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Cody Holdaway Math 505 Assignment 2 January 20, 2010 Solution to problem 5 Cody Holdaway (1) Let G be a finite abelian group. Prove that there exists an abelian extension of Q with the Galois group G. Proof. If G is a finite abelian group then we know that we can write ∼ G = Z/n1 × Z/n2 ×···× Z/nk . By Dirichlet’s theorem in number theory we know that for any fixed positive integer n, there exists an infinite amount of prime numbers of the form kn +1 where k is a positive integer. This shows that for each ni, we can choose distinct primes pi such ∗ that pi = kini + 1. In other words, ni|pi − 1. Since (Z/pi) is a cyclic group of order ∗ pi − 1 we can find a subgroup Hi of order (pi − 1)/ni. Therefore, (Z/pi) /Hi is a cyclic group of order ni and is therefore isomorphic to Z/ni. This shows that we can write ∼ ∗ ∗ ∗ G = (Z/p1) /H1 × (Z/p2) /H2 ×···× (Z/pk) /Hk ∼ ∗ ∗ = ((Z/p1) ×···× (Z/pk ) )/H1 ×···× Hk. ∗ ∗ Let n = p1 ··· pk, then since the pi are distinct we know that (Z/p1) ×···×(Z/pk) ∼= ∗ ∗ (Z/n) . Let H be the subgroup of (Z/n) that corresponds to H1 ×···× Hk, then we see that ∗ G =∼ (Z/n) /H. Let ξ be a primitive n-th root of unity. Then we know that Gal(Q(ξ)/Q)=(Z/n)∗. Let E = Q(ξ)H be the fixed field of H. Then E is an extension of Q and since (Z/n)∗ is abelian, we know that H is a normal subgroup which implies that E/Q is Galois. Moreover, we know that ∗ Gal(E/Q) =∼ Gal(Q(ξ)/Q)/H ∼= (Z/n) /H =∼ G. This shows that any finite abelian group is the Galois group of some extension of Q. (2) Let K/Q be a finite extension of Q. Let G be a non-trivial finite abelian group. Prove that there exist infinitely many abelian extensions of K with Galois group G. Proof. Just as in part (a) above, write G = Z/n1 × Z/n2 ×···× Z/nk. Then by Dirichlet’s theorem we know that we can find an infinite number of primes pi such that ni|pi − 1 and we can therefore construct an infinite number of k-tuples 0 0 0 {(p1, p2,...,pk)} such that (p1, p2,...,pk) ∩ (p1, p2,...,pk) = ∅. Then we saw that we can construct a Galois extension E with group G such that Q ⊂ E ⊂ Q(ξp1···pk ). where ξp1···pk is a p1 ··· pk-th primitive root of unity. Since this can be done for any of the k-tuple of primes chosen above, we potentially have an infinite number of distinct 1 2 0 0 extensions of Q with Galois group G. Let(p1,...,pk) and (p1,...,pk) be two distinct 0 0 0 k-tuple of primes and let E and E be the subextensions of Q(ξ 1··· ) and Q(ξ ··· ) p pk p1 pk respectively. Since Q(ξn) ∩ Q(ξm)= Q for (n, m) = 1 we see that since all of the pi are different than the pj that 0 0 0 Q ⊂ E ∩ E ⊂ Q(ξ 1··· ) ∩ Q(ξ ··· )= Q p pk p1 pk which shows that E and E0 are distinct extensions of Q with Galois groups G. Let {Eα} be the set of extensions of Q with Galois group G constructed above using the k-tuples of primes. K/Q is a finite extension and since we are in characteristic 0 we know that it is separable. Therefore, there are only a finite number of fields L such that Q ⊂ L ⊂ K. Consider the following diagram, KEα w HH ww HH ww HH ww HH ww H K Eα HH v HH vv HH vv HH vv H vv K ∩ Eα Q Since Eα/Q is Galois we know that KEα/K is Galois and moreover, Gal(KEα/K) = Gal(Eα/K ∩ Eα). K ∩ Eα is a subfield of K for each Eα and we know that there are only finitely many subfields of K/Q. Also, since (K ∩ Eα) ∩ (K ∩ Eβ)= K ∩ (Eα ∩ Eβ)= K ∩ Q = Q we see that only a finite number of the Eα can have intersection with K to be more than Q. Therefore, there is an infinite number of the Eα such that K ∩ Eα = Q. For these Eα we see then that Gal(KEα/K) = Gal(Eα/K ∩ Eα) = Gal(Eα/Q)= G which shows that there are an infinite number of distinct Galois extensions of K that have Galois group G. Addition (Julia). We nee to take care of the following detail pointed out by Jim: prove that there are infinitely many αi such that KEαi , KEαj are different extensions of K. Lemma 0.1. Let m, n, ` be mutually relatively prime. Then Q(ξmn) ∩ Q(ξm`) = Q(ξm). Proof. exercise Lemma 0.2. Let K/Q be a finite extension, and let {m1,...,mi,...} be an infinite sequence of positive integers such that (a) (mi,mj) = 1 for any pair of distinct indices (i, j). 3 (b) K ∩ Q(ξmi )= Q for any mi. Then there exist only finitely many pairs i, j such that K ∩ Q(ξmimj ) =6 Q. Proof. Let Ls be a non-trivial subextension of K, Q ⊂ Ls ⊂ K. Suppose Ls ⊂ Q(ξmimj ) and Ls ⊂ Q(ξmk ml ). Then Ls ⊂ Q(ξmimj ) ∩ Q(ξmkml ). If all four numbers mi,mj,mk,m` are different, then (mimj,mkm`) = 1, and, hence, Q(ξmimj )∩Q(ξmk ml )= Q, a contradiction. Without loss of generality, we may assume that mi = mk. Then Ls ⊂ Q(ξmimj ) ∩ Q(ξmim` )= Q(ξmi ) where the last equality holds by Lemma 0.1. This implies that K ∩ Q(ξmi ) contains Ls and, hence, is nontrivial. This contradicts our hypothesis (b). Hence, for any Ls there is at most one pair (i, j) such that Ls ⊂ Q(ξmimj ). Since K/Q is separable, there are only finitely many different subfields Ls, which finishes the proof of the lemma. Lemma 0.2 implies that by removing finitely many k-tuples (p1,...,pk) we can 0 0 ensure that K ∩ Q(ξp1···pk ξp1···p ) = Q for any two k-tuples in our set. Let m = 0 0 k p1 ··· pk, n = p1 ··· pk. We shall show that K(ξn) ∩ K(ξm) = K which would imply that the corresponding abelian extensions KEα, KEβ (for Eα ⊂ Q(ξn), Eβ ⊂ Q(ξm) are different. Consider the diagram K(ξmξn) L ww LL ww LL ww LLL ww LL ww K H Q(ξmξn) HH q HH qq HH qq HH qqq HH qqq Q q Since K ∩ Q(ξmξn)= Q, and Q(ξmξn)/Q is Galois, we get [K(ξmξn): K]=[Q(ξmξn): Q]= φ(mn) Now consider a tower of extensions K(ξmξn) K(ξm) K The degree of the bottom extension is φ(m) (it follows as above from the fact that K ∩ Q(ξm)= Q). The degree of K(ξmξn)/K equals φ(mn). Hence, [K(ξmξn): K(ξm)] = φ(m) 4 and similarly [K(ξmξn): K(ξn)] = φ(n) Now consider another diagram K(ξmξn) K φ(n) ss KKK φ(m) sss KK ss KKK sss K K(ξm) K(ξn) LL LL rrr LLL rrr φ(m) LL rr φ(n) LL rrr K Since K(ξn)/K is Galois, we have [K(ξmξn): K(ξm)]=[K(ξn): K(ξm) ∩ K(ξn)]. It follows that [K(ξn): K(ξm) ∩ K(ξn)] = φ(n)=[K(ξn): K]. Hence, K(ξm) ∩ K(ξn)= K. .
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