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For distinct or equal prime numbers $\ell ,\lambda$, the polynomial $X^{\ell }-\lambda$ is Eisenstein at $\lambda$. Its splitting field over $\mathbb{Q}({\zeta }_{\ell })$ has Galois group a subgroup of $C_{\ell }$, which is nontrivial because a degree-$\ell$ radical cannot belong to a degree-$(\ell -1)$ cyclotomic field. Thus the full Galois group has order $\ell (\ell -1)$. On roots indexed by $j\in {\mathbb{F}}_{\ell }$, all automorphisms act as $j\mapsto uj+b$. Equality of orders identifies the group with the full affine group over a finite field.

// Target: algebra.bigb

Let $V_{\infty }$ be the norm limit of local principal units in the $p$-power cyclotomic field tower, for odd $p$. The Coleman power series $f_u$ gives the Frobenius-corrected logarithm $p^{-1}\log (f_u^p/\varphi f_u)$, whose inverse Amice transform is supported on ${\mathbb{Z}}_p^{\times }$. This is the middle map in the displayed exact sequence; the last map is the first moment $\mu \mapsto \int xd\mu$. Its kernel consists of norm-compatible $p$-power roots of unity. The logarithmic difference equation has one resonant degree, namely degree one in $\log (1+T)$, accounting for the last term. Complex conjugation acts by $-1$ on both copies of ${\mathbb{Z}}_p(1)$. Taking its even part therefore identifies the local principal-unit limit in the real tower with ${\mathbb{Z}}_p[[{\mathbb{Z}}_p^{\times }/\{\pm 1\}]]$.

For a prime $p=n^2+1$, this real number field is a Galois extension of degree four with cyclic Galois group. If $\theta =\sqrt{p+\sqrt{p}}$, its minimal polynomial of an algebraic element is $X^4-2p{X}^2+p^2-p$, irreducible by Eisenstein's criterion at $p$. The other positive root $\beta =\sqrt{p-\sqrt{p}}$ belongs to the same field because $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$. Thus all four roots lie in the field, proving normality. The Galois automorphism $\theta \mapsto \beta$ sends $\sqrt{p}$ to its negative and consequently sends $\beta$ to $-\theta$. It has order four, so the group is $C_4$, rather than a product of two order-two groups.

The prime condition is used for irreducibility, while the square condition on $p-1$ supplies the missing conjugate inside the field. The example $p=2$ yields $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, the real subfield of the sixteenth cyclotomic field.

For the cyclotomic field $\mathbb{Q}({\zeta }_n)$, arithmetic Frobenius automorphisms at primes $\ell \nmid n$ send ${\zeta }_n$ to ${\zeta }_n^{\ell }$: reduction distinguishes the nth roots of unity, and Frobenius raises residues to their $\ell$th powers. Multiplicativity therefore sends a fractional ideal with positive rational generator $a$, prime to $n$, to the exponent $a\mathrm{mod}n$. Its kernel consists exactly of positive generators congruent to one modulo $n$. This proves Artin reciprocity for the ray class group of modulus $n\infty$ without assuming the general reciprocity theorem.

For a primitive $m$th root of unity ${\zeta }_m$, every Galois automorphism sends ${\zeta }_m$ to a unique ${\zeta }_m^a$ with $a$ invertible modulo $m$. The assignment $\sigma \mapsto a$ is a group homomorphism, since composition multiplies the exponents, and is injective because ${\zeta }_m$ generates the cyclotomic field. The irreducibility of cyclotomic polynomials gives degree $\phi (m)$, equal to the order of the target, so the map is an isomorphism. The cyclotomic character on an absolute Galois group factors through this finite quotient and generally has a nontrivial kernel.

A cyclotomic integer is an element of $\mathbb{Z}[{\zeta }_n]$, where ${\zeta }_n$ is a root of unity. For a primitive $n$-th root, this is the full ring of integers of the cyclotomic field $\mathbb{Q}({\zeta }_n)$.

For $a\ge 1$ and $p\nmid m$, the identity ${\Phi }_{p^{a}m}(X)={\Phi }_m(X^{p^a})/{\Phi }_m(X^{p^{a-1}})$ gives the displayed reduction, since taking a $p$th power equals substitution $X\mapsto X^p$ in characteristic $p$. The reduction of ${\Phi }_m$ is squarefree because it divides the separable polynomial $X^m-1$. The Kummer-Dedekind theorem therefore gives ramification index of a prime ideal $\phi (p^a)$ in the cyclotomic field. In particular, a conductor factor $2$ occurring only to the first power has index one and creates no ramification by itself.

Cyclotomic units include roots of unity and suitable ratios $(1-{\zeta }^a)/(1-\zeta )$ in cyclotomic fields. Compatible roots $\zeta$ make these ratios norm-compatible along cyclotomic towers. Their closed norm-limit module is a concrete submodule of global units, and its local images generate the power series in the Iwasawa main conjecture.

A Gaussian period is a sum of roots of unity indexed by a multiplicative coset in the group of nonzero residues modulo a prime number. Such sums describe subfields of a cyclotomic field and occur as values of induced characters of certain semidirect products.

For $m\ge 3$, put $\eta ={\zeta }_m+{\zeta }_m^{-1}$. Since $\eta$ is real and ${\zeta }_m$ is nonreal but satisfies $T^2-\eta T+1=0$, the cyclotomic field $\mathbb{Q}({\zeta }_m)$ has degree two over $\mathbb{Q}(\eta )$. Complex conjugation fixes $\eta$ and is nontrivial. The Fundamental theorem of Galois theory therefore identifies its fixed field with $\mathbb{Q}(\eta )$. That fixed field is exactly $\mathbb{Q}({\zeta }_m)\cap \mathbb{R}$, proving maximality among real subfields.

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

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

The factorization form of Kummer-Dedekind theorem is as follows. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial $f$, and $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. ThenThe index condition is essential. All integral bases used below make the index one, so it applies to every prime.

For a squarefree quadratic $d\equiv 2,3(\mathrm{mod}4)$, take $\theta =\sqrt{d}$, $f=X^2-d$ and field discriminant $D=4d$. At an odd prime, a repeated factor occurs exactly when $p\mid d$, in which case $\bar{f}=X^2$ and $p{\mathcal{O}}_K=(p,\sqrt{d}{)}^2$. At $p=2$, reduction is $X^2$ if $d$ is even and $(X+1{)}^2$ if $d$ is odd, so $2$ is totally ramified in both cases. Thus the totally ramified primes are exactly those dividing $4d$.

For $d\equiv 1(\mathrm{mod}4)$, take $\theta =(1+\sqrt{d})/2$, with polynomial $X^2-X+(1-d)/4$ and discriminant $D=d$. At an odd prime it has a double root exactly when $p\mid d$, giving a single linear factor squared and total ramification. At $p=2$, its derivative is $1$, so no repeated factor occurs. Hence again

For the prime cyclotomic field, use ${\Phi }_q(X)=1+X+\cdots +X^{q-1}$. Modulo $q$ this is $(X-1{)}^{q-1}$, so $q{\mathcal{O}}_K=(q,\zeta -1{)}^{q-1}$, proving total ramification. For $p\ne q$, the discriminant has no prime factor $p$, so the reduction is squarefree and $p$ is unramified. There is an actual exponent error in the printed discriminant: the correct value is $(-1{)}^{(q-1)/2}{q}^{q-2}$, not $q^{q-1}$. To check it, at a nontrivial root ${\zeta }^j$,and multiplying these derivatives and using ${\prod }_{j=1}^{q-1}(1-{\zeta }^j)=q$ gives the exponent $q-2$. The prime support is still just $\{q\}$, so the requested ramification conclusion is unaffected.

For $\theta =\sqrt[3]{2}$, the integral basis gives $f=X^3-2$ and $D=-108$. Modulo $2$, $\bar{f}=X^3$; modulo $3$, $\bar{f}=(X+1{)}^3$. Dedekind factorization gives $(2)=(2,\theta {)}^3$ and $(3)=(3,\theta +1{)}^3$. For other primes the polynomial is squarefree because its discriminant is not divisible by the prime. Therefore the equivalence also holds in this cubic field. This is a conclusion for these specific fields, not the false general assertion that every ramified prime in every higher-degree number field is totally ramified.

Let $K=\mathbb{Q}(\alpha )$ with $\alpha$ an algebraic integer, let $m\in \mathbb{Z}[X]$ be its monic minimal polynomial of an algebraic element, and write $A=\mathbb{Z}[\alpha ]$. The Kummer-Dedekind theorem applies to a rational prime $p$ not dividing the finite index $[{\mathcal{O}}_K:A]$. Ifwith distinct monic irreducible polynomials $g_j$, choose monic integer lifts ${\tilde{g}}_j$. Then the distinct prime ideals above $p$ areIn particular, $p$ is unramified precisely when $\overline{m}$ has no repeated irreducible factor. The exclusion of primes dividing the index is essential.

Here is the quotient-ring proof of the Kummer-Dedekind theorem. The finite group ${\mathcal{O}}_K/A$ has order coprime to $p$. Localizing at the integers coprime to $p$ therefore gives $A\otimes {\mathbb{Z}}_{(p)}={\mathcal{O}}_K\otimes {\mathbb{Z}}_{(p)}$, and reduction yieldsThe second isomorphism follows from $A\simeq \mathbb{Z}[X]/(m)$. The Chinese remainder theorem givesEach factor is a local ring with unique maximal ideal generated by $g_j$ and residue field ${\mathbb{F}}_p[X]/(g_j)$. Pulling its maximal ideal back to ${\mathcal{O}}_K$ gives exactly ${\mathfrak{p}}_j$; these exhaust the primes above $p$. This proves the asserted residue degrees and their independence of the chosen lifts.

It remains to prove the exponents rather than simply identify the primes. Unique prime ideal factorization in the Dedekind domain ${\mathcal{O}}_K$ writes $(p)={\prod }_j{\mathfrak{p}}_j^{b_j}$. At ${\mathfrak{p}}_j$, its discrete valuation ring has $p=u{\pi }^{b_j}$, with $u$ a unit. Thus the maximal ideal of ${\mathcal{O}}_{K,{\mathfrak{p}}_j}/p{\mathcal{O}}_{K,{\mathfrak{p}}_j}$ has nilpotence index of an ideal $b_j$: its $b_j$th power is zero and its $(b_j-1)$st is nonzero. In ${\mathbb{F}}_p[X]/(g_j^{e_j})$ this index is $e_j$, since $g_j^{e_j}=0$ but $g_j^{e_j-1}\ne 0$. The displayed product isomorphism identifies these local factors, hence $b_j=e_j$. This completes the theorem's proof. The same local argument works over a base number field whenever ${\mathcal{O}}_K[\alpha ]$ equals the full relative ring of integers after localizing at the chosen base prime.

For the required cyclotomic field, the allowed ring of integers formula makes the index one. Its cyclotomic polynomial isFor $p\notin \{2,5\}$, the polynomial $X^{80}-1$ has derivative $80X^{79}$ coprime to it modulo $p$. Its divisor ${\overline{\Phi }}_{80}$ is therefore squarefree, so the Kummer-Dedekind theorem gives no ramification.

The cyclotomic reduction at a conductor prime is explicit at the remaining two primes:The quartic over ${\mathbb{F}}_2$ has no linear factor. Its only possible irreducible quadratic divisor would be $X^2+X+1$, whose roots have order three, whereas its own roots have order five. Hence it is irreducible. The two quartics over ${\mathbb{F}}_5$ are also irreducible: their roots have multiplicative order sixteen because their fourth powers are $2$ or $3$, both of order four; the finite-field Frobenius automorphism has orbit length four on an element of order sixteen, since $5^4\equiv 1(\mathrm{mod}16)$ and neither $5$ nor $5^2$ is $1$ modulo sixteen.

Consequently the prime ideal factorization above $2$ has one prime with ramification index eight and residue degree four. Above $5$ it has two primes, each with ramification index four and residue degree four. In particular,

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

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

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

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

Write $K=\mathbb{Q}({\mu }_m)$, $L=\mathbb{Q}({\mu }_n)$. For coprime $m,n$, Bézout's identity and powers of a primitive $mn$-th root give $KL=\mathbb{Q}({\mu }_{mn})$. The degrees of these cyclotomic fields are $\phi (m)$, $\phi (n)$ and $\phi (mn)=\phi (m)\phi (n)$. The natural restriction mapis injective and, by these orders, bijective. Under this product description the subgroups fixing $K$ and $L$ are the two complementary factors. Together they generate the whole group, so their common fixed field is $\mathbb{Q}$. By the Fundamental theorem of Galois theory, $K\cap L=\mathbb{Q}$. This also includes $m=1$ or $n=1$.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.

A cyclotomic field over $\mathbb{Q}$ is a field $\mathbb{Q}({\zeta }_m)$ generated by an $m$th root of unity of exact order $m$. It contains all roots of $X^m-1$, and is its splitting field. In characteristic zero this polynomial is separable, so the extension is Galois.

Every automorphism sends ${\zeta }_m$ to ${\zeta }_m^a$ with $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$, and is determined by that image. Composition multiplies the exponents modulo $m$, yielding an injective homomorphism into an abelian group. Thus the Galois group is abelian. Using irreducibility of the cyclotomic polynomial, every unit exponent occurs andIf “cyclotomic” is used more broadly for a subfield of such a field, it is still Galois and abelian, because every subgroup of an abelian Galois group is normal.

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

The multiplicative group of $L={\mathbb{F}}_{q^N}$ is cyclic of order $m=q^N-1$. For a primitive element $\zeta$, $L={\mathbb{F}}_q(\zeta )$, so this is a cyclotomic extension. The cyclotomic character isIts image is the cyclic subgroup $⟨q⟩$, of order $N$.