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Section 14.5. Today We Will Once Again Discuss Cyclotomic Fields And Section 14.5. Today we will once again discuss cyclotomic fields and figure out which regular n-gons can be constructed with straightedge and compass. a Theorem 1. Given a primitive n-th root of unity ζn, define σa 2 Gal(Q(ζn)=Q) by σa(ζn) = ζn. Then × (Z=nZ) ' Gal(Q(ζn)=Q); a 7−! σa: Proof. First of all, let us note that the map a 7! σa is a well-defined bijection. On one hand Q(ζn)=Q is a Galois extension of degree φ(n), on the other, an element σ 2 Gal(Q(ζn)=Q) is completely defined by σ(ζn) and has to permute roots of the cyclotomic polynomial Φn(x) a which are precisely fζng for (a; n) = 1. Therefore, σa defined above is indeed an element of Gal(Q(ζn)=Q), moreover it is clear that σa only depends on the class of a in Z=nZ. The fact that the map a 7! σa is a homomorphism (and hence an isomorphism) is a trivial exercise. Let us now focus on the case Q(ζp) for p a prime number. Then, 2 p−1 ζp; ζp ; : : : ; ζp are precisely the roots of the cyclotomic polynomial Φp(x), that is the primitive p-th roots of unity, and give a basis for Q(ζp) over Q. Note that any element of the Galois group Gal(Q(ζp)=Q) simply permutes these roots. Let now H ⊂ Gal(Q(ζp)=Q) be a subgroup, defining αH 2 Q(ζp) to be X αH = σ(ζp); σ2H one can express the fixed subfield of H as H Q(ζp) = Q(αH ): The latter statement is left as an exercise. × Let us look at a specific example, say p = 13. The Galois group Gal(Q(ζ13)=Q) ' (Z=13Z) 2 is generated by σ = σ2 which maps ζ = ζ13 to ζ . The nontrivial subgroups of Gal(Q(ζ)=Q) correspond to divisors of 12, namely, 2, 3, 4, and 6, and are generated by σ6, σ4, σ3, and σ2 respectively. By the above, the generator of the first of these subgroups can be written as ζ + σ6(ζ) = ζ + ζ−1: Similarly, generators of the other three are ζ + ζ3 + ζ9; ζ + ζ5 + ζ8 + ζ12; ζ + ζ3 + ζ4 + ζ9 + ζ10 + ζ12 respectively. The corresponding lattice of extensions looks as follows: (ζ) 2 Q −1 3 Q(ζ + ζ ) 2 (ζ + ζ3 + ζ9) 3 Q 5 8 12 Q(ζ + ζ + ζ + ζ ) 2 (ζ + ζ3 + ζ4 + ζ9 + ζ10 + ζ12) 3 Q 2 Q 1 2 a1 ak Corollary 2. Let n = p1 : : : pk be the prime factor decomposition. Then we have k \ Q(ζ ai ) = Q and Q(ζ a1 ) ::: Q(ζ ak ) = Q(ζn): pi p1 pk i=1 The latter implies the isomorphism Gal(Q(ζn)=Q) ' Gal(Q(ζ a1 )=Q) × · · · × Gal(Q(ζ ak )=Q); p1 pk which is equivalent to the statement of the Chinese Remainder Theorem: × a1 × ak × (Z=nZ) ' (Z=p1 Z) × · · · × (Z=pk Z) : Proof. The proof is a straightforward application of results we proved during the last two lectures and is left as an exercise. Definition 3. The extension K=F is abelian if so is its Galois group. It is an open problem to determine which groups arise as Galois groups of Galois extensions of Q, however we can now show that every finite abelian group can be realized this way. Before we prove it, let us do some preparatory work. Lemma 4. Let n be a positive integer, m its proper divisor, and p be an integer which does m not divide n. Then Φn(x) and x − 1 do not share roots modulo p. m Proof. Suppose α is a common root of Φn(x) and x − 1. Let us write n Y x − 1 = Φn(x) Φd(x); d j n d<n and observe that the polynomial xm − 1 divides the last product in the right hand side of the above equality. Then α is a double root of xn − 1 modulo p. On the other hand, we have αm = 1 modulo p which implies (α; p) = 1. The condition that α is a double root of xn − 1 implies that α is a root of the derivative nxn−1, that is p j nαn−1, which leads to a contradiction since (p; α) = 1 and p does not divide n. Theorem 5. For any positive integer n there are infinitely many primes congruent to 1 modulo n. Proof. Once again we will give a proof by contradiction. Suppose p1; : : : ; pN are all primes which are congruent to 1 modulo n. Now, let us choose a positive integer l, and set a = lp1 : : : pN and M = Φn(a): Since Φn(x) is monic, choosing l to be large enough we can guarantee that M > 1 and hence there is a prime p such that p j M. Note that (p; a) = 1. Indeed, otherwise p would divide a and therefore divide all non- constant terms of Φn(a), at the same time the constant term of Φn(x) is equal to 1 which would imply that p does not divide M. Now, we have Φn(a) ≡ 0 modulo p which implies p j (an − 1). By the previous lemma, there is no proper divisor m of n such that p j (am − 1). Therefore, the order of a mod p is n which yields n j (p − 1) since ap−1 ≡ 1 mod p. But then we have found another prime p, not equal to any of p1; : : : ; pN with that property, which leads to a contradiction. Corollary 6. Let G be a finite abelian group. Then there is a subfield K of a cyclotomic field with Gal(K=Q) ' G. 3 Proof. Any finite abelian group G can be written as G ' Z=n1Z × · · · × Z=nkZ for some integers n1; : : : ; nk. By the previous theorem we can choose distinct primes p1; : : : ; pk × such that pi − 1 is divisible by ni for all i. Then each group (Z=piZ) ' Z=(pi − 1)Z has × a subgroup Hi of order (pi − 1)=n, and the quotient subgroup (Z=piZ) =Hi is cyclic. Now, setting n = p1 : : : pk and H = H1 × · · · × Hk we get × G ' (Z=nZ) =H: Therefore, there is a subfield K of Q(ζn) with Gal(K=Q) ' G. The following theorem provides a converse statement to the last Corollary, however its proof is beyond the scope of this class. Theorem 7 (Kronecker { Weber). Any finite abelian extension of Q is contained in some cyclotomic extension Q(ζn). We conclude this lecture with a discussion of which regular n-gons can be constructed with straightedge and compass. Lemma 8. If 2k + 1 is prime, k must be a power of 2. Proof. Assume k is not a power of 2, then we have k = rs, and one of the factors, say s, must be odd. Then one can check that (2r + 1) j (2rs + 1) = 2k + 1: Definition 9. Prime numbers of the form 22s + 1 are called Fermat primes. The first several Fermat primes are 3 = 21 + 1; 5 = 22 + 1; 17 = 24 + 1; 257 = 28 + 1; 65537 = 216 + 1; however 232 + 1 is divisible by 641 and it is not even known whether there are infinitely many Fermat primes. Proposition 10. A regular n-gon can be constructed by a straightedge and a compass if and only if n is a product of a power of 2 and distinct Femrat primes. Proof. First of all, note that the primitive n-th roots of unity sit at the vertices of a regular n-gon, and therefore the problem of constructing a regular n-gon is equivalent to that of constructing a primitive n-th root of unity ζn. The latter, in turn, is the same as constructing the x-coordinate of ζn which is simply equal to its real part: ζ + ζ−1 x = n n : 2 Now, we see that ζ satisfies the quadratic equation 2 ζn − 2xζn + 1 = 0; 4 but since x 2 R we conclude that jQ(ζ): Q(x)j = 2. The latter implies that φ(n) j (x): j = : Q Q 2 Recall now that an element α is constructible over Q if and only if the field Q(α) is contained in field obtained from Q by a series of quadratic extensions. Hence, if a regular n-gon can be constructed with a straightedge and compass then φ(n) must be a power of 2. Conversely, if φ(n) is a power of 2, then m+1 m j Gal(Q(ζn): Q])j = 2 =) j Gal(Q(x): Q])j = 2 : It is easy to see that an abelian group G of order 2m has a chain of subgroups f1g = G0 ⊂ G1 ⊂ · · · ⊂ Gm = G with jGi+1 : Gij = 2 for all i = 1; : : : ; m − 1. This shows that a regular n-gon can be constructed by a straightedge and a compass if and only if φ(n) is a power of 2. Decomposing n into prime factors, we see that the latter condition is satisfied if and only if k n = 2 p1 : : : pr where pr − 1 is a power of 2 for all i = 1; : : : ; r. Now, the result follows from the previous lemma. .
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