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Algebraic Number Theory (V): Cyclotomic Fields

While developing any theory, it is always helpful to have explicit examples at hand. We have previously encountered the family of quadratic fields, for which it is possible to work out many of their properties (eg. generators of the number ring). In this post, we introduce the cyclotomic fields, which form another such family of fields where explicit computations are viable.

Cyclotomic fields

For an integer m1m\geq1, let ω=e2πi/mC\omega=e^{2\pi i/m}\in\bb C be a primitive mm-th root of unity. Then Q[ω]\bb Q[\omega] is called the mm-th cyclotomic field.

We have already encountered some examples of cyclotomic fields. For instance, the mm-th cyclotomic field is Q\bb Q for m=1,2m=1,2, Q[1+32]\bb Q\left[\frac{1+\sqrt{-3}}2\right] for m=3,6m=3,6, and Q[i]\bb Q[i] for m=4m=4.

Proposition: Let m1m\geq1 be odd. Then the mm-th cyclotomic field is equal to the 2m2m-th cyclotomic field, ie. Q[e2πi/m]=Q[e2πi/2m]\bb Q[e^{2\pi i/m}]=\bb Q[e^{2\pi i/2m}].

Proof: \subseteq follows from e2πi/m=(e2πi/2m)2e^{2\pi i/m}=\left(e^{2\pi i/2m}\right)^2, and \supseteq follows from e2πi/2m=(e2πi/m)(m+1)/2e^{2\pi i/2m}=-\left(e^{2\pi i/m}\right)^{(m+1)/2}.  \qed

The above result essentially follows from the fact that 1Q-1\in\bb Q is a primitive square root of unity.

Our first task is to determine the degree of the extension [Q[ω]:Q][\bb Q[\omega]:\bb Q], which is equal to the number of conjugates of ω\omega in C\bb C, or the degree of the minimal polynomial of ω\omega over Q\bb Q. Note that this minimal polynomial divides Xm1X^m-1, but does not divide Xn1X^n-1 for any 1n<m1\leq n< m (since ωm=1\omega^m=1 but ωn1\omega^n\neq1). Hence the only possible conjugates of ω\omega are ωk\omega^k for kk coprime to mm. (Note that this immediately implies that Q[ω]\bb Q[\omega] is Galois over Q\bb Q.) We prove the converse statement below.

Proposition: For any kk with gcd(k,m)=1\gcd(k,m)=1, we have ωk\omega^k is conjugate to ω\omega.

Proof: We will prove the following statement: for any θ=ωl\theta=\omega^l with gcd(l,m)=1\gcd(l,m)=1, and any prime pp not dividing mm, we have θp\theta^p is conjugate to θ\theta. Then since conjugacy is transitive, by induction we have ω\omega is conjugate to ωk\omega^k for any kk which is a product of primes not dividing mm, ie. for any k1k\geq1 coprime to mm, and we are done.

Let ff be the minimal polynomial of θ\theta over Q\bb Q, so Xm1=f(X)g(X)X^m-1=f(X)g(X) for some monic gQg\in\bb Q. By Gauss’s lemma, we have f,gZ[X]f,g\in\bb Z[X].

Assume for sake of contradiction that θp\theta^p is not conjugate to θ\theta. Then f(θp)0f(\theta^p)\neq0, so g(θp)=0g(\theta^p)=0. Thus θ\theta is a root of g(Xp)g(X^p), so f(X)g(Xp)f(X)\mid g(X^p) in Q[X]\bb Q[X], and by Gauss’s lemma again we have f(X)g(Xp)f(X)\mid g(X^p) in Z[X]\bb Z[X].

Therefore, under the mod pp homomorphism :Z[X]Fp[X]\overline\cdot:\bb Z[X]\to\bb F_p[X], we have f(X)g(Xp)=g(X)p\overline f(X)\mid\overline g(X^p)=\overline g(X)^p in Fp[X]\bb F_p[X]. Now Fp[X]\bb F_p[X] is a UFD, so f\overline f and g\overline g have a nontrivial common factor, say hh.

Thus h2fg=Xm1h^2\mid\overline{fg}=X^m-1, so hh divides the derivative mXm1mX^{m-1}. By unique factorisation, we have hh is a multiple of some XbX^b, b1b\geq1, but this contradicts hXm1h\mid X^m-1, and we are done.  \qed

Recall that the Euler totient function φ:Z>0Z>0\varphi:\bb Z_{>0}\to\bb Z_{>0} is defined by φ(n)=(Z/nZ)×\varphi(n)=\lvert(\bb Z/n\bb Z)^\times\rvert; equivalently, φ(n)\varphi(n) is the number of integers in {1,2,,n}\{1,2,\ldots,n\} which are coprime to nn.

Corollary: [Q[ω]:Q]=φ(m)[\bb Q[\omega]:\bb Q]=\varphi(m).

Proof: By the above result, the conjugates of ω\omega are precisely

{ωk:1km,gcd(k,m)=1},\{\omega^k\,:\,1\leq k\leq m,\,\gcd(k,m)=1\},

which is a set with φ(m)\varphi(m) many elements. Hence the minimal polynomial of ω\omega has degree φ(m)\varphi(m).  \qed

Corollary: The Galois group of Q[ω]\bb Q[\omega] over Q\bb Q is isomorphic to (Z/mZ)×(\bb Z/m\bb Z)^\times. More precisely, each k(Z/mZ)×k\in(\bb Z/m\bb Z)^\times is mapped to the automorphism sending ωωk\omega\mapsto\omega^k.

Proof: It suffices to check that the composition law for automorphisms corresponds to multiplication mod mm: performing ωωk\omega\mapsto\omega^k then ωωl\omega\mapsto\omega^l yields

ωωk(ωl)k=ωkl,\omega\mapsto\omega^k\mapsto(\omega^l)^k=\omega^{kl},

and we are done.  \qed

Corollary: The only roots of unity in Q[ω]\bb Q[\omega] are the mm-th roots of unity for mm even, and the 2m2m-th roots of unity for mm odd.

Proof: Since the mm-th and 2m2m-th cyclotomic fields are equal for mm odd, it suffices to prove the claim for mm even. Suppose that Q[ω]\bb Q[\omega] contains a primitive kk-th root of unity θ=e2πih/k\theta=e^{2\pi ih/k}, with gcd(h,k)=1\gcd(h,k)=1. Let gcd(k,m)=d\gcd(k,m)=d, so that gcd(k,hm)=d\gcd(k,hm)=d. Hence there exists a,bZa,b\in\bb Z with ak+bhm=dak+bhm=d. Then Q[ω]\bb Q[\omega] contains

ωaθb=e2πid/km,\omega^a\theta^b=e^{2\pi id/km},

which is a primitive cmcm-th root of unity (where c=kdc=\frac kd). Hence Q[ω]\bb Q[\omega] contains the cmcm-th cyclotomic field, so φ(m)φ(cm)\varphi(m)\geq\varphi(cm). By well-known properties of the Euler totient function, this implies that c=1c=1, so d=kd=k divides mm; hence θ\theta is an mm-th root of unity, as desired.  \qed

Corollary: The mm-th cyclotomic fields, for even mm, are pairwise non-isomorphic.

Proof: Since all such fields are Galois over Q\bb Q, and any such isomorphism is an embedding in C\bb C, it suffices to show that these fields are pairwise distinct.

Now if Q[e2πi/m]=Q[e2πi/m]\bb Q[e^{2\pi i/m}]=\bb Q[e^{2\pi i/m’}], then the above result implies that e2πi/me^{2\pi i/m} is an mm’-th root of unity, so mmm’\mid m; similarly, we have mmm\mid m’, so m=mm=m’.  \qed

Discriminant of an element

We want to investigate the ring of algebraic integers in cyclotomic fields. However, we first need some properties of the discriminant of nn-tuples with a special form.

For an algebraic number α\alpha of degree nn over Q\bb Q, we will write disc(α)\disc(\alpha) to denote disc(1,α,,αn1)\disc(1,\alpha,\ldots,\alpha^{n-1}).

Lemma: Let K=Q[α]K=\bb Q[\alpha], and let α1,,αn\alpha_1,\ldots,\alpha_n be the conjugates of α\alpha over Q\bb Q. Then

disc(α)=r<s(αrαs)2=(1)n(n1)2NK(f(α)), \disc(\alpha)=\prod_{r< s}(\alpha_r-\alpha_s)^2=(-1)^{\frac{n(n-1)}2}N^K(f'(\alpha)),

where ff is the minimal (monic irreducible) polynomial of α\alpha over Q\bb Q.

Proof: Let σ1,,σn\sigma_1,\ldots,\sigma_n be the nn embeddings of KK into C\bb C such that σk(α)=αk\sigma_k(\alpha)=\alpha_k. Then

det(σi(αj1))i,j=det(αij1)i,j=r<s(αrαs), \det(\sigma_i(\alpha^{j-1}))_ {i,j}=\det(\alpha_i^{j-1})_ {i,j}=\prod_{r< s}(\alpha_r-\alpha_s),

from the formula for Vandemonde determinants. Squaring both sides gives the first equality.

Next, note that for each rr, we have

f(αr)=limxαrs(xαs)xαr=sr(αrαs).f'(\alpha_r)=\lim_{x\to\alpha_r}\frac{\prod_s(x-\alpha_s)}{x-\alpha_r}=\prod_{s\neq r}(\alpha_r-\alpha_s).

Now fQ[X]f’\in\bb Q[X] implies

NK(f(α))=rσr(f(α))=rf(αr)=rs(αrαs), \begin{aligned} N^K(f'(\alpha))&=\prod_r\sigma_r(f'(\alpha))\\ &=\prod_rf'(\alpha_r)\\ &=\prod_{r\neq s}(\alpha_r-\alpha_s), \end{aligned}

where the last product is over all n(n1)n(n-1) ordered pairs (r,s)(r,s) with rsr\neq s. Then note that

r<s(aras)2=(1)n(n1)/2rs(aras), \prod_{r< s}(a_r-a_s)^2=(-1)^{n(n-1)/2}\prod_{r\neq s}(a_r-a_s),

which yields the second equality.  \qed

Algebraic integers in cyclotomic fields

We now consider the ring of algebraic integers AQ[ω]\bb A\cap\bb Q[\omega] in Q[ω]\bb Q[\omega]. (Recall that ω=e2πi/m\omega=e^{2\pi i/m} is a primitive mm-th root of unity.) We first observe that ω\omega is a root of Xm1X^m-1, and is hence an algebraic integer. This implies Z[ω]AQ[ω]\bb Z[\omega]\subseteq\bb A\cap\bb Q[\omega]. In this section, we show that this is in fact an equality.

Lemma: disc(ω)\disc(\omega) divides mφ(m)m^{\varphi(m)}.

Proof: Let ff be the minimal polynomial for ω\omega over Q\bb Q. As before, we have Xm1=f(X)g(X)X^m-1=f(X)g(X) with f,gZ[X]f,g\in\bb Z[X] monic. Differentiating both sides yields

mXm1=f(X)g(X)+f(X)g(X).mX^{m-1}=f'(X)g(X)+f(X)g'(X).

Now multiply both sides by XX, and substitute X=ωX=\omega to get

m=ωf(ω)g(ω).m=\omega f'(\omega)g(\omega).

Taking norms yields

mφ(m)=±disc(ω)N(ωg(ω)),m^{\varphi(m)}=\pm\disc(\omega)N(\omega g(\omega)),

and we finish by noting that N(ωg(ω))ZN(\omega g(\omega))\in\bb Z (since ω\omega and g(ω)g(\omega) are algebraic integers).  \qed

Lemma: Let m3m\geq3. Then Z[1ω]=Z[ω]\bb Z[1-\omega]=\bb Z[\omega], and disc(1ω)=disc(ω)\disc(1-\omega)=\disc(\omega).

Proof: The first statement is clear. For the second statement, note that if αi\alpha_i are the conjugates of ω\omega, then 1αi1-\alpha_i are the conjugates of 1α1-\alpha. Hence

disc(ω)=r<s(αrαs)2=r<s((1αr)(1αs))2=disc(1ω).   \begin{aligned} \disc(\omega)&=\prod_{r< s}(\alpha_r-\alpha_s)^2\\ &=\prod_{r< s}((1-\alpha_r)-(1-\alpha_s))^2\\ &=\disc(1-\omega).\ \qed \end{aligned}

We will first determine the ring of integers in the mm-th cyclotomic field Q[ω]\bb Q[\omega] in the case where mm is a power of a prime.

Lemma: Let pp be a prime, and let m=prm=p^r. Then

1km,pk(1ωk)=p.\prod_{1\leq k\leq m,\,p\nmid k}(1-\omega^k)=p.

Proof: Let

f(x)=xpr1xpr11=1+xpr1+x2pr1++x(p1)pr1.f(x)=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}=1+x^{p^{r-1}}+x^{2p^{r-1}}+\cdots+x^{(p-1)p^{r-1}}.

Then for all pkp\nmid k, we have ωk\omega^k is a root of Xpr1X^{p^r}-1 but not Xpr11X^{p^{r-1}}-1. Thus all the ωk\omega^k are roots of ff. Furthermore, we have

#{1km:pk}=φ(m)=pmpm1=degf,\#\{1\leq k\leq m\,:\,p\nmid k\}=\varphi(m)=p^m-p^{m-1}=\deg f,

so the ωk\omega^k are all the roots of ff. Thus

f(X)=1km,pk(Xωk),f(X)=\prod_{1\leq k\leq m,\,p\nmid k}(X-\omega^k),

and the result follows by substituting X=1X=1.  \qed

Theorem: Let ω=e2πi/m\omega=e^{2\pi i/m}, where m=prm=p^r for some prime pp. Then AQ[ω]=Z[ω]\bb A\cap\bb Q[\omega]=\bb Z[\omega].

Proof: We write K=Q[ω]K=\bb Q[\omega], R=AKR=\bb A\cap K, and n=φ(pr)=[K:Q]n=\varphi(p^r)=[K:\bb Q]. Also, let ddisc(1ω)=disc(ω)d\coloneqq\disc(1-\omega)=\disc(\omega). Note that d0d\neq0, since

{1,ω,,ωn1}\{1,\omega,\ldots,\omega^{n-1}\}

is a basis for KK over Q\bb Q. Hence

{1,1ω,,(1ω)n1}\{1,1-\omega,\ldots,(1-\omega)^{n-1}\}

is also a basis of KK over Q\bb Q. Thus by a theorem in the previous post, every αR\alpha\in R can be written as

α=1dj=0n1mj(1ω)j,\alpha=\frac1d\sum_{j=0}^{n-1}m_j(1-\omega)^j,

where mjZm_j\in\bb Z. Also, by a previous lemma, we have dmφ(m)d\mid m^{\varphi(m)}, so dd is a power of pp.

Assume for the sake of contradiction that RZ[ω]=Z[1ω]R\neq\bb Z[\omega]=\bb Z[1-\omega]. Then there exists αR\alpha\in R such that not all mjm_j above are divisible by dd. By multiplying a suitable power of pp to all the mjm_j, we may assume that dpmj\frac dp\mid m_j; by subtracting some initial terms, we see that RR contains an element of the form

β=1pj=j0n1μj(1ω)j\beta=\frac1p\sum_{j=j_0}^{n-1}\mu_j(1-\omega)^j

for some j0n1j_0\leq n-1, where μjZ\mu_j\in\bb Z and pμj0p\nmid\mu_{j_0}.

Now note that

p=1km,pk(1ωk)=(1ω)n1km,pk(1+ω++ωk1), \begin{aligned} p&=\prod_{1\leq k\leq m,\,p\nmid k}(1-\omega^k)\\ &=(1-\omega)^n\prod_{1\leq k\leq m,\,p\nmid k}(1+\omega+\cdots+\omega^{k-1}), \end{aligned}

so p(1ω)nZ[ω]\frac p{(1-\omega)^n}\in\bb Z[\omega]. Thus

j=j0n1μj(1ω)jj01=βp(1ω)j0+1R.\sum_{j=j_0}^{n-1}\mu_j(1-\omega)^{j-j_0-1}=\frac{\beta p}{(1-\omega)^{j_0+1}}\in R.

Hence μj01ωR\frac{\mu_{j_0}}{1-\omega}\in R, so

p=N(1ω)N(μj0)=μj0n,p=N(1-\omega)\mid N(\mu_{j_0})=\mu_{j_0}^n,

which is our desired contradiction.  \qed

For the case of general mm, we need the following corollary of the last theorem in the previous post:

Corollary: Let K,LK,L be number fields, and let R,S,TR,S,T be the number rings of K,L,KLK,L,KL respectively. If [KL:Q]=[K:Q][L:Q][KL:\bb Q]=[K:\bb Q][L:\bb Q] and gcd(discR,discS)=1\gcd(\disc R,\disc S)=1, then T=RST=RS.

We can now prove the main theorem for this section.

Theorem: Let ω=e2πi/m\omega=e^{2\pi i/m}. Then AQ[ω]=Z[ω]\bb A\cap\bb Q[\omega]=\bb Z[\omega].

Proof: We proceed by strong induction on mm. We have already solved the case when mm is a prime power. If mm is not a prime power, then we can write m=m1m2m=m_1m_2 for coprime integers m1,m2>1m_1,m_2>1.

Let K=Q[ω]K=\bb Q[\omega], R=AKR=\bb A\cap K. For k=1,2k=1,2, write ωk=e2πi/mk\omega_k=e^{2\pi i/m_k}, Kk=Q[ωk]K_k=\bb Q[\omega_k], and Rk=AKk=Z[ωk]R_k=\bb A\cap K_k=\bb Z[\omega_k] by inductive hypothesis.

Let a,bZa,b\in\bb Z satisfy am1+bm2=gcd(m1,m2)=1am_1+bm_2=\gcd(m_1,m_2)=1. Then ω=ω1aω2b\omega=\omega_1^a\omega_2^b implies KK1K2K\subseteq K_1K_2 and Z[ω]R1R2\bb Z[\omega]\subseteq R_1R_2. Note that the reverse inclusions are clear, so K=K1K2K=K_1K_2 and Z[ω]=R1R2\bb Z[\omega]=R_1R_2.

Hence it suffices to show that R=R1R2R=R_1R_2, so we are left to check the conditions of the previous corollary. The degree condition follows from φ(m)=φ(m1)φ(m2)\varphi(m)=\varphi(m_1)\varphi(m_2), since m1,m2m_1,m_2 are coprime. The discriminant condition follows since discR1\disc R_1 divides a power of m1m_1, and similarly for discR2\disc R_2, so m1,m2m_1,m_2 coprime implies discR1,discR2\disc R_1,\disc R_2 coprime.  \qed

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