity of order $\displaystyle {lcm}(m,n)$ . Hence the compositum of finitely many cyclotomic fields is cyclotomic. We conclude that every finite abelian extension $K/\mathbb Q$ is contained in $\mathbb Q(\zeta_N)$ . for some for some $N.$ This is the Kronecker–Weber theorem.

It is worth returning once more to the beginning, because the proof has a simpler interpretation than its later steps might suggest. We started with an element $\displaystyle \alpha$ and wrote its conjugates in Fourier coordinates. In the cubic case, the important coordinate was $\displaystyle R$ , satisfying $\displaystyle \sigma(R)=\omega R$ . Its cube belonged to $\displaystyle \mathbb Q(\omega)$ and could be factored. The exponents of prime factors in $\displaystyle R^3$ showed where the cube-root equation could ramify. At a prime different from three, any exponents not divisible by three could be matched by a Gauss sum built from roots of unity. Dividing by that Gauss sum removed the local obstruction. At the prime three, the argument was different. After the other prime factors were removed, the norm relation forced the remaining cube equation to involve only units. Those units led directly to ninth roots of unity.

In higher degrees, the same distinction remains. At a prime $\displaystyle q$ different from the degree prime $\displaystyle p$ , the movement of conjugate roots is controlled by one multiplicative residue character. A Gauss sum supplies a matching character, so the movement can be cancelled. At the prime $\displaystyle p$ , the conjugates may remain extremely close even after their first-order movement has disappeared. The derivative $\displaystyle f'(\varpi)$ measures how much the conjugates collide. Its divisibility cannot be arbitrary, because the coefficients of $\displaystyle f$ are rational integers. This restriction eventually forces the remaining extension to agree with a prime-power cyclotomic extension. In other words, the proof is built around one repeated operation: replace an unknown expression by a better one whose conjugates have less troublesome behavior, while retaining enough information to recover the original expression.

The cubic proof used two fortunate arithmetic facts. First, $\displaystyle \mathbb Z[\omega]$ has unique factorization, so we could track the prime exponents in $\displaystyle R^3$ very explicitly. Second, after removing all ordinary prime factors, there was essentially nothing left except units and the prime above three. The higher-degree proof uses a related property of $\displaystyle \mathbb Q$ : there is no nontrivial number field unramified at every finite prime. Once all ramification has been removed, no further extension can remain hidden.

For an arbitrary number field, neither simplification is guaranteed. Prime ideals may fail to be generated by single elements, and there may be nontrivial everywhere-unramified abelian extensions. Moreover, residue fields can have $\displaystyle q^f$ elements rather than just $\displaystyle q$ , so the required local characters need not come from ordinary $\displaystyle q$ th roots of unity. The basic Fourier construction still works: abelian symmetries can still be separated into one-dimensional characters. What changes is the arithmetic needed to realize and cancel those characters. Class field theory replaces the simple factorization of rational numbers by ideal classes, congruence classes, and local norm conditions. But the question remains the same as at the beginning: given an algebraic number and its conjugates, which combinations of their symmetries can be reproduced by explicit arithmetic constructions, and what global restrictions prevent any additional combinations from occurring? Kronecker–Weber answers that question completely for the rational numbers.

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