Firstly we note that F is also the field of invariants of the closure $\bar{\Delta}$ of A in Γ, and that $\bar{\Delta}$ is a closed normal subgroup of Γ. By V, p. 68, Cor. 4, F is thus a Galois extension of K. Further, the Galois group of F over K is isomorphic to $\Gamma / \bar{\Delta}$ and hence is commutative. Hence every subextension of F is abelian. Conversely let L be an abelian extension of K contained in E, and let Π be the Galois group of E over L. Since L is Galois, Π is a normal subgroup of Γ and the Galois group of L over K is isomorphic to $\Gamma / \Pi$ (V, p. 68, Cor. 4). Therefore $\Gamma / \Pi$ is commutative and Π contains A, whence $L \subset F$.

In the first place E is a Galois extension of K (V, p. 57, Prop. 1). If the field F is defined as in Prop. 1, then we have $E_i \subset F$ for all $i \in I$, whence $E = F$.

An extension E of K is said to be an abelian closure of K if it is an abelian extension of K and every abelian extension of K is isomorphic to a subextension of E. Prop. 1 implies the existence of an abelian closure of K: for let $K_s$ be a separable closure of K, with Galois group $\Gamma$ and let $(\overline{\Gamma}, \overline{\Gamma})$ be the closure of the derived group of $\Gamma$; denote by $K_{ab}$ the field of invariants of $(\overline{\Gamma}, \overline{\Gamma})$; since every separable algebraic extension of K is isomorphic to a subextension of $K_s$ (V, p. 45, Cor.). Prop. 1 shows that $K_{ab}$ is an abelian closure of K. The Galois group of $K_{ab}$ over K is canonically isomorphic to $\Gamma / (\overline{\Gamma}, \overline{\Gamma})$. Let us now show the uniqueness of abelian closures: let E, E' be two abelian closures of K; by definition there exist K-homomorphisms $u : E \to E'$ and $v : E' \to E$ and Prop. 1 (V, p. 52) implies that $v(u(E)) = E$ and $u(v(E')) = E'$, hence $u$ is a K-isomorphism of E onto E'. Any other K-isomorphism of E onto E' is of the form $u_1 = \sigma_0 \circ u$ with $\sigma_0 \in \mathrm{Gal}(E'/K)$; since $\mathrm{Gal}(E'/K)$ is commutative, the isomorphism $\sigma \mapsto u \circ \sigma \circ u^{-1}$ of $\mathrm{Gal}(E/K)$ onto $\mathrm{Gal}(E'/K)$ is independent of $u$; it is called the canonical isomorphism of $\mathrm{Gal}(E/K)$ onto $\mathrm{Gal}(E'/K)$.

It is enough to prove the first assertion of the theorem which is a consequence of the following more precise lemma:

Consider G as Z-module; we have $mx = 0$ for all $x \in G$, hence the annihilator of G is an ideal of the form $r\mathbf{Z}$, where the integer $r \geq 1$ divides $m$. We thus have $G \subset \mu_r(K)$. By Lemma 4 (V, p. 74) applied with $A = \mathbf{Z}$ and $M = G$ there exists an element $x$ of G of order $r$; let $G'$ be the cyclic subgroup of G generated by $x$. We have $G' \subset \mu_r(K)$, Card $(G') = r$ and Card $\mu_r(K) \leq r$; therefore we have

G' = μ_r(K) ⊃ G and G is cyclic of order r, hence equal to μ_r(K). Since G has order m, we have m = r and the lemma follows.

We have denoted by Z[1/p] the subring of Q generated by 1/p, that is, the set of rational numbers of the form a/p^n with a ∈ Z and n ≥ 1; thus we have Z[1/p] = Z if K is of characteristic 0.

Let (v_n), be the strictly increasing sequence consisting of all the integers which are not divisible by p, if p ≠ 1; put X = v_1 v_2 ... v_n and denote by H_n the group of λ_n-th roots of unity; we have H_{n+1} ⊃ H_n and μ_∞(K) = U H_n. Since H_n is cyclic of order X, (Th. 1), there exists a sequence (a_n), of roots of unity such that a_n generates H_n and a_n = α_n^{v_n + 1}.

Further let β_n be the residue class mod Z[1/p] of 1/λ_n and let H'_n be the cyclic subgroup of Q/Z[1/p] generated by β_n. It is clear that β_n = v_{n+1} β_{n+1} and H'_n is of order λ_n because X is not divisible by p if p ≠ 1. Thus there exists for all n ≥ 1 an isomorphism φ_n : H_n → H'_n such that φ_n(α_n) = β_n and the relations α_n = α_{n+1}^{v_n + 1}, β_n = v_{n+1} β_{n+1}, show that φ_n+, extends φ_n. Finally there exists a unique isomorphism φ of μ_∞(K) onto Q/Z[1/p] extending the isomorphisms φ_n, that is, φ(α_n) = β_n for all n ≥ 1.

a) the residue class of $k$ mod $n$ is invertible in the ring $\mathbf{Z}/n\mathbf{Z}$; b) the residue class of $k$ mod $n$ generates the cyclic group $\mathbf{Z}/n\mathbf{Z}$; c) the integers $k$ and $n$ are relatively prime (I, p. 112).

Each of the conditions a) and b) means that there exists an integer $x$ such that $kx \equiv 1 \pmod{n}$, that is, there exist two integers $x$ and $y$ such that $kx + ny = 1$. This latter condition just means that $k$ and $n$ are relatively prime.

Since $G$ is isomorphic to $\mathbf{Z}/n\mathbf{Z}$, the number of generators of $G$ is equal to $\varphi(n)$ by Prop. 3. Let $g$ be a generator of $G$; then the elements $h$ of $G$ such that $h^d = 1$ constitute the subgroup $H$ of $G$ generated by $g^{n/d}$; this group is cyclic of order $d$ and the elements of order $d$ of $G$ are the generators of $H$, hence their number is $\varphi(d)$.

With the notation of Cor. 1, every element of $G$ has a finite order which is a divisor $d$ of $n$ and for a fixed $d$ there are $\varphi(d)$ such elements.

The calculation of $\varphi(n)$ is based on the two formulae:

(2) $\varphi(mn) = \varphi(m)\varphi(n)$ if $m$ and $n$ are relatively prime , (3) $\varphi(p^a) = p^{a-1}(p-1)$ ($p$ prime, $a \geq 1$).

The first follows at once from the fact that the rings $\mathbf{Z}/mn\mathbf{Z}$ and $(\mathbf{Z}/m\mathbf{Z}) \times (\mathbf{Z}/n\mathbf{Z})$ are isomorphic (I, p. 112), and that $(A \times B)^* = A^* \times B^*$ for two rings $A$ and $B$. To prove (3) we note that the positive divisors of $p^a$ are $1, p, p^2, \ldots, p^a$; hence the integer $k$ has no common divisor with $p^a$ other than $1$ if and only if it is not divisible by $p$; since there are $p^a - 1$ multiples of $p$ between $0$ and $p^a - 1$, we find indeed (3).

The formulae (2) and (3) show at once that

where $p$ runs over the set of prime divisors of $n$.

^1 The relation $d \mid n$ between integers > 0 means « d divides n » (VI, p. 5).

An n-th root of unity is said to be primitive if it is of order $n$; if there exists such a root $\zeta$, the group $\mu_n(K)$ is of order $n$ and is generated by $\zeta$. \* For example, the primitive n-th roots of unity in $\mathbf{C}$ are the numbers $e^{2\pi i k / n}$ with $0 \leq k < n$ and $k$ prime to $n$. \* Cor. 1 of Prop. 3 now implies the following result.

Let Q be an algebraic closure of the field Q of rational numbers, and let n be an integer. Then the cyclotomic polynomial $\Phi_n$ of level $n$ is defined by

where $S_n$ is the set of primitive n-th roots of unity in $Q$. The polynomial $\Phi_n$ is of degree $\varphi(n)$ (Prop. 4). It is clear that $\Phi_n(X)$ is invariant under all automorphisms of $\bar{\mathbf{Q}}$, and so belongs to $\mathbf{Q}[X]$. Since every element $\zeta$ of $S_n$ is a root of the polynomial $X^n - 1$, the polynomial $\Phi_n(X)$ divides $X^n - 1$, and the following lemma shows that $\Phi_n(X)$ is a monic polynomial with integer coefficients.

Let $a$ (resp. $b$) be the least of the integers $\alpha \geq 3$ (resp. $\beta \geq 1$) such that $\alpha g$ (resp. $\beta h$) has integer coefficients; we put $g' = ag$ and $h' = bh$ and show by reductio ad absurdum that $a = b = 1$. If this were not so, there would be a prime divisor $p$ of $ab$. If $u \in \mathbf{Z}[X]$, let us write $\bar{u}$ for the polynomial with coefficients in the field $\mathbf{F}_p$ obtained by reduction mod $p$ of the coefficients of $u$. We have $g'h' = abf$, whence $\bar{g}'\bar{h}' = 0$; since the ring $\mathbf{F}_p[X]$ is an integral domain (IV, p. 9, Prop. 8) we thus have $\bar{g}' = 0$ or $\bar{h}' = 0$. In other words, $p$ divides all the coefficients of $g'$ or all those if $h'$ and this contradicts the hypothesis.

We have the relation

For we have $X^n - 1 = \prod_{\zeta \in \mu_n(\mathbf{Q})} (X - \zeta)$ and the sets $S_d$ for $d$ dividing $n$ form a partition of $\mu_n(\mathbf{Q})$.

The formula (6) determines $\Phi_n(X)$ when we know $\Phi_d(X)$ for all the divisors $d < n$ of $n$; since $\Phi_1(X) = X - 1$, we thus have a recursive procedure for calculating $\Phi_n$. For example for prime $p$ we have

whence

and

Let us list the values of the polynomials $\Phi_n(X)$ for $1 \leq n \leq 12$:

The values of $\Phi_1, \Phi_2, \Phi_3, \Phi_4, \Phi_5, \Phi_7, \Phi_8, \Phi_9$ and $\Phi_{11}$ follow directly from (7); now we have $\Phi_1 \Phi_2 \Phi_3 \Phi_6 = X^6 - 1$ and $\Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12} = X^{12} - 1$, whence $\Phi_4 \Phi_{12} = \frac{X^{12} - 1}{X^6 - 1} - X^6 + 1$ and finally $\Phi_{12} = \frac{X^6 + 1}{X^2 + 1} = X^4 - X^2 + 1$. The cases $n = 6$ and $n = 10$ may be treated similarly.

Denote by $S$, the set of roots of $\Phi_n$ in $K_s$. By Formula (6), the set $\mu_n(K_s)$ is the union of the $S_d$ for $d$ dividing $n$. Every primitive $n$-th root of unity thus belongs to $S_n$ and the lemma now follows from Prop. 4 (V, p. 81).

By Lemma 3 we have $R_n(K) = K(\zeta)$ for each root $\zeta$ of $\Phi_n(X)$ and hence $\Phi_n(X)$ is irreducible in $K[X]$ if and only if the degree $\varphi(n)$ of $\Phi_n(X)$ is equal to $[R_n(K):K]$. Further, the Galois group of $R_n(K)$ over $K$ is of order $[R_n(K):K]$ and it is isomorphic to the subgroup of $(\mathbf{Z}/n\mathbf{Z})^*$ which is the image of $\chi_n$. Now Prop. 6 follows from the fact that $(\mathbf{Z}/n\mathbf{Z})^*$ is of order $\varphi(n)$.

a) The cyclotomic polynomial $\Phi_n(X)$ is irreducible in $Q[X]$.

b) The degree of $R_n(Q)$ over $Q$ is $\varphi(n)$.

c) The homomorphism $\chi_n$ of $\mathrm{Gal}(\bar{Q}/Q)$ into $(\mathbf{Z}/n\mathbf{Z})^*$ is surjective and defines by passage to quotients an isomorphism of $\mathrm{Gal}(R_n(Q)/Q)$ onto $(\mathbf{Z}/n\mathbf{Z})^*$.

Taking account of Prop. 6, we need only prove c). Every integer $r$ prime to $n$ is a product of prime numbers $p_1, \ldots, p_s$ not dividing $n$; so it is enough to show that for every prime number $p$ not dividing $n$, the mapping $x \mapsto x^p$ of $\mu_n(Q)$ into itself extends to an automorphism of $R_n(Q)$. It suffices to prove that if $\zeta$ is a primitive $n$-th root of unity, the minimal polynomial $f$ of $\zeta$ over $Q$ is equal to the minimal polynomial $g$ of $\zeta^p$ over $Q$.

We shall assume that $f \neq g$ and argue by contradiction. The polynomials $f$ and $g$ are monic irreducible in $Q[X]$ and divide $X^n - 1$, so there exists $u \in Q[X]$ such that $X^n - 1 = fg u$ (IV, p. 13, Prop. 13). Lemma 2 (V, p. 82) shows that $f, g$ and $u$ have integer coefficients. Let us denote by $\bar{v}$ the polynomial with coefficients in $\mathbf{F}_p$ obtained from a polynomial $v \in \mathbf{Z}[X]$ by reduction mod $p$. We thus have $X^n - 1 = \bar{f} \bar{g} \bar{u}$ in $\mathbf{F}_p[X]$.

Further we have $g(\zeta^p) = 0$ and so $g(X^p)$ is a multiple of $f(X)$ in $\mathbf{Q}[X]$. By Lemma 2 there exists $h \in \mathbf{Z}[X]$ such that $g(X^p) = f(X) \cdot h(X)$. Now we have $v(X^p) = v(X)^p$ for every polynomial $v \in \mathbf{F}_p[X]$. By reduction mod $p$ we thus obtain $\bar{g}^p = \bar{f} \bar{h}$. If $v$ is an irreducible polynomial in $\mathbf{F}_p[X]$ dividing $\bar{f}$, it must then divide $\bar{g}$. Since $\bar{f} \bar{g}$ divides $X^n - 1$, we conclude that $v^2$ divides $X^n - 1$ in $\mathbf{F}_p[X]$. This is absurd because the polynomial $X^n - 1$ is separable in $\mathbf{F}_p[X]$.

It may be shown that for every abelian extension $E$ of finite degree over $Q$ there exists an integer $n \geq 1$ such that $E$ is isomorphic to a subextension of $R_n(Q)$. \* In other words, the field $Q(\mu_{\infty}(C))$ is an abelian closure of $Q$. *(« Kronecker-Weber theorem »).

a) For an element $x \in E$ to satisfy $N_{E/K}(x) = 1$ it is necessary and sufficient that $y \in E^*$ exist such that $x = y / \sigma(y)$; every $y_1 \in E^*$ with $x = y_1 / \sigma(y_1)$ is then of the form $\lambda y$ where $\lambda \in K^*$.

b) For an element $x \in E$ to satisfy $\mathrm{Tr}_{E/K}(x) = 0$ it is necessary and sufficient that $z \in E$ exist with $x = z - \sigma(z)$; then every $z_1 \in E$ with $x = z_1 - \sigma(z_1)$ is of the form $z + \mu$, where $\mu \in K$.

Let us first prove a lemma.*

Put $u(f) = f(a)$ for $f \in Z$ and $t(m) = \sum_{\tau \in \Gamma} \tau . m$ for $m \in M$. Then the sequence

is exact.

Let $f \in \mathbf{Z}$; taking $\tau = \tau' = 1$ in (10) we find that $f(1) = 0$. Further, by induction on $m \geqslant 0$ we deduce the relation

We have $\sigma^n = 1$, whence $f(\sigma^n) = 0$; the preceding relation with $m = n$ is equivalent to the equality $t(u(f)) = 0$, whence $\operatorname{Im}\ u \subset \operatorname{Ker}\ t$. Moreover, from (11) it follows that $\operatorname{Ker}\ u = 0$.

Let $m \in M$ be such that $t(m) = 0$, that is, $m + \sigma \cdot m + \cdots + \sigma^{n-1} \cdot m = 0$. Let us define the mapping $f$ of $\Gamma$ into $M$ by

for $0 \leqslant j \leqslant n - 1$. It may be left to the reader to establish the relation (10). We clearly have $m = f(a)$, whence $\operatorname{Im}\ u \supset \operatorname{Ker}\ t$.

With the lemma now proved, let us take $y \in E^*$ and $x = y/\sigma(y)$; we have $N_{E/K}(\sigma(y)) = N_{E/K}(y)$, whence $N_{E/K}(x) = 1$. Conversely, let $x$ in $E^*$ be such that $N_{E/K}(x) = 1$; by Lemma 4, applied to $M = E^*$, there exists a family of elements $(c_\tau)_{\tau \in \Gamma}$ of $E^*$ satisfying the relation $c_{\tau \tau'} = c_\tau \cdot \tau(c_{\tau'})$ for $\tau, \tau'$ in $\Gamma$ and $c_\sigma = x$. By Cor. 1 of Prop. 9 (V, p. 65) there exists $y \in E^*$ with $c_\tau = y/\tau(y)$ for all $\tau \in \Gamma$, whence in particular $x = c_\sigma = y/\sigma(y)$. If $y_1 \in E^*$ satisfies $x = y_1/\sigma(y_1)$, then we have

hence $y_1 y^{-1}$ belongs to $K^*$ because $a$ generates the Galois group of $E$ over $K$. This proves a).

The assertion b) follows in similar fashion from Cor. 2 of Prop. 9 (V, p. 65).

For let $f : H \to T$ be a homomorphism; we shall prove that there exists a homomorphism of $G$ into $T$ extending $f$. Suppose first that $G$ is cyclic, generated by an element $g$ of order $r$ dividing $n$; denote by $t$ a generator of $T$. There exists a divisor s of r such that H is generated by $g^s$ (I, p. 50, Prop. 19), and for every $x \in \mathbf{Z}$ we have $f(g^{sx}) = t^{ax}$, where a is an integer such that n divides $ar/s$. Then $a/s = (ar/ns)(n/r)$ is an integer and the homomorphism $g^x \mapsto t^{(a/s)x}$, $x \in \mathbf{Z}$, of G into T extends f. In the general case consider the set of pairs $(H', f')$ where $H'$ is a subgroup of G containing H and $f'$ is a homomorphism of $H'$ into T extending $f$, and let us order this set by the relation $(H', f') \leq (H'', f'')$ if $H' \subset H''$ and the restriction of $f''$ to $H'$ is $f'$. By Set Theory, III, p. 154, Def. 3 and Th. 2, this set has a maximal element $(H_1, f_1)$ and it suffices to prove that $H_1 = G$; if this is not the case, there exists $g \in G, g \notin H_1$ and it is enough to prove that $f_1$ may be extended to a homomorphism T of the subgroup of G generated by $H_1$ and g. Now if C denotes the cyclic group generated by g, the restriction of $f_1$ to $C \cap H_1$ extends to a homomorphism $f_2$ of C into T and the homomorphism $xy \mapsto f_1(x) f_2(y), x \in H_1, y \in C,$ of $H_1C$ into T is the desired mapping.

For it suffices to note that if H is a cyclic subgroup of G distinct from $\{1\}$, then $\mathrm{Hom}(H, T) \neq \{1\}$, and to apply Prop. 7.

If G is cyclic of order r, with generator g, then the mapping $f \mapsto f(g)$ is a bijection of $\mathrm{Hom}(G, T)$ onto the set of elements t of T such that $t^r = 1$, hence the assertion follows in this case. On the other hand, if H is a cyclic subgroup of G, we have $\mathrm{Card}(G) = \mathrm{Card}(H) \cdot \mathrm{Card}(G/H)$; further, we have an exact sequence

(II, p. 227, Th. 1 and Prop. 7 above), hence

Since $\mathrm{Card}(\mathrm{Hom}(H, T)) = \mathrm{Card}(H)$, it comes to the same to prove the corollary for G or for $G/H$. Now the result follows by induction on $\mathrm{Card}(G)$.

Now let G and H be two groups and $f : G \times H \to T$ a bimultiplicative mapping, that is, such that for any $g, g' \in G, h, h' \in H$ we have

We define group homomorphisms

by $s_f(g)(h) = d_f(h)(g) = f(g, h)$ (cf. II, p. 268, Cor. to Prop. 1, when G and H are commutative).