The 2-Power Degree Subfields of the Splitting Fields of Polynomials with Frobenius Galois Groups

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1 COMMUNICATIONS IN ALGEBRA Õ Vol. 31, No. 10, , 2003 The 2-Power Degree Subfields of the Slitting Fields of Polynomials with Frobenius Galois Grous Blair K. Searman, 1 Kenneth S. Williams, 2, * and Qiduan Yang 1 1 Deartment of Mathematics and Statistics, Okanagan University College, Kelowna, British Columbia, Canada 2 School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada ABSTRACT Let f (x) be an irreducible olynomial of odd degree n > 1 whose Galois grou is a Frobenius grou. We suose that the Frobenius comlement is a cyclic grou of even order h. Let 2 t h. For each i ¼ 1, 2,..., t we show that the slitting field L of f (x) has exactly one subfield K i with [K i : Q] ¼ 2 i. These subfields form a tower of normal extensions Q K 1 K 2 K t with [K i : K i 1 ] ¼ 2(i¼1, 2,..., t) and K 0 ¼ Q. Our main result in this aer is an exlicit formula for ffiffiffiffi an element a i in K i 1 such that K i ¼ Qð a iþ (i ¼ 1, 2,..., t). This result is alied to DeMoivre s quintic x 5 5ax 3 þ 5a 2 x b, solvable *Corresondence: Kenneth S. Williams, School of Mathematics and Statistics, Carleton University, Ottawa, Ontario K1S 5B6, Canada; williams@ math.carleton.ca DOI: /AGB Coyright # 2003 by Marcel Dekker, Inc (Print); (Online)

2 4746 Searman, Williams, and Yang quintic trinomials x 5 þ ax þ b, as well as to some numerical olynomials of degrees 5, 9, and 13. Key Words: Frobenius grou; Subfields of slitting field; Galois grou. 1. INTRODUCTION A finite grou G is said to be a Frobenius grou if there exists a transitive G-set X such that every g 2 G nf1g has at most one fixed oint ð1þ and there is some g 2 G nf1g that does not have a fixed oint. ð2þ It can be roved (Rotman, 2002, Proosition 8.161) that a finite grou G is a Frobenius grou if and only if it contains a roer nontrivial subgrou H such that H \ ghg 1 ¼ f1g for all g62h: ð3þ Such a subgrou H of G is called a Frobenius comlement of G. Let N ¼ f1g[ G n [ ghg!!: 1 g2g N is called the Frobenius kernel of G. Frobenius roved using character theory the following result (Rotman, 2002, Theorem 8.164): Let G be a Frobenius grou with comlement H and kernel N: Then N is a normal subgrou of G with N \ H ¼ f1g and G ¼ NH: ð4þ Furthermore, we have (Robinson, 1982, Ex ) h j n 1; where h ¼ jhj and n ¼ jnj: ð5þ By (4), G is the semi-direct roduct of N and H, written G ¼ N e H. Note that there is a natural G-action on N: for s in G, f s (v) ¼ svs 1, n 2 N. We state the following result without roof.

3 2-Power Degree Subfields 4747 The semi-direct roduct G ¼ N e H is a Frobenius grou with kernel N and comlement H if and only if the action of H nf1g on N nf1g is fixed-oint free; that is; if s 2 H; u 2 N nf1g and sus 1 ¼ u imly s ¼ 1: In this aer, we consider irreducible olynomials f (x) 2 Z[x] with Galois grou G ¼ Gal( f ) satisfying the following three conditions: G ¼ NeH is a Frobenius grou with kernel N and comlement H; ð6þ ð7aþ H is a cyclic grou with even degree h, hence N is abelian, ð7bþ degðf ðxþþ is odd, greater than 1, and equal to n, the order of N: ð7cþ In (7b) the fact that N is abelian follows from Robinson (1982, Ex. 10.5). We define the ositive integer t by 2 t k h; ð8þ and the odd ositive integer h 1 by h 1 ¼ h=2 t : We denote the slitting field of f (x) byl so that GalðL=QÞ ¼ Galðf Þ¼ G ¼ NeH: For each j ¼ 1, 2,..., t we show that L has exactly one subfield K j with [K j : Q] ¼ 2 j. These subfields form a tower of normal extensions Q K 1 K 2 K t with [K i : K i 1 ] ¼ 2(i¼1, 2,..., t) where K 0 ¼ Q. Our objective in this aer is to give an exlicit element a i 2 K i 1 such ffiffiffiffi that K i ¼ Qð a iþ (i ¼ 1, 2,..., t). This determination is given in Sec. 3 after some reliminary results are roved in Sec. 2. In Sec. 4 we aly our results to certain classes of olynomials. Remark 1. Let K be a subfield of C. Let y 1, y 2,..., y n be the roots in C of f (x) 2 K[x]. The discriminant of f (x) is defined by D f ¼ Yn i; j¼1 i <j ðy i y j Þ 2 : ð9þ

4 4748 Searman, Williams, and Yang If the roots of f (x) are distinct, we fix some ordering of the roots and view the Galois grou G of f (x) as a subgrou ffiffiffiffiffiffi of the symmetric grou S n. Galois theory tells us that the field Kð D f Þ is always a subfield of the slitting field of f (x), ffiffiffiffiffiffi and that G is a subgrou of the alternating ffiffiffiffiffiffi grou A n if and only if D f 2 K. Therefore the field extension Kð D f Þ=K is quadratic if and only if G contains odd ermutations on fy 1, y 2,..., y n g. In this aer, we shall see that when G is ffiffiffiffiffiffi not contained in A n, the quadratic extension K 1 =K is reroducing Kð D f Þ=K. It is worth noting that even when G is not a subgrou of A n, a quadratic tower over K can still be constructed. Definition 1. Let y 1, y 2,..., y n be the roots in C of f (x) 2 K[x]. The discriminant olynomial of f (x) is defined to be gðxþ ¼ Yn ðx ðy i y j ÞÞ: ð10þ i;j¼1 i 6¼j It is clear that g(x) 2 K[x] and deg g(x) ¼ n(n 1). We now state our main result. Theorem. Let f (x) 2 Z[x] be an irreducible olynomial. Let the roots of f (x) in C be y 1, y 2,..., y n. Let L ¼ Q(y 1, y 2,..., y n ) be the slitting field of f (x), and G ¼ Gal(f)¼Gal(L=Q) be the Galois grou of f (x). Assume that f (x) and G satisfy the following four conditions: (a) G ¼ N e H is a Frobenius grou with kernel N and comlement H. (b) H is a cyclic grou with even degree h. (c) deg(f (x)) is odd, greater than 1, and equal to n the order of N. (d) The discriminant olynomial of f (x) is squarefree. Define t and h 1 as in (8) and (9) resectively. Then L contains exactly one normal subfield K j with [K j : Q] ¼ 2 j for each j ¼ 1, 2,..., t. These subfields satisfy Q K 1 K 2 K t ð11þ with K i =Q a cyclic extension of degree 2 i for i ¼ 0, 1,..., t. Further, for i ¼ 0, 1,...,t 1, gðxþ ¼ 2i ðn 1Þ=h Y j¼1 g ij ðxþ; ð12þ

5 2-Power Degree Subfields 4749 where each g ij (x) 2 K i [x] is monic, irreducible, of degree nh=2 i, and even. Finally, for any j 2f1, 2,...,2 i (n 1)=hg, we have qffiffiffiffiffiffiffiffiffiffiffi K iþ1 ¼ Qð g ij ð0þþ; for i ¼ 0; 1; 2;...; t 2; ð13þ and K t ¼ Qð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g t 1 j ð0þþ: ð14þ Remark 2. The existence of a quadratic tower of the form (11) follows from Galois theory. Let L N be the subfield of L fixed by N. Then the Galois grou of L N over Q, Gal(L N =Q), is isomorhic to G=N, hence to H, which is cyclic of order 2 t h 1. Gal(L N =Q) has a unique sequence of subgrous (each of which is normal since Gal(L N =Q) is abelian) P t C P t 1 C C P 1 C P 0 ¼ GalðL N =QÞ; such that [P i 1 : P i ] ¼ 2, i 2f1, 2,..., tg. Corresondingly, G ¼ Gal(L N =Q) has a unique sequence of normal subgrous M t C M t 1 C M 1 C M 0 ¼ G; ð15þ such that [M i 1 : M i ] ¼ 2, i 2f1, 2,..., tg, and N M i, i 2f0, 1,..., tg, by the Corresondence Theorem (Rotman, 2002, Proosition 2.76). A quadratic tower of the form (11) thus exists in which each K i is the fixed field of M i for i 2f1, 2,..., tg. Moreover, we claim that every subfield of L of degree 2 j over Q must be a field in this tower. Such a subfield, written as L M, is fixed by a subgrou M of G such that [G : M] ¼ 2 j, j 2f1, jmnj 2,..., tg. We notice that jmj is a ower of 2, as it is a factor of [G : M]. On the other hand jmnj jmj ¼ jnj jm\nj is odd since jnj is odd. Hence jmnj M j j ¼ 1. This shows that N M. Therefore M must be the subgrou M j in (15) and it follows that the subfield L M is the field K j in (11). This imlies the uniqueness of the tower (11). The following diagram illustrates the Galois corresondence between some subgrous of G and some subfields of L. 1 C N C M t C C M 2 C M 1 C M 0 ¼ G l l l l l l L L N K t K 2 K 1 K 0 ¼ Q

6 4750 Searman, Williams, and Yang 2. SOME PRELIMINARY RESULTS We recall and reorganize some basic facts about Frobenius grous in Cangelmi 2000 and Robinson (1982) for our uroses. Let f (x) 2 Z[x ] satisfy all the assumtions of the Theorem. Let fy 1, y 2,..., y n g be the roots of f (x) in C. We may relace Q by a number field K. For a fixed i 2f1, 2,..., ng, let H i be the stabilizer of y i in G, that is, H i ¼fs 2 G : s(y i ) ¼ y i g. Then the subfield of the slitting field L fixed by H i is K(y i ). As f (x) is irreducible over K, we have ½G : H i Š¼½Kðy i Þ : KŠ ¼degðf ðxþþ ¼ jnj ¼½G : HŠ: It follows that jh i j¼jhj, hence N \ H i ¼f1g, i ¼ 1, 2,..., n, since jh i j¼h and jnj¼n are corime by (5). The natural rojection s 2 G! sn restricted to the subgrou H i must be one-to-one because the kernel of the ma is N \ H i ¼f1g. Therefore H i ffi G=N ffi H as grous. As G is transitive on the set fy 1, y 2,..., y n g, for any j 2f1, 2,..., ng with j 6¼ i there exists g 2 G such that g(y i ) ¼ y j. Then the grou gh i g 1 (a conjugate of H i ) is the stabilizer H j of the root y j. Thus H i has exactly n conjugates including itself, and each of these fixes exactly one root of f (x). The stabilizer of two distinct roots of f (x) is the trivial subgrou f1g of G, since H i \ H j ¼f1g for i 6¼ j. It is clear that (3) is satisfied and G is a Frobenius grou with comlement H i for any i 2f1, 2,..., ng. From the orders of N, H i and G, it is not hard to verify that N ¼ f1g[ G n [ ghg!!: 1 g2g Thus N is the Frobenius kernel with resect to the comlement H i of G. The following is a summary of the above discussion. Lemma 1. Let G ¼ N e H be a Frobenius grou serving as the Galois grou of an irreducible olynomial f (x) over a number field K, such that deg( f (x)) ¼ n ¼jNj. Let fy 1, y 2,..., y n g be the set of all roots of f (x) in C. Then (i) (ii) G ¼ N e H i, where H i ¼fs2G:s(y i ) ¼ y i g,i2f1, 2,...,ng. The set N nf1g contains all elements in G that do not have a fixed oint in fy 1, y 2,..., y n g. (iii) If s 2 G and s(y r ) ¼ y r, s(y s ) ¼ y s for r, s 2f1, 2,...,ng with r 6¼ s, then s ¼ 1. The following result is an easy corollary of Lemma 1.

7 2-Power Degree Subfields 4751 Proosition 1. Kee the assumtions in Lemma 1. Let i be a fixed integer in f1, 2,...,ng. If H is a cyclic grou then there exists a 2 G such that G ¼ N e hai and a(y i ) ¼ y i. Proof. The subgrou H i is cyclic since H i ffi H. Let a be a generator of H i and the statement follows. & Now we turn to some roerties of the Frobenius kernel N. Proosition 2. For any i 2f1, 2,...,ng, N is a comlete set of left coset reresentatives of H i in G. Proof. Assume that n 1 2 N, n 2 2 N and n 1 H i ¼ n 2 H i so that n 1 1 n 2 2 H i. Hence n 1 1 n 2 ¼ 1 since N \ H i ¼f1g. Thus n 1 ¼ n 2. The roosition now follows from the fact jnj¼[g : H i ]. & Proosition 3. The Frobenius kernel N acts transitively on the set of roots fy 1, y 2,..., y n g of f (x). Proof. For r, s 2f1, 2,..., ng, r 6¼ s, there exists s 2 G, such that s(y r ) ¼ y s, since G acts transitively on the set fy 1, y 2,..., y n g. By Proosition 2, s 2 nh r for some n 2 N. Thus s ¼ nz for some Z 2 H r. Now we have nðy r Þ¼nZðy r Þ¼sðy r Þ¼y s ; comleting the roof. & Next we consider the subgrous of G of the form N e ha 2m i, m 2f0, 1, 2,..., tg. Proosition 4. For m 2f0, 1, 2,...,tg, we have Proof. (i) N e ha 2m i is a subgrou of G containing N. (ii) The index of N e ha 2m i in G is 2 m. (iii) N e ha 2m i acts transitively on fy 1, y 2,..., y n g, the set of roots of f (x). (iv) N e ha 2m i is a Frobenius grou with Frobenius kernel N and comlement ha 2m i. (i) is obvious. (ii) follows from the calculations ½Nehai : Neha 2m iš ¼ jnjjaj jnjja 2m j ¼ h h=2 m ¼ 2m :

8 4752 Searman, Williams, and Yang To rove (iii) we notice that, by Proosition 3, N acts transitively on the set fy 1, y 2,..., y n g. So does N e ha 2m i. Now conditions (1) and (2) in Sec. 1 are satisfied when fy 1, y 2,..., y n g is considered as the N e ha 2m i-set. This roves (iv). & In Remark 2, we observed that G ¼ Gal(L=Q) has a unique sequence of normal subgrous M t C M t 1 C M 1 C M 0 ¼ G; such that [M i 1 : M i ] ¼ 2, i 2f1, 2,..., tg, and N M i, i 2f0, 1, 2,..., tg. Combining this observation and Proosition 4, we obtain Proosition 5. (i) M m ¼ N e ha 2m i,m2f0, 1, 2,...,tg. (ii) K m is the subfield of L fixed by M m ¼ N e ha 2m i, m2 f0, 1, 2,...,tg. Proosition 6. For r, s 2f1, 2,...,ng with r 6¼ s, there exists t 2 G such that t(y r ) ¼ y s and t(y s ) ¼ y r. Proof. For any i 2f1, 2,..., ng the subgrou H i ¼fs2G : s(y i ) ¼ y i g is cyclic of even order. Denote the unique element of order 2 in H i by t i. If t 2 G is of order 2, then t lies in H i for some i 2f1, 2,..., ng, since G ¼ ð S n i¼1 H iþ[n and jnj is odd. Thus t ¼ t i for some i and ft 1, t 2,..., t n g is the comlete set of order 2 elements in G. Each t i (i 2f1, 2,..., ng) fixes exactly one root y i of f (x), hence t i is a roduct of (n 1)=2 transositions. We oint out that no two of these order 2 elements can have a transosition in common. Otherwise, say that the transosition (y r, y s ), for some r 6¼ s, occurs in both t i and t j, for some i 6¼ j. Then t i t j ðy r Þ¼t i ðy s Þ¼y r ; t i t j ðy s Þ¼t i ðy r Þ¼y s : It follows from Lemma 1(iii) that t i t j ¼ 1, hence t i ¼ t j, a contradiction. Now assume that r, s 2f1, 2,..., ng and r 6¼ s. Then there are (n 2) order 2 elements in G which fix neither y r nor y s. Let t k be such an order 2 element. Then k 6¼ r and k 6¼ s. t k contains a transosition (y r, t k (y r )), where t k (y r ) 2fy 1, y 2,..., y n gnfy r, y k g, which is a set of (n 2) elements containing y s. Therefore there exists t 2 G, such that t(y r ) ¼ y s and t(y s ) ¼ y r. & In the rest of this section we assume the following set of conditions.

9 2-Power Degree Subfields 4753 Condition Set. (i) (ii) K is a subfield of C and y 1, y 2,..., y n are the roots in C of an irreducible olynomial f (x) 2 K[x]. The discriminant olynomial of f (x) gðxþ ¼ Yn i;j¼1 i6¼j ðx ðy i y j ÞÞ (iii) (iv) (v) is squarefree. L ¼ K(y 1, y 2,..., y n ) is the slitting field of f (x). G ¼ Gal(L=K) is a Frobenius grou with Frobenius kernel N and comlement H, such that H is a cyclic grou with order jh j¼2 m h 1, where m is a ositive integer and h 1 is an odd ositive integer. The degree of f (x) is odd, greater than 1, and equal to n, the order of N. Let g(x) be an irreducible factor of g(x) over K. We have the following observations. Proosition 7. The grou G ¼ Gal(L=K) acts transitively on the set of roots of g(x). Moreover, G acts regularly on the set of roots of g(x), that is, the stabilizer of any root of g(x) in G is the trivial subgrou f1g. Proof. The first statement is clear. A root of g(x) is of the form y r y s, for some r 6¼ s, r, s 2f1, 2,..., ng. Ifs 2 G and s(y r y s ) ¼ y r y s, then s(y r ) ¼ y r and s(y s ) ¼ y s, since g(x) is squarefree. Thus s ¼ 1 by Lemma 1(iii). & Corollary. The degree of g(x) is equal to jg j. We note that the discriminant olynomial g(x) is the olynomial R( 1, f)(x) in Cangelmi (2000,. 852). A more general treatment can be found in Cangelmi (2000, Theorem 3.1). Proosition 8. (i) If y r y s is a root of g(x), for some r, s 2f1, 2,...,ngwith r 6¼ s, so is y s y r. (ii) g(x) ¼ h(x 2 ) for some h(x) 2 K[x].

10 4754 Searman, Williams, and Yang Proof. By Proosition 6, there exists t 2 G, such that t(y r ) ¼ y s and t(y s ) ¼ y r. Thus t(y r y s ) ¼ y s y r is a root of g(x) ify r y s is a root of g(x). Over L, whenever x (y r y s ) is a linear factor of g(x), so is x (y s y r ). Therefore g(x) is a roduct of quadratic factors of the form x 2 (y r y s ) 2 for some r, s 2f1, 2,..., ng with r 6¼ s. This roves (ii). & We note that d ¼jG j=2 is the degree of h(x). Next we label the roots x 1,..., x d, x dþ1,..., x 2d of g(x) in such a way that x k ¼ x kþd, k ¼ 1, 2,..., d. We observe that gðxþ ¼ Yd k¼1 gð0þ ¼ð 1Þ d Yd hðxþ ¼ Yd D h ¼ k¼1 Then we have D g ¼ ðx x k Þðx þ x k Þ¼ Yd k¼1 x 2 k ; ðx x 2 k Þ; Y 1k<ld Y 1k<l2d ðx 2 k x2 l Þ2 : ðx k x l Þ 2 k¼1 ðx 2 x 2 k Þ; " Y # " 2 ¼ ðx k x l Þ Y # " 2 d Y # 2 ð2x k Þ 2 ðx k þ x l Þ 2 1k<ld k¼1 " # 2 Y ¼ ðx 2 k x2 l Þ2 ð2 2d Þ Yd 1k<ld ¼ 2 2d D 2 h ð 1Þd gð0þ: It follows that ffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi D g ¼2 d D h ð 1Þ d gð0þ : x 2 k k¼1 1k<ld Noting that D h 2 K we have roved the following result. ffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Proosition 9. Kð D g Þ¼Kð ð 1Þ d gð0þ Þ,whered ¼ 1 2 j G j ¼ 1 2 degðgðxþþ.

11 2-Power Degree Subfields 4755 Proosition 10. Assume Condition Set holds. If g(x) is an irreducible factor of g(x) over K, then the field extension Kð D g Þ¼Kð ð 1Þ d gð0þ ffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Þ over K has degree 2. Proof. It suffices to show that G, viewed as a ermutation grou on the roots of g(x), contains an odd ermutation. Fix a root x of g(x). Then the ma s 2 G 7! sx is a one-to-one corresondence from G onto the set of roots of g(x), by Proosition 7. Thus we just need an element of G acting as an odd ermutation when G acts on itself by left multilication. Let r be an element of H of order 2 m and m be an element of H of order h 1. Then H is the direct roduct of the two cyclic subgrous generated by r and m resectively. We also notice that G ¼ NH ¼ H N since N is a normal subgrou of G. Thus each element in G can be reresented uniquely as r i m j n for some n 2 N, i 2f0, 1,...,2 m 1g and j 2f0, 1,..., h 1 1g. We now claim that left multilication by r, denoted r L : s 2 G 7! rs 2 G, serves as an odd ermutation on the set G. For fixed j 2 f0, 1,..., h 1 1g and n 2 N, the action of r L is r i m j n 7! r iþ1 m j n for i 2f0, 1,...,2 m 2g and r 2m 1 m j n 7! m j n. Therefore the cycle of length 2 m j;n ¼ðm j n; rm j n; r 2 m j n;...; r 2m 1 m j nþ occurs in the reresentation of r L as the roduct of disjoint cycles, and r L ¼ Yh 1 1 j; n : j¼0 n2n As each j,n is an odd ermutation and h 1 n is an odd integer, r L is an odd ermutation on G. & 3. PROOF OF THE THEOREM We verify that for all i 2f0, 1,..., t 1g, K i ¼ K satisfies all five conditions in the Condition Set. f (x) is irreducible over K 0 ¼ Q by assumtion. To show that f (x) is irreducible over K i, i 2f1, 2,..., t 1g, it suffices to show that the Galois grou Gal(L=K i ) acts transitively on the set of roots of f (x). But Gal(L=K i ) is, by Proosition 5, M i ¼ N e ha 2i i, which acts on fy 1,..., y n g transitively by Proosition 4(iii). Hence (i) of the Condition Set holds. It is clear that gðxþ ¼ Yn ðx ðy i y j ÞÞ i;j¼1 i6¼j

12 4756 Searman, Williams, and Yang is squarefree over K i, and L ¼ K(y 1, y 2,..., y n ) is the slitting field of f (x). Thus (ii) and (iii) of the Condition Set hold. The Galois grou Gal(L=K i ) ¼ M i ¼ N e ha 2i i is a Frobenius grou with kernel N and comlement h a 2i i, which is a cyclic grou of even order 2 t i h 1, where t i is a ositive integer and h 1 is an odd ositive integer. This verifies (iv) of the Condition Set. Finally, the degree of f (x) is n ¼jNj by assumtion. Thus (v) of the Condition Set is valid. Recall that the degree of g(x) isn(n 1). According to Proosition 7 and its corollary, each irreducible factor of g(x) overk i is of degree jg j¼2 t i h 1 n ¼ nh=2 i. Therefore g(x) hasn(n 1)=jG j¼2 i (n 1)=h irreducible factors over K i. Hence over K i we have gðxþ ¼ 2i ðn 1Þ=h Y j¼1 g ij ðxþ; ð16þ where each g ij (x) 2 K i [x] is monic, irreducible, and of degree jg j¼2 t i h 1 n ¼ nh=2 i. By Proosition 10, the field extension qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K i ð 1Þ d i g ij ð0þ =K i has degree 2, where d i ¼ deg(g ij (x))=2. It is now clear that for i 2f0, 1,..., t 1g, the degree of the element qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 1Þ d i g ij ð0þ over the rational field Q is 2 iþ1. By the uniqueness of the quadratic tower (11) (Remark 2), we have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K iþ1 ¼ Qð ð 1Þ d i g ij ð0þþ; i 2f0; 1;...; t 1g: When i 2f0,..., t 2g, d i ¼ deg(g ij (x))=2 ¼ 2 t 1 i h 1 n is even, and it qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi follows that ð 1Þ d i g ij ð0þ ¼ ffiffiffiffiffiffiffiffiffiffiffi g ij ð0þ. When i ¼ t 1, d t 1 ¼ deg(g t 1j (x))=2 ¼ 2 t 1 (t 1) h 1 n ¼ h 1 n is odd, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi hence we have ð 1Þ d t 1 g t 1 j ð0þ ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g t 1 j ð0þ. The roof is now comlete since both (13) and (14) are established by (15) and the notes above. & 4. EXAMPLES Our theorem gives a ractical way of determining the normal subfields K i of degree 2 i of the slitting field of L of f since the olynomial g(x) can be conveniently comuted using resultants (see Soicher, 1981) and factored over a number field using for examle a ackage such as

13 2-Power Degree Subfields 4757 MAPLE. If g(x) has reeated factors it is necessary to change the olynomial f (x) by a Tschirnhausen transformation. Examle 1. Let f (x) ¼ x 5 5ax 3 þ 5a 2 x b 2 Z[x] be irreducible. Then 4a 5 b 2 6¼ 0, otherwise there exists an integer c such that a ¼ c 2, b ¼ 2c 5, and f (x) has the linear factor x 2c. The Galois grou G of f is the Frobenius grou F 20. Here n ¼ 5, h ¼ 4, (n 1)=h ¼ 1 and t ¼ 2. The olynomial f (x) is known as DeMoivre s quintic. Set gðxþ ¼ Resultantðf ðx þ XÞ; f ðxþþ x 5 : MAPLE gives g(x) as a olynomial of degree 20 with constant term g(0) ¼ 5 5 (4a 5 b 2 ) 2 ¼ g 01 (0). By our theorem the unique quadratic subfield K 1 of L is ffiffiffiffiffiffiffiffiffi ffiffi K 1 ¼ Qð gð0þ Þ¼Qð 5 Þ: ffiffiffi Next we factor g(x) inqð ffiffiffi 5 Þ½xŠ. MAPLE gives two monic olynomials g 11 (x) and g 12 (x) inqð 5 Þ½xŠ of degree 10 such that gðxþ ¼g 11 ðxþg 12 ðxþ: ffiffi By our theorem these olynomials are irreducible in Qð 5 Þ½xŠ. Evaluating them at x ¼ 0, MAPLE gives g 11 ð0þ ¼ 1000a5 250b 2 25 þ 11 ffiffi ; g 12 ð0þ ¼ 1000a5 250b ffiffiffi ; 5 and our theorem yields the unique quartic subfield K 2 of L as s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! K 2 ¼ Q 1000a5 250b 2 25 þ 11 ffiffi : 5 Since 1000a5 250b 2 25 þ 11 ffiffi ¼ 5 þ 5 ffiffi! 2 5 ð4a 5 b 2 Þð5 þ 2 ffiffi 5 Þ; 5 2 we have qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 2 ¼ Q ð4a 5 b 2 Þð5 þ 2 ffiffi 5 Þ in agreement with Searman and Williams (1999, Theorem).

14 4758 Searman, Williams, and Yang Examle 2. We choose f ðxþ ¼x 5 þ ax þ b 2 Z½xŠ to be a solvable, irreducible quintic trinomial with ab 6¼ 0. Let r be the unique rational root of the resolvent sextic of x 5 þ ax þ b (Searman and Williams, 1994,. 988). Set 3r 16a c ¼ 3r 16a 4r þ 12a ; e ¼ sgn ; e ¼ 5be 4r þ 12a 2r þ 4a ; so that c ð 0Þ 2Q; e ¼1; e ð6¼ 0Þ 2Q: Then (see, for examle, Searman and Williams, 1994, Theorem,. 987) we have a ¼ 5e4 ð3 4ecÞ c 2 þ 1 The Galois grou G of f is D 5 ; if 5ðc 2 þ 1Þ 2Q 2 ; F 20 ; if 5ðc 2 þ 1Þ 62 Q 2 ; ; b ¼ 4e5 ð11e þ 2cÞ c 2 : þ 1 where D 5 is the dihedral grou of order 10 and F 20 is the Frobenius grou of order 20 (Searman and Williams, 1994,. 990). We note that D 5 is a Frobenius grou. We just treat the case when G ¼ F 20 as the case G ¼ D 5 is simler. Here n ¼ 5, h ¼ 4, (n 1)=h ¼ 1 and t ¼ 2. Set Resultantðf ðx þ XÞ; f ðxþþ gðxþ ¼ x 5 : MAPLE gives g(x) as a olynomial of degree 20 with constant term gð0þ ¼g 01 ð0þ ¼ ec 3 84c ec 122 ðc 2 þ 1Þ 5 : By the theorem we obtain ffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 1 ¼ Q g 01 ð0þ ¼ Q 5ðc 2 þ 1Þ ; in agreement with Searman et al. (1995,. 16). Next we use MAPLE to factor g(x) over K 1. MAPLE gives g(x) as the roduct of two monic olynomials g 11 (x) and g 12 (x)

15 2-Power Degree Subfields 4759 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi in Qð 5ðc 2 þ 1ÞÞ½xŠ of degree 10 such that gðxþ ¼g 11 ðxþg 12 ðxþ: MAPLE gives qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g 11 ð0þ ¼ðsquareÞ 25ðc 2 þ 1Þþð5þ10eÞ 5ðc 2 þ 1Þ ¼ðsquareÞ 5ðc 2 þ 1Þ r ffiffiffiffiffiffiffiffiffiffiffiffiffi! 5 5 þð1þ2eþ c 2 : þ 1 By the theorem we have 0sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi 1 K 2 ¼ Q@ 5 5 þð1þ2eþ A c 2 þ 1 in agreement with Searman et al. (1995, Theorem,. 17). We conclude by giving brief details of four numerical examles. Examle 3. f ðxþ ¼x 5 70x 3 140x 2 þ 385x þ 28; G ¼ F 20 ; n ¼ 5; h ¼ 4; ðn 1Þ=h ¼ 1; t ¼ 2; g 01 ð0þ ¼ ; ffiffiffiffiffi K 1 ¼ Qð 10 Þ; g 11 ð0þ ¼ ð 650 þ 201 ffiffiffiffiffi 10 Þ; g 12 ð0þ ¼ ð ffiffiffiffiffi 10 Þ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 2 ¼ Q 650 þ 201 ffiffiffiffiffi 10 0vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 17 þ 4 ffiffiffiffiffi 1! u 2 B 10 ¼ Q t ð10 þ ffiffiffiffiffi 10 Þ A 3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q 10 þ ffiffiffiffiffi 10 :

16 4760 Searman, Williams, and Yang Examle 4. f ðxþ ¼x 9 3x 8 þ 3x 7 15x 6 þ 33x 5 3x 4 þ 24x 3 þ 6x 2 4; see (Cangelmi, 2000,. 856); G ¼ðZ 3 Z 3 ÞeZ 4 ; n ¼ 9; h ¼ 4; ðn 1Þ=h ¼ 2; t ¼ 2; g 01 ð0þ ¼ ; g 02 ð0þ ¼ ; ffiffi K 1 ¼ Qð 5 Þ; g 11 ð0þ ¼ ð5 þ 2 ffiffi 5 Þ; g 12 ð0þ ¼ ð5 2 ffiffi 5 Þ; g 13 ð0þ ¼ ð5 ffiffiffi 5 Þ; g 14 ð0þ ¼ ð5 þ ffiffiffi 5 Þ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 2 ¼ Q ð15 þ 6 ffiffiffi 5 Þ 0vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q 1 þ ffiffi 1! u 2 Bt 5 ð30 þ 6 ffiffi 5 Þ A 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q ð30 þ 6 ffiffiffi 5 Þ : Examle 5. f ðxþ ¼x 9 72x 7 þ 1464x 5 960x x 3 þ 13440x x 2560: The MAGMA database gives G ¼ T 15 (notation of Butler and McKay, 1983), jgj ¼ 72: The grou T 15 has one normal subgrou N ¼ Z 3 Z 3 of order 9 as well as nine conjugate subgrous of order 8, each of which is cyclic. These conjugate subgrous intersect only trivially so G is a Frobenius grou and is the semidirect roduct (Z 3 Z 3 ) e Z 8.

17 2-Power Degree Subfields 4761 n ¼ 9; h ¼ 8; ðn 1Þ=h ¼ 1; t ¼ 3; g 01 ð0þ ¼ ; ffiffiffi K 1 ¼ Qð 2 Þ; g 11 ð0þ ¼ ð þ ffiffi 2 Þ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 2 ¼ Q þ ffiffiffi 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q 10 5 ffiffi 2 ; g 21 ð0þ ¼ ð þ b b b 3 Þ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where b ¼ 10 5 ffiffiffi 2; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 3 ¼ Q þ b b b 3 : Since! þ b b b 3 3b b þ 5 ¼ 9450 þ 5820b 531b 2 336b 3 2; we have 0sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 K 3 ¼ Q@ 30 3b þ 3b2 A 5 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q þ 5 ffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ ffiffi! 2 : Examle 6. f ðxþ ¼x 13 26x x 8 þ 143x 7 910x 6 þ 585x x 4 þ 4472x x 2 þ 520x 131: MAPLE gives the discriminant of f (x) as ffiffiffiffiffi so that the quadratic subfield of L is Qð 13 Þ.Ifais

18 4762 Searman, Williams, and Yang ffiffiffiffiffi any root of f (x) MAPLE factors f (x) over Qða; 13 Þ. There are six irreducible quadratics and one linear olynomial in the factorization. Hence ½L : QŠ ¼2 k 13 for some k 2 Z þ. Therefore f (x) is solvable so Gal( f ) ¼ F 13l, where l j 12. It is known that [L : Q] ¼ 13l, where l 6¼ 1asLhas a quadratic ffiffiffiffiffi subfield and l 6¼ 2asf does not factor into linear factors over Qða; 13 Þ. Hence l ¼ 4 and Gal( f ) ¼ F 52. We remark that a theorem of Cangelmi (2000, Theorem 3.17,. 851) rovides an alternative way of verifying that Gal( f ) ¼ F 52. n ¼ 13; h ¼ 4; ðn 1Þ=h ¼ 3; t ¼ 2; g 01 ð0þ ¼ ; g 02 ð0þ ¼ ; g 03 ð0þ ¼ ; ffiffiffiffiffi K 1 ¼ Qð 13 Þ; g 11 ð0þ ¼ ð þ ffiffiffiffiffi 13 Þ; r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K 2 ¼ Q 1 2 ð þ ffiffiffiffiffi! 13 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Q 13 2 ffiffiffiffiffi 13 : ACKNOWLEDGMENTS The authors are grateful to the referee for his=her invaluable suggestions based on his=her thorough reading and understanding of our work. The research of the first two authors was suorted by grants from the Natural Sciences and Engineering Research Council of Canada. REFERENCES Butler, G., McKay, J. (1983). The transitive grous of degree u to eleven. Comm. Algebra 11: Cangelmi, L. (2000). Polynomials with Frobenius Galois grous. Comm. Algebra 28:

19 2-Power Degree Subfields 4763 Robinson, D. J. S. (1982). A Course in the Theory of Grous. Graduate Text in Mathematics 80. New York: Sringer-Verlag. Rotman, J. (2002). Advanced Modern Algebra. Prentice-Hall. Soicher, L. (1981). M. Com. Sci. thesis, Concordia University, Montréal. Searman, B. K., Williams, K. S. (1994). Characterization of solvable quintics x 5 þ ax þ b. Amer. Math. Monthly 101: Searman, B. K., Williams, K. S. (1999). DeMoivre s quintic and a theorem of Galois. Far East J. Math. Sci. (FJMS) 1: Searman, B. K., Searman, L. Y., Williams, K. S. (1995). The subfields of the slitting field of a solvable quintic trinomial. J. Math. Sci. 6: Received November 2001 Revised December 2002