Matematisk Institut Mat 3AL 4.1 Chapter IV. Cyclotomic fields and applications
ROOTS OF UNITY AND CYCLOTOMIC POLYNOMIALS.
In this chapter we shall consider an important class of normal extensions of the rational number field Q. Historically these were the first algebraic extensions of Q which were the subject of thorough investigations. First some remarks about roots of unity. n 2πi k The complex solutions of the equation x = 1, i.e. the numbers e n ,0 k Proof. Just an easy exercise in group theory.(Cf. Remark 1.82) 2πi k In particular e n , 1 ≤ k Theorem 4.3. Fn(x) is a polynomial in Z[x]. To prove this we need the following n Lemma 4.4. x − 1= d|n Fd(x),whered runs through the positive divisors of n. n Proof. The monic polynomials x − 1and d|n Fd(x) have no multiple roots (note that Fd1 (x)andFd2 (x) have no common roots if d1 = d2). Therefore it suffices to n show that the two polynomials x − 1and d|n Fd(x) have exactly the same roots. Matematisk Institut Mat 3AL 4.2 Indeed, an n-th root of unity is a primitive d-th root of unity for exactly one divisor d of n.Conversely,ifε is a primitive d-th root of unity for a divisor d of n,thenε is certainly an n-th root of unity. Proof of Theorem 4.3. By induction on n.SinceF1(x)=x − 1 the assertion is clear for n = 1. Assume it has been proved that Fm(x) ∈ Z[x] for all m We now give an explicit formula for Fn(x). n μ(d) Theorem 4.5. Fn(x)= d|n(x d − 1) ,whereμ denotes the M¨obius function introduced in Chap. II. Proof. By lemma 4.4 and Theorem 2.83 in Chap. II we get: n μ(d) μ(d) (x d − 1) = ( Fδ(x)) = n d|n d|n δ| d P μ(d) d| n μ(d) Fδ(x) = Fδ(x) δ = Fn(x) . d,δ,d·δ|n δ|n Since Fn(x) has degree ϕ(n), where ϕ(n)isEulersϕ-function, by comparison of degrees we obtain n 1 Corollary 4.6. ϕ(n)= d|n d μ(d)=n · p|n(1 − p ),wherep runs through the distinct prime divisors of n. n μ(d) d − Proof. By identifying the degrees of Fn(x)andof d|n(x 1) we see that n n 1 ϕ(n)= d|n d μ(d). It remains then to show that d|n d μ(d)=n · p|n(1 − p ). Since μ(d)=0ifd is divisible by the square of a prime, it suffices to consider a1 ar the square-free divisors of n.Letn = p1 ···pr be the prime decomposition of n where p1,...,pr are the distinct prime divisors of n. We consider first the divisor 1, then the prime divisors of n, then all products of two distinct prime divisors, then all products of three distinct prime divisors, etc. For the sum we then find n n n n 1 μ(d)=n − + − + ... = n · (1 − ). where the pi’s d|n d p i pipj pipj pk pi|n p i are the distinct prime divisors of n,thepipj ’s are the products of two distinct prime divisors, pipjpk’s are the products of three distinct prime divisors etc. Matematisk Institut Mat 3AL 4.3 Some remarks concerning the coefficients of the cyclotomic polynomials. Remark 4.7. For n>2 the degree of Fn(x) is an even number and the constant term is 1. Remark 4.8. The coefficient of the next highest term (i.e. the coefficient of xϕ(n)−1) is −μ(n), since the sum of the primitive n-th roots of unity is μ(n) (this is an exercise provable by induction). Remark 4.9. Looking at the first cyclotomic polynomials one might have the temp- tation to conjecture that all the coefficents of the cyclotomic polynomials were 0 or 1or−1. This, however, is not true. The first counterexample is F105 where the coefficient of x7 is -2.But it can be proved that if n is divisible by at most two dis- tinct odd prime numbers, then the coefficients of Fn(x)are0,1or−1. (The proof is elementary, but not trivial.) i Remark 4.10. Fn(x) is ”reciprocal” for n>1, i.e. if ai is the coefficient of x then ai = aϕ(n)−i for 0 ≤ i ≤ ϕ(n). p p−1 Remark 4.11. For a prime number p one has Fp(x)=(x − 1)/(x − 1) = x + ···+ x +1. 2 Numerical examples 4.12. F1(x)=x − 1,F2(x)=x +1,F3(x)=x + x + 2 4 3 2 2 1,F4(x)=x +1,F5(x)=x + x + x + x +1,F6(x)=x − x +1,F8(x)= 4 6 3 4 3 2 4 2 x +1,F9(x)=x +x +1,F10(x)=x −x +x −x+1,F12(x)=x −x +1,F15(x)= x8 − x7 + x5 − x4 + x3 − x +1. Explicit computations of some roots of unity. 2πi Let εn be the n-th root of unity e n . It is clear that ε1 =1andε2 = −1andε4√= i. Similarly it is straightforward to 2π 2π check that ε8 =cos(8 )+isin( 8 )=(1+i)/ 2. ε3 is√ the root of F3(x) lying in the upper half complex plane. Therefore ε3 = (−1+i 3)/2. ε6 is the√ root of F6(x) lying in the first quadrant of the complex plane; hence ε6 =(1+i 3)/2. As for ε5 we write −2 2 1 −1 −2 2 x F5(x)=x + x +1+x + x =(x +1/x) +(x +1/x) − 1 2π −1 which implies that 2 cos( 5 )=ε5 + ε5 is the (positive) root of z2 + z − 1=0 √ √ 2π − 2π − hence 2 cos( 5 )=( 5 1)/2andcos(5 )=( 5 1)/4. √ 2π − 2 2π 1 Now sin( 5 )= 1 cos ( 5 )= 4 10 + 2 5sothat 1 √ √ ε5 = 5 − 1+i 10 + 2 5 . 4 Matematisk Institut Mat 3AL 4.4 CYCLOTOMIC POLYNOMIALS ARE IRREDUCIBLE. For the proof of the irreducibility of cyclotomic polynomials in Q[x] we need the following lemmas. Lemma 4.13.. If f(x) and g(x) are monic polynomials in Q[x] for which f(x)·g(x) ∈ Z[x],thenf(x) ∈ Z[x] and g(x) ∈ Z[x]. Proof. There are natural numbers a and b such that af(x)andbg(x) are primi- tive polynomials in Z[x]. According to Gauss’s lemma the product af(x)bg(x)= abf(x)g(x) is also a primitive polynomial in Z[x]. But then ab must be +1, and therefore a and b must also be 1. This implies that f(x)andg(x) are polynomials in Z[x]. Lemma 4.14. If p is a prime number and g(x) is a polynomial in Zp[x]=Fp[x] then g(xp)=(g(x))p. Proof. This follows from Freshmans Dream and the fact that αp = α for every α in the finite field Zp with p elements. Theorem 4.15. Fn(x) is irreducible in Q[x]. Proof. Let f(x) be a monic irreducible polynomial in Q[x]. We prove: 1◦ If ε is a primitive n-th root of unity and p is a prime number that does not divide n, then: f(ε)=0⇒ f(εp)=0. Proof of 1◦:Sincef(x) = Irr(ε, Q) the polynomial f(x) must divide xn − 1inside Q[x]. By lemma 4.13 it follows that f(x) must be a polynomial in Z[x]. We now consider g(x) = Irr(εp, Q). As before we see that g(x) ∈ Z[x]. The polynomial g(xp)hasε as a root. Hence f(x)|g(xp)insideQ[x] and thus as before inside Z[x]. Therefore g(xp)=f(x) · k(x) , (∗) where k(x) a priori is in Q[x]andthenasbeforeinZ[x]. Assume now f(εp) =0. Then f(x)andg(x) would be two non-associate irreducible polynomials in Q[x]. Both of them divide xn − 1. Since Q[X]isaUFD,f(x) · g(x)|xn − 1insideQ[x]and thus as before inside Z[x], consequently xn − 1=f(x) · g(x) · h(x) , (∗∗) where h(x) ∈ Z[x]. For the polynomials f(x) , g(x) , h(x)andk(x)iZp[x] obtained by applying the homomorphism Z[x] → Zp[x], we conclude from (∗),and (∗∗) g(x)p = f(x) k(x) , (∗∗∗) Matematisk Institut Mat 3AL 4.5 xn − 1= f(x) g(x) h(x) , (∗∗∗∗) wherewehaveusedlemma4.14in(∗∗∗). From (∗∗∗) we see that every irreducible factor π(x)off(x) also must appear n in g(x), so that the equation (∗∗∗∗) implies that x − 1(insideZp[X]) must be divisible by the square of the polynomial π(x) of positive degree: n 2 x − 1=π(x) · q(x) ,π(x),q(x) ∈ Zp[X] . By taking formal derivatives we get nxn−1 = π(x)[2π(x)q(x)+π(x)q(x)] . −1 Here n =0in Zp since p n; therefore n has an inverse n in Zp and from the above we get −1 1=π(x) [2π (x)q(x)+π(x)q (x)] x · n − π(x)q(x) ∈ Zp[X] , which gives the desired contradiction since π(x) has positive degree. 2◦ If a monic irreducible polynomial f(x) ∈ Q[x] has some primitive n-th root of unity ε as a root then all primitive n-th roots of unity will be roots of f(x). Proof of 2◦: Every primitive n-th root of unity has the form εk,where(k, n)=1 (cf. lemma 4.1). k can be written as a product of (not necessarily distinct) prime numbers. None of these prime factors divides n. The assertion 2◦ now follows by successive application of 1◦. ◦ 3 Fn(x) is irreducible in Q[x]. Proof of 3◦: Let ε be a primitive n-th root of unity and let f(x) = Irr(ε, Q). Then ◦ f(x) divides Fn(x). By 2 all primitive n-th roots of unity will be roots of f(x). This implies that degree(f(x)) ϕ(n)=degreeFn(x) and hence Fn(x)=f(x), since these polynomials are monic. But f(x) is by definition irreducible and therefore Fn(x) is also irreducible. −−−−−−−−−−−−−−− DIRICHLETS THEOREM ABOUT PRIME NUMBERS IN ARITHMETIC PROGRESSIONS. We make a little digression, where we use that Fn(x) ∈ Z[X]. Dirichlet’s famous theorem about prime numbers in arithmetic progressions says that for every pair (a, n) of relatively prime natural numbers there exist infinitely many prime numbers that are ≡ a (mod n). We shall prove this theorem in an important special case: Matematisk Institut Mat 3AL 4.6 Theorem 4.16. Dirichlets Theorem for a =1. For every natural number n there are infinitely many prime numbers that are ≡ 1(modn). Proof. The statement “For every natural number n there exists a prime number that is ≡ 1(modn)” implies the statement: “For every natural number n there exist infinitely many prime numbers that are ≡ 1(modn)”. Indeed, let p1,...,pt be primes that are ≡ 1(modn). By assumption there exists a prime number p that is ≡ 1(modnp1 ···pt). This prime number p is ≡ 1(modn) and definitely = pi for 1 i t. Therefore it suffices to show that for every natural number n there is a prime number that is ≡ 1(modn). We may, of course, assume that n>2. The desired theorem is a consequence of the following two assertions: Assertion 1. |Fn(n)| > 1 for every natural number n>2 and therefore Fn(n) is divisible by at least one prime number. Assertion 2. Every prime divisor p of Fn(n) is ≡ 1(modn). 2πi k Proofofassertion1. Since Fn(n)= (n − e n ) and every factor for n>2has 1≤k Proof of assertion 2. Let p be a prime divisor of Fn(n), where n>2. Since Fn(x)has constant term 1 it follows that Fn(n) ≡ 1(modn) and therefore p does not divide n. n From lemma 4.4 we know that x − 1= Fd(x) and by setting x = n we see d|n that nn − 1 is divisible by p. Therefore the group theoretical order t of the residue class n modulo p divides n. We claim that t = n. Indeed, assume that t n t n n − 1 (n ) t − 1 t n −1 t n −2 t = =(n ) t +(n ) t + ...+ n +1 nt − 1 nt − 1 shows that nn − 1 n ≡ 1+···+1= (mod p) nt − 1 t n t terms Matematisk Institut Mat 3AL 4.7 n The above would therefore imply that p should divide t and thereby also n contra- dicting our first observation. Consequently t = n, and since the group theoretical order of every element in the multiplicative group (Zp \{0}, ·)isadivisorofp − 1 it follows that p ≡ 1(modn). CYCLOTOMIC FIELDS. 2πi We now consider the field Qn = Q e n , which is called the n-th cyclotomic 1 n field. Qn is the splitting field for x − 1overQ.SoQn/Q is a normal extension. Here is 2πi 2πi [Qn : Q]=degree Irr e n , Q =degree(Fn(x)) = ϕ(n). Let ε be e n .Ifσ is an a a automorphism in Gal(Qn/Q) then (by Lemma 3.4) σ(ε)mustbe=ε ,whereε is a arootofFn(x), i.e. ε is a primitive n-th root of unity. Hence (a, n)=1wherea is determined modulo n. Consequently we get a well defined map: ψ- ∗ ∗ Gal(Qn/Q) Zn , (where Zn are the residue classes modulo n prime to n) by ψ(σ)= a (modulo n), if σ(ε)=εa . Here ψ is injective since σ is uniquely determined by its value on ε. Because ∗ | Gal(Qn/Q)| =[Qn : Q]=ϕ(n)=|(Zn)| the mapping ψ is also surjective. The ∗ residue classes prime to n modulo n form a multiplicative group (notice that Zn consists of the invertible elements in Zn). Furthermore ψ is a homomorphism: ψ(σ1σ2) ψ(σ1σ2) determined by σ1σ2(ε)=ε ψ(σ2) ψ(σ2) ψ(σ2) σ2(ε)=ε ; σ1(σ2(ε)) = σ1 ε =(σ1(ε)) = ψ(σ2) ψ(σ1) ψ(σ1)ψ(σ2) ε = ε i.e.: ψ(σ1σ2)=ψ(σ1)ψ(σ2) . Thus we have proved ∗ Theorem 4.17. Gal(Qn/Q) Zn (= the multiplicative group of the prime residue classes modulo n). In particular, Gal(Qn/Q) is abelian. It is clear [cf. the fundamental theorem of Galois theory 4) and 5)] that every subfield K ⊆ Qn is normal over Q with abelian Galois group, namely Gal(K/Q)( Gal(Qn/Q)/ Gal(Qn/K)). 1”Beloved child has many names”: There are several notations for the n-th cyclotomic field. (n) Some authors use Q(ζn)orQ(μn), others Q[n]orQ and the list comprises at least 20 other names. We have chosen Qn since it is the shortest. Matematisk Institut Mat 3AL 4.8 A classical (very deep) theorem gives a characterization of the normal extensions of Q with abelian Galois group. Theorem 4.18. Kronecker–Webers Theorem. Let K/Q be a finite normal extension. Then: K/Q is normal with abelian Galois group ⇔ K ⊆ Qn for a suitable n. We have already proved ⇐. The other implication ⇒ is very hard to prove. We shall just prove (the hard implication of) the theorem in the very special case where Gal(K/Q) is cyclic of order 2. First a quite elementary lemma. Lemma 4.19. Let K be a field of charactistic 0 and L a quadratic extension of K, i.e. [L : K]=2. √ i) There exists an element a in K such that L = K( a). ii) If K = Q we can choose a as a rational square-free integer (i.e. a ∈ Z and a is not divisible by the square of any prime number). Proof. ad i) Any α ∈ L \ K generates L over K, i.e. L = K(α). The polynomial 2 Irr(α, K) can be written x + k1x + k2,wherek1 and k2 belong to K.Asa one may 2 use the discriminant k1 − 4k2. ad ii) The assertion follows from the fact, that for every rational number q =0 2 there exists a rational number q1 such that qq1 is a square-free integer. Proof of Kronecker-Webers theorem for quadratic extensions of Q. √ Because of lemma 4.19 it suffices to show that Q( a) Q4|a| for every square-free integer a. This will be done in 4 steps: √ √ √ √ √ 1. We observe that Q( −1) = Q4 og Q( −1, 2) = Q( −1, −2) = Q8. 2. For a natural number n the cyclotomic field Qn is the splitting field over Q for the polynomial xn − 1. By Theorem 2.84 the square root discrim(xn − 1) lies n (n−1)(n−2) in Qn. In Theorem 2.88 it was shown that this discriminant is n (−1) 2 . n−1 2 n−1 When n is odd, the discriminant can be written (n 2 ) · (−1) 2 · n. Therefore n n−1 Q( discrim(x − 1)) = Q( n(−1) 2 )whichis √ Q(√n), if n ≡ 1(mod4)and Q( −n), if n ≡ 3(mod4). 3. If n and m are natural numbers and n divides m then obviously Qn Qm. √ 4. If a is a is a square-free integer, then Q( a) Q4|a|. There are two possibilities: i) a is odd and ii) a is even. Matematisk Institut Mat 3AL 4.9 ad i) We distinguish between the case i1), where a is positive and the case i2), where a er negative. √First case i1): For a ≡ 1 mod 4, the above assertions 2. and 3. imply that Q Q Q ( a) a 4|a|. √ √ For a ≡ 3 mod 4, the above assertions 1. 2. and 3. imply that Q( −1, −a) Q 4|a|. √ √ Second case i2): Clearly Q( a) Q( −1, | a√|). By the assertions 1. and 2. and the above treated case i1) we conclude that Q( a) Q4|a|. √ Q √ad ii)√ Here a =2u,whereu is odd since a is square-free. Obviously ( a) Q( 2, u) which [according to assertions 1., 3. and the above case i)] is contained in Q8Q4|u| Q4|a|. Example 4.20. If p is an odd prime number the Galois group Gal(Qp/Q) is cyclic of order p − 1, since the multiplicative group of the non-zero elements in the finite field Zp = Fp is cyclic. Therefore Gal(Qp/Q) contains exactly one subgroup of index 2. By Q the fundamental theorem of Galois theory p thus contains exactly√ one quadratic Q ≡ subfield. As the√ above proof shows, this quadratic subfield is ( p)whenp 1 (mod 4) and Q( −p)whenp ≡ 3(mod4). RELATIONS BETWEEN CYCLOTOMIC FIELDS. We first give some useful applications of the formula for ϕ(n) from the Corollary 4.6. Lemma 4.21. Let m and n be positive integers and let d,resp.f,bethegreatest common divisor of m and n, resp. the least common multiple of m and n.Then i) mn = df ii) ϕ(m)ϕ(n)=ϕ(d)ϕ(f). Proof. Consider the prime decompositions of m and n mp np m = p ,mp ≥ 0andn = p ,np ≥ 0 p p where p runs through the primes dividing m or n. Then the prime decompositions of d and f are d = pmin(mp,np) and f = pmax(mp,np). p p This immediately shows i). By the formula for Euler’s ϕ-function we have Matematisk Institut Mat 3AL 4.10 ϕ(m)= ϕ(pmp )and ϕ(n)= ϕ(pnp ) p p as well as ϕ(d)= ϕ(pmin(mp,np))andϕ(f)= ϕ(pmax(mp,np)), p p wherewehavedefinedϕ(1) to be 1. The above expressions for ϕ(m), ϕ(n), ϕ(d)andϕ(f) immediately yield ii). Lemma 4.22. Let m be a positive integer and p a prime number. Then ϕ(mp)= ϕ(m) ⇔ m is odd and p =2. Proof. The ”⇐” part is an immediate consequence of the formula for Euler’s ϕ- function. The ”⇒”part:Ifp divides m the formula for Euler’s ϕ-function shows that ϕ(mp)=ϕ(m) · p which is = ϕ(m). If p does not divide m the formula for Euler’s ϕ-function shows that ϕ(mp)= ϕ(m) · (p − 1) which is = ϕ(m)forp = 2. Hence m must be odd and p must be 2 if ϕ(mp)=ϕ(m). Euler’s ϕ-function is not multiplicative in the sense that it unconditionally sends products into products, but it has the weaker property that ϕ(m) divides ϕ(n)ifm divides n. This follows either from the formula for Euler’s ϕ-function or from the fact that the natural homomorphism from (Z/n)∗ to (Z/m)∗ is surjective when m divides n. We formulate this as Lemma 4.23. Let m and n be natural numbers. If m divides n,thenϕ(m) divides ϕ(n). Lemma 4.24. Let m and n be natural numbers. If m | n,thenϕ(m)=ϕ(n),ifand only if either m = n or m is odd and n =2m. Proof. The ”if” part is an immediate consequence of the formula for Euler’s ϕ- function. As for the ”only if” part assume that m Theorem 4.25. Let m and n be positive integers and let d,resp.f,bethegreatest common divisor of m and n, resp. the least common multiple of m and n. Then the compositum QmQn = Qf and the intersection Qm ∩ Qn = Qd. Proof. By lemma 4.21 the greatest common divisor of m and n is mn/f, hence there exist integers a and b such that am + bn = mn/f and thus a/n + b/m =1/f. Therefore 2πi a 2πi b 2πi (e n ) (e m ) = e f which implies that Qf is contained in the compositum QmQn. Since clearly Qm and Qn are contained in Qf it follows that QmQn = Qf . Clearly Qd ⊆ Qm and Qd ⊆ Qn, hence Qd ⊆ Qm ∩ Qn. The translation theorem (Theorem 3.46) applied on Qm QmQn = Qf Qm ∩ Qn Qn Q yields [Qm :(Qm ∩ Qn)] = [Qf : Qn]=ϕ(f)/ϕ(n)=ϕ(m)/ϕ(d) where we have used Lemma 4.21 to obtain the last equality. Therefore [(Qm ∩Qn):Q]=ϕ(d). Since [Qd : Q]=ϕ(d) this together with above inclusion Qd ⊆ Qm ∩ Qn shows that Qm ∩ Qn = Qd. In view of Theorem 3.48 about the compositum of finite normal extensions The- orem 4.25 implies the following Corollary 4.26. Let n1,...,nt be pairwise coprime natural numbers. If we set ··· Q Q Q Q Q n = n1 nt then n is the compositum of the fields n1 ,... nt and Gal( n/ ) Q Q ×···× Q Q is isomorphic to the direct product Gal( n1 / ) Gal( nt / ). Moreover, if Q ··· for each i, 1 i t, Ki is a subfield of ni then the compositum K1 Kt is a normal extension of Q whose Galois group is isomorphic to the direct product Gal(K1/Q) ×···×Gal(Kt/Q). Theorem 4.27. Let m n be natural numbers. Then Qm = Qn if and only if either m = n or m is odd and n =2m. Proof. Since ϕ(m)=ϕ(2m) for any odd natural number m the ”if” part is clear. Indeed, in this case Qm ⊆ Qn and [Q2m : Q]=[Qm : Q]. Matematisk Institut Mat 3AL 4.12 As for the ”only if” part let d,resp. f, be the greatest common divisor of m and n resp. the least common multiple of m and n. By the preceding theorem we conclude that Qd = Qm = Qn = Qf .Sinced divides f we deduce from [Qd : Q]=ϕ(d)and [Qf : Q]=ϕ(f)thatϕ(d)=ϕ(f). If m Theorem 4.28. Let m and n be natural numbers. Then Qm ⊆ Qn if and only if either m divides n or m =2u for some odd divisor u of n. Proof. The ”if” part follows from the previous theorem since Q2u = Qu for an odd u and Qu ⊆ Qn if u divides n. The ”only if” part: Let as usual d be the greatest common divisor of m and n. From Theorem 4.25 we get Qd = Qm.Sinced divides m the previous theorem implies that either d = m or d is odd and m =2d. In the first case m must divide n and in the second case d is an odd divisor of n,sothatm is 2·(an odd divisor of n). √ Corollary 4.29. Qn contains the number i (= −1) if and only if 4 divides n. THE DEGREES OF COS(2π/n) AND SIN(2π/n). 2π thedegreeofcos n . 2π Since cos n is rational for n = 1 or 2 we may restrict ourselves to the case where n>2. 2πi 2π 2π n Consider the primitive n-th root of unity εn = e = cos n + isin n . 2π −1 2π −1 Clearly Q(cos n )=Q(εn + εn ), since 2cos n = εn + εn . −1 Let τ be the automorphism (complex conjugation) sending εn to εn . 2π Obviously the fixed field F(τ)(=Qn ∩R) contains the field Q(cos n ). Since Qn con- tains non-real numbers for n>2, the automorphism τ has order 2. The fundamental theorem of Galois theory implies that [Qn : F(τ)] = 2. −1 2 2π Since εn is a root of the polynomial (x − εn)((x − εn )=x − 2cos n x + 1 having Q 2π Q Q 2π Q Q 2π ≤ coefficients in (cos n ) it follows that [ n : (cos n )] = [ (εn): (cos n )] 2. 2π 2π Because F(τ) ⊇ Q(cos n ) we conclude that F(τ)=Q(cos n ), hence Q 2π Q 2π [ (cos n ): ]=ϕ(n)/2. In other words cos n is an algebraic number of degree ϕ(n)/2forn>2. 2π thedegreeofsin n . Matematisk Institut Mat 3AL 4.13 2π As before we may assume that n>2sincesin n is rational for n = 1 or 2. 2π We first determine the degree of isin n . With the notations from above we have 2π −1 2isin n = εn − εn . 2π Q 2π Clearly isin n lies in n. To find the degree of isin n we determine the auto- −1 morphisms in Gal(Qn/Q)thatfixεn − εn . Every automorphism of Gal(Qn/Q)isoftheformσa defined by a σa(εn)=εn, where 1 ≤ a 2(a +1)=n (∗) If n is odd (∗) has no solutions in a. If n =2·(an odd number) a solution of (∗)ina must be even, so (a, n) > 1. −1 Therefore, if n is not divisible by 4, σ1 is the only automorphism fixing εn − εn , 2π 2π hence Q(isin n )=Qn and the degree of isin n is ϕ(n). Q Q 2π By Corollary 4.29 n = (isin n ) does not contain i when n is not divisible by 2π 4. We conclude from Example 3.50 (Chap. III) that the degree of sin n is ϕ(n). 2π If n is divisible by 4 we first discard the case n =4:Hereisin 4 = i has the 2π degree 2 and sin 4 =1hasdegree1. So we now assume n>4andn is divisible by 4. If n is divisible by 4 the equation (∗) has exactly one solution in a,where1≤ a< n n − − n n, (a, n)=1,namelya = 2 1. Since n>4, a = 2 1 =1.Thusσ1 and σ 2 −1 are − −1 Q Q 2π the only automorphisms fixing εn εn . Hence [ n : (isin n )] = 2, so the degree 2π of isin n is ϕ(n)/2. Qn contains i when n is divisible by 4. In view of Example 3.50 (Chap. III), 2π to find the degree of sin n we have to determine the values of n for which i lies 2π 2π in Q(isin n ). This boils down to finding the n’s for which Q(i) ⊆ Q(isin n ). By the main theorem of Galois theory this is equivalent to finding the n’s for which 2π 2π Q ⊇ Q Q { n } T ( (i)) T ( (isin n )). We know that T ( (isin n )) = σ1,σ2 −1 .Since n n n n 4 4 ( 2 −1) −1 n n n n 2 σ 2 −1(i)=σ 2 −1(ε )=ε = i n the automorphism σ 2 −1 fixes i exactly if n − 1 ≡ 1(mod4)⇔ n − 2 ≡ 2(mod8)⇔ n ≡ 4(mod8). 2 Matematisk Institut Mat 3AL 4.14 2π Consequently, if n ≡ 4 (mod 8), then i lies in Q(isin n ) and therefore (by Example 2π 2π 3.50, Chap. III) the degree of sin n is (degree of isin n )/2=ϕ(n)/4. 2π If, however, n ≡ 0(mod8)theni does not lie in Q(isin n ) and therefore (by 2π 2π Example 3.50, Chap III) sin n and isin n have the same degree, namely ϕ(n)/2. Let us summarize the above results in the following Theorem 4.30. Apart from the degenerate cases n =1, 2 or 4, then: 2π cos n has the degree ϕ(n)/2 for all n. 2π sin n has the degree ϕ(n) if n is not divisible by 4. 2π sin n has the degree ϕ(n)/2 if n ≡ 0 (mod 8). 2π sin n has the degree ϕ(n)/4 if n ≡ 4 (mod 8). 2π 2π Exercise 4.31. Show that Q(sin n ) ⊆ Q(cos n ) if and only if n =1, 2orn ≡ 0 (mod 4). Q 2π Q 2π ≡ Show that (sin n )= (cos n ) if and only if n =1, 2, 4orn 0(mod8). Matematisk Institut Mat 3AL 4.15 CONSTRUCTION OF REGULAR POLYGONS BY COMPASS AND STRAIGHTEDGE. In this section we use cyclotomic fields to answer some classical problems dating back to Euclid. These problems concern the construction of one geometrical segment from an- other, using only an (unmarked) straightedge and a (collapsible) compass. With the straightedge we can draw the line through through two given points and with the compass we can draw the circle with a given point as centre and a given radius. We assume that we have a starting figure in the real Euclidean plane, consisting of the points (0, 0) and (1, 0). A point is called constructible, if it can be obtained from the starting figure by successive applications of the following operations: 1) draw the straight line through two given or already constructed points. 2) draw the circle with a given or already constructed point as its centre and the distance between two given or already constructed points as its radius. 3) Add intersection points between two constructed straight lines, between a constructed straigth line and a constructed circle or between two constructed circles. By straightforward computation one shows Theorem 4.32. Assume a construction successively give the following points P0(= (0, 0)), P1(= (1, 0)), P2, P3,...,Pn, Pn+1,.... If the coordinates of P0,P1,...,Pn belong to some real number field√ K the coordinates of Pn+1 either belong to K or a number field of the form K( d),whered is a positive real number which is not the square of a number in K. Definition 4.33.Acomplexnumbera + ib is called constructible,ifthepoint(a, b) is constructible according to the above definition. Theorem 4.32 then yields: Theorem 4.34. For every complex constructible√ number z there√ exists a sequence of quadratic√ extensions K = Q, K1 = Q(i)√, (i = −1), K2 = K1( d1), (d1 ∈ K1), K3 = K2( d2), (d2 ∈ K2),...,Kt = Kt−1( dt−1), dt−1 ∈ Kt−1, such that z ∈ Kt. Since we may assume that K1 K2 K3 etc., the dimension [Kt : Q] (by the transitivity theorem, Theorem 2.47 in Chap. II) is 2t. This implies (again using the transitivity theorem, Theorem 2.47 in Chap.II): z constructible number ⇒ [Q(z):Q] is a power of 2 . Matematisk Institut Mat 3AL 4.16 Furthermore classical constructions (formerly known from high school mathtem- atics) show Theorem 4.35. The set of all constructible√ complex number is a field K closed under formation of square roots (i.e. α ∈K⇒ α ∈K.) 2πi Definition 4.36. The regular n-gon is called constructible,ife n is a constructible number. It was known already to Euclid that a regular n-gon is constructable if n is either apower2t,t > 1, of 2, a number of the form 3 · 2t,t ≥ 0, a number of the form 5 · 2t,t≥ 0, or a number of the form 15 · 2t,t≥ 0. For more than two thousand years it was an open question whether there are more regular n-gons than those, which are constructible . In 1796 Gauss (who at that time was 19 years old) showed that the regular 17-gon is constructible. Later he found a complete characterization of the n’s for which the regular n-gon is constructible. This was published in his famous Disquisitiones Arithmeticae (called the bible of number theory). Theorem 4.37. Gauss’ Theorem. (Disquisitiones Arithmeticae 1801). The a regular n-gon is constructible ⇔ n has the form n =2 · p1 ···pr,wherea is a non- negative integer and p1,... ,pr are distinct odd prime numbers which all of them are of the form: (a power of 2) +1. Remark 4.38.If2k + 1 is a prime number, k is necessarily a power of 2. Indeed, otherwise k = u · s, u = being an odd number > 1, and: 2k +1 = (2s)u +1 = (2s)u −(−1)u =(2s −(−1)). ((2s)u−1 +(2s)u−2(−1)+···+(−1)u−1). In other words 2s + 1 divides 2k +1.Sinceu>1thenumbers is ProofofGauss’theorem.Since the regular n-gon is constructible if and only if the regular 2n-gon is constructible (”bisection of angles”) it suffices to show Gauss’ the- orem for an odd number n>1. a a “⇒”: Every odd number n>1 can be written n = p1 1 ...pr r ,wherep1,... ,pr are distinct odd prime numbers and the exponents a1,...,ar are natural numbers. 2πi Now by Theorem 4.17 [Q(e n ):Q]=ϕ(n), where ϕ(n) denotes Euler’s ϕ-function. By Theorem 4.34, ϕ(n) must be a power of 2. By Corollary 4.6 we know that 1 a1−1 ar −1 ϕ(n)=n · p|n(1 − p )=p1 (p1 − 1) ···pr (pr − 1). If this number is a power of 2 each of the exponents a1,...,ar must be 1, and the numbers p1 − 1,...,pr − 1 must be powers of 2. But this means exactly that n has the form indicated in Gauss’theorem. “⇐”Ifn has the form indicated in the theorem the computation in the first half of the proof shows that [Qn : Q] is a power of 2. The Galois group G =Gal(Qn/Q) is then an abelian 2-group. In particular, G is solvable so that the factors in a Matematisk Institut Mat 3AL 4.17 composition series G ⊃ G1 ⊃ G2 ⊃ ··· ⊃ Gk = {e} are cyclic of order 2,(a well- known theorem in group theory): [G : G1]=[G1 : G2]=···=[Gk−1 : Gk]=2. We have the following situation: Gk = {e} F(Gk)=Qn 2 2 Gk−1 F(Gk−1) 2 2 2 2 G2 F(G2) 2 2 G1 F(G1) 2 2 G F(G)=Q F(G1) is a quadratic extension of F(G) and therefore (by Lemma 4.19) obtainable by adjunction of the square root of a number in F(G). Furthermore F(G2)isa quadratic extension of F(G1) and therefore obtainable by adjunction of the square root of a number in F(G1)etc. Since the field K of the constructibe numbers is closed under formation of square roots (Theorem 4.35) and the rational numbers are constructible, it follows that Qn 2πi is contained in K.Inparticulare n lies in the field of constructible numbers. Remark 4.39 concerning the prime numbers appearing in Gauss’ theorem. These prime numbers are called the Fermat prime numbers.21 +1=3,22 +1=5, 24 + 1 = 17, 28 + 1 = 257, 216 + 1 = 65537 are Fermat prime numbers. More Fermat n prime number than those are not known. (A criterion for 22 + 1 to be a prime number can be found in Chap. VI:) Remark 4.40 about the ”trisection of an arbitrary angle”. From Gauss’ theorem it in particular follows that a regular 9-gon cannot be constructed by compass and straightedege. This gives a negative solution of the classical problom whether every angle can be trisected by compass and straightedge. Matematisk Institut Mat 3AL 4.18 AN APPLICATION TO THE ‘INVERSE PROBLEM OF GALOIS THEORY‘. We conclude this chapter by yet another application of cyclotomic fields. It is a famous problem (dating back to Hilbert (1892)), whether every finite group can be realized as the Galois group of a finite normal extension of the rational number field Q. This problem, the inverse problem of Galois theory, is still unsolved and has a central place in current research in Galois theory. We shall here prove, that every finite abelian group can be realized as a Galois group over Q. We first deal with the cyclic case. Theorem 4.41. Let Zn be the cyclic group of order n and let p be a prime number such that p ≡ 1(modn). Then there exists a subfield of Qp which is normal over Q with Zn as its Galois group. ∗ Proof. Since Gal(Qp/Q) Zp (Zp−1, +) there exists a (uniquely determined) sub- Q Q p−1 F group H of Gal( p/ )oforder n . The fixed field (H) is according to the funda- mental theorem of Galois theory is a normal extension of Q with Gal(Qp/Q)/H Zn as its Galois group. We are now in a position to prove Theorem 4.42. Every finite abelian group can be realized as the Galois group for a finite normal extension of Q. Proof. Let A be a finite abelian group. By a well-known theorem A is a direct product of cyclic groups Z × Z ×···×Z A n1 n2 nt By the earlier proved special case of Dirichlet’s theorem about prime numbers in arithmetic progressions there exist distinct prime numbers p1,...,pt such that p1 ≡ 1(modn1) ,p2 ≡ 1(modn2),...,pt ≡ 1(modnt) . ⊆ Q ⊆ Q ⊆ Q By Theorem 4.41 there are subfields of K1 p1 , K2 p2 ,...,Kt pt which are normal over Q and Q Z Q Z Q Z Gal(K1/ ) n1 , Gal(K2/ ) n2 ,...,Gal(Kt/ ) nt . Combining the above theorem, Corollary 4.26 and Corollary 3.48 we conclude that K1K2 ...Kt is normal over Q and Q Z × Z ×···×Z Gal(K1K2 ...Kt/ ) n1 n2 nt A.