arXiv:0709.2868v1 [math.NT] 18 Sep 2007
ON GALOIS GROUPS OF PRIME DEGREE POLYNOMIALS
WITH COMPLEX ROOTS
OZ BEN-SHIMOL
f be an irredu ible polynomial of prime degree p ≥ 5 over
Q, with pre isely k pairs of omplex roots. Using a result of Jens Hö hsmann
(1999), we show that if p ≥ 4k + 1 then Gal(f /Q) is isomorphi to Ap or Sp .
Abstra t. Let
This improves the algorithm for omputing the Galois group of an irredu ible
polynomial of prime degree, introdu ed by A. Bialosto ki and T. Shaska.
If su h a polynomial
f is solvable by radi als then its Galois group is a
Frobenius group of degree p. Conversely, any Frobenius group of degree p and
of even order, an be realized as the Galois group of an irredu ible polynomial
of degree
p over Q having omplex roots.
1.
Introdu tion
A lassi al theorem in Galois theory says that an irredu ible polynomial f of
prime degree p ≥ 5 over Q whi h has pre isely one pair of omplex (i.e., non-real)
roots, has the symmetri group Sp as its Galois group over Q (see e.g., Stewart[18℄).
It is natural then to ask the following question: let k be a positive integer and f an
irredu ible polynomial of prime degree p with pre isely k pairs of omplex roots.
What is its Galois group Gal(f /Q)?. If one tries to imitate the proof of the lassi al
theorem (i.e., the ase k = 1), one would nd, onstru tively, the subgroup of Sp
whi h is generated by the p- y le (1 2 ... p) and an involution (a1 a2 ) · · · (a2k−1 a2k ).
My unsu essful attempts (so far) to solve the problem in this way indi ated that
the dieren e between the degree p and the number 2k of the omplex roots, need
not be "large" in order to obtain the alternating group Ap at least (i.e., Gal(f /Q)
is isomorphi to Ap or Sp ).
More general observations on su h permutation groups brings us to a well-known
problem in the theory of permutation groups: let G be a 2-transitive permutation
group of degree n whi h does not ontain the alternating group An , and let m be
its minimal degree. Find the inmum for m in terms of n.
If f is an irredu ible polynomial of prime degree p with k > 0 pairs of omplex
roots, where p > 2k + 1, then its Galois group Gal(f /Q) is 2-transitive of degree
p, with minimal degree at most 2k . Therefore, if B(p) is a lower bound for the
minimal degree, then Gal(f /Q) ne essarily ontains the alternating group Ap when
2k ≤ B(p). Thus, as B(p) approa hes the inmum, the dieren e p − 2k gets
smaller, as required.
Returning to the group-theoreti problem stated√above (for degree n, not ne essarily a prime), Jordan [10℄ showed that B(n) = n − 1 + 1 is a lower bound for
the minimal degree. A substantial improvement of this bound is due to Bo hert [3℄
who showed that B(n) = n/8, and if n > 216 then one has an even better bound,
namely B(n) = n/4. Proofs for the Jordan and Bo hert estimates an be found
also in Dixon & Mortimor [7℄, Theorem 3.3D and Theorem 5.4A, respe tively. More
1
2
OZ BEN-SHIMOL
re ently, Liebe k and Saxl [11℄, using the lassi ation of nite simple groups, have
proved B(n) = n/3.
Finally, Hö hsmann [8℄, using a on ept suggested by W.Knapp whi h renes
the notion of minimal degree in a natural way, namely, r-minimal degree mr (G),
where r is a prime divisor of the order of the group G, gave some better estimates,
whi h in the worst ase meet Liebe k and Saxl's bounds. Sin e the group we are
dealing with is of prime degree, and we have information about its 2-minimal degree,
Ho hmann's result serves us better than that of Liebe k and Saxl.
The paper of A.Bialosto ki and T.Shaska [2℄ fo uses on the pra ti al aspe ts of
this theoreti al problem, in the pro ess of omputing the Galois group of prime
degree polynomials over Q: 1. The existing te hniques, whi h are mainly based
on a theorem of Dedekind (see Cox [6, Theorem 13.4.5℄), are expensive and many
primes p might be needed in the pro ess. 2. Polynomials in general have plenty of
omplex roots. 3. Che king whether a polynomial has omplex roots is very e ient
sin e numeri al methods an be used. Therefore, he king rst if the polynomial
has omplex roots, and then use a "good" bound for the dieren e between the
polynomial's degree and the number of its omplex roots, makes the omputation
of its Galois group mu h easier. However, they make a use of estimate due to
Jordan (summarized in Wielandt [19, page 42℄), whi h is not sharp at all (as the
authors point out in their paper). In fa t, Jordan's bound holds for any primitive
group of any nite degree - not ne essarily 2-transitive of prime degree. In the
present paper, we improve their algorithm and dis uss some theoreti al aspe ts of
the subje t.
2.
Galois groups of prime degree polynomials with omplex roots
A Frobenius group is a transitive permutation group whi h is not regular, but in
whi h only the identity has more then one xed point. In other words, a Frobenius
group G is a transitive permutation group on a set Ω in whi h Gα 6= 1 for some
α ∈ Ω, but Gα ∩ Gβ = 1 for all α, β ∈ Ω, α 6= β . It an be shown that the
set of elements xing no letters of Ω, together with the identity, form a normal
subgroup K alled the Frobenius kernel of G. Frobenius groups are hara terized
as non-trivial semi-dire t produ ts G = K ⋊ H su h that no element of H \ {1}
ommutes with any element of K \ {1}. Basi examples of Frobenius groups are the
subgroups of AGL1 (F ) - the group of the 1-dimensional ane transformations of a
eld F , i.e. the group onsisting of the permutations of the form tα,β : ζ 7→ αζ + β ,
α ∈ F ∗ , β, ζ ∈ F . Clearly, AGL1 (F ) ∼
= F ⋊ U , where U is a non-trivial subgroup
of F ∗ . Identifying U with {0} ⋊ U , it is easy to verify that no nontrivial subgroup
of U is normal in AGL1 (F ). In parti ular, if F = Fp - the eld of p elements (p
prime), then AGL1 (p) := AGL1 (Fp ) ∼
= Fp ⋊ U , where U is a subgroup of F∗p (so U
is a y li of order n, where n 6= 1 and n divides p − 1), is a Frobenius group of
degree p. The stru ture of a Frobenius group of degree p ≥ 5 is des ribed in the
following theorem.
Theorem 1.
(Galois)
Let G be a transitive permutation group of prime degree
p ≥ 5, and of order > p. Then the following statements
i. G has a unique p-Sylow subgroup.
ii. G is a solvable group.
iii. G is isomorphi to a subgroup of AGL1 (p).
iv. G is a Frobenius group of degree p.
are equivalent:
ON GALOIS GROUPS OF PRIME DEGREE POLYNOMIALS WITH COMPLEX ROOTS
Proof.
See Huppert [9℄.
3
Let G ∼
= Fp ⋊ U , U y li of order n, n 6= 1, n|p − 1, be a Frobenius group of
degree p. Then it is ustomary to denote G = Fpn . For example, the dihedral group
D2p = Fp·2 is a Frobenius group of degree p. The Frobenius groups Fp(p−1) appear
as Galois groups of the polynomials X p − a ∈ Q[X], where a ∈ Q∗ \ (Q∗ )p . For
onstru tive realization of Frobenius groups of degree p, see A.A.Bruen, C.Jensen
and N.Yui [4℄.
If f is an irredu ible polynomial of degree p ≥ 5 over Q, then its Galois group
G = Gal(f /Q), as a permutation group a ting on the p-set onsisting of the p roots
of f , is a transitive group of order p (if and only if G ontains a p- y le). Complex
onjugation is a Q-automorphism of C and, therefore, indu es a Q-automorphism
of the splitting eld of f . This leaves the real roots of f xed, while transposing the
omplex roots. Therefore, if f has a pair of omplex roots, then |G| > p. Furthermore, if, in addition, f has more then one real root, then the omplex onjugation
has more then one xed point. In parti ular, G is not a Frobenius group of degree
p. By Theorem 1, G is not solvable, thus, f is not solvable by radi als. So we have
Corollary 1. Let
whi h has a pair of
f
be an irredu ible polynomial of prime degree
omplex roots. If
Frobenius group of degree
p,
and
f
f
is solvable by radi als then
has exa tly one real root.
p ≥ 5 over Q,
Gal(f /Q) is a
Let f be an irredu ible polynomial of prime degree p ≥ 5 and with k > 0 pairs
of omplex roots. By Corollary 1, if p > 2k + 1 then G = Gal(f /Q) is not solvable.
Our purpose is to show that if p ≥ 4k + 1 then G ontains the alternating group
(i.e., G isomorphi to Ap or to Sp ).
Theorem 2.
gree is
(Burnside)
A non-solvable transitive permutation group of prime de-
2-transitive.
Therefore, a transitive permutation group of prime degree is either 2-transitive
or a Frobenius group (see Theorem 1).
Proof.
See, Burnside [5℄, or Dixon & Mortimor [7, Corollary 3.5B℄.
Re all that the minimal degree m(G) of a permutation group G a ting on a
set Ω is the minimum of the supports of the non-identity elements: m(G) :=
min{| supp(x)| : x ∈ G, x 6= 1}. Hen e, G is a Frobenius group if and only if it is
a transitive permutation group with minimal degree |Ω| − 1, and by Theorem 1, a
transitive permutation group of prime degree p ≥ 5 and of order > p is not solvable
if and only if it has minimal degree < p − 1.
Now, for every prime divisor r of |G| we dene the minimal r-degree mr (G) of
G to be the minimum of the supports of the non-identity r-elements (that is, the
non-identity elements whose order is a power of r). Using elementary properties of
the minimal r-degrees and together with results based on the lassi ation of the
nite simple groups, J. Hö hsmann [8℄ has proved
Theorem 3.
not
(Hö hsmann)
Let
G
be a
mr (G) ≥ r−1
r · n or
m
m
ii. G ≥ PSL(2, 2 ), r = 2 − 1 ≥ 7
i.
2-transitive group of degree n whi h
r be a prime divisor of |G|. Then
is a Mersenne prime and
or
iii.
does
ontain the alternating group, and let
G = P Sp(2m, 2), n = 2m−1 · (2m − 1)
with
m > 2, r = 2
mr (G) = r = n − 2
and
4
OZ BEN-SHIMOL
mr (G) =
In any
2m−1 −1
2m −1
ase
· n ≥ 73 · n.
mr (G) ≥ r−1
r+1 · n.
An immediate onsequen e (in fa t, a spe ial ase) of this theorem is
Corollary 2. Let
G be a 2-transitive group of prime degree p whi
m2 (G) ≥ p2 .
h does not
the alternating group. Then
ontain
f be an irredu ible polynomial of prime degree p ≥ 5 over Q.
f has pre isely k > 0 pairs of omplex roots. If p ≥ 4k + 1 then
G = Gal(f /Q) is isomorphi to Ap or to Sp . Clearly, if k is odd then G ∼
= Sp .
Theorem 4. Let
Suppose that
Proof. Complex onjugation has support 2k , hen e m2 (G) ≤ 2k . By Corollary 1,
G is not solvable (f has more than one real root). By Theorem 3, G is 2-transitive
and, by Corollary 2, G ne essarily ontains the alternating group.
Therefore, the algorithm given in [2℄ for omputing the Galois group of an irredu ible prime degree polynomial, an be improved:
Input:
An irredu ible polynomial f (x) ∈ Q[x] of prime degree p.
The Galois group Gal(f /Q).
Output:
begin
r:=NumberOfRealRoots(f(x));
k:=(p-r)/2;
if k > 0 and p ≥ 4k + 1 then
if k is odd then
Gal(f /Q) = Sp ;
else
if ∆(f ) is a omplete square then
Gal(f /Q) = Ap ;
else
Gal(f /Q) = Sp ;
endif;
endif;
else
Redu tionMethod(f(x));
endif
end;
∆(f ) denotes the dis riminant of f (x). It is well known that if f is
a polynomial of degree n with oe ients in a eld K , har(K )6= 2, then ∆(f ) is
a perfe t square in K if and only if Gal(f /K) is isomorphi to a subgroup of An .
See e.g., Stewart [18, Theorem 22.7℄.
Remark 1.
A short dis ussion on the redu tion modulo p method, an be found
in [2℄ and in Cox [6, page 401℄.
Remark 2.
Remark 3. Corollary 1 in [2℄ an also be improved: (repla e their r with our k the number of pairs of the omplex roots of a given irredu ible polynomial of prime
degree p). (i) k = 2 and p > 7. (ii) k = 3 and p > 11. (iii) k = 4 and p > 13. (iv)
k = 5 and p > 19.
ON GALOIS GROUPS OF PRIME DEGREE POLYNOMIALS WITH COMPLEX ROOTS
3.
non-real realization of
5
Fpn
As stated in Corollary 1, an irredu ible solvable polynomial of prime degree
p ≥ 5 over Q, whi h has omplex roots, has a Frobenius group of degree p (and of
even order, of ourse) as its Galois group over Q. We shall prove that the related
"inverse problem" has a positive answer - any Frobenius group of degree p and of
even order appears as Galois group of an irredu ible polynomial of degree p over Q
having omplex roots.
(Diri hlet) Let k ,h be integers su h that k > 0 and (h, k) = 1. Then
there are innitely many primes in the arithmeti progression nk+h, n = 0, 1, 2, . . ..
Theorem 5.
Proof. See e.g., Serre [15℄ or Apostol [1℄.
Lemma 1. Let l be a positive integer, and let ζ be a primitive l -th root of unity.
Then 1, ζ, . . . , ζ ϕ(l)−1 form a Z-basis for the ring of integers of Q(ζ).
Proof. See e.g., Neukir h [12, Chapter I, Proposition 10.2℄.
(Galois) Let f be a polynomial of prime degree over Q. Then, f is
solvable by radi als if and only if any two distin t roots of f generate its splitting
eld.
Lemma 2.
Proof. See Cox [6, Theorem 14.1.1℄.
Theorem 6. (S holz) A splitting embedding problem has a proper solution over
number elds. (That is, let K be a number eld and let M/K be a Galois extension
with Galois group H . Suppose that H a ts on an abelian group A. Then, there exist
a Galois extension L/K whi h ontains M/K su h that Gal(L/K) ∼
= A ⋊ H ).
Proof. See S holz [14℄.
Let Fpn be a Frobenius group of degree p and of even order. Then
Fpn o urs as Galois group of an irredu ible polynomial f of degree p over Q having
omplex roots. Furthermore, the splitting eld of f is Q(a, ib) for every omplex root
a + ib of f .
Theorem 7.
Proof. By Theorem 5, there exist a prime q su h that q ≡ 1( mod n) and (q − 1)/n
is odd number. Indeed, for every natural number k , write 1 + (2k − 1)n = (1 − n) +
(2n)k . So, (1 − n, 2n) = 1 sin e n is even. Thus, su h a prime q does exist. Let m
be a primitive root modulo q (that is, a generator of F∗q ). Consider the sum
(1)
n
αn = ζq + ζqm + ζqm
2n
+ . . . + ζqm
−1)n
( q−1
n
,
where ζq is a primitive q -th root of unity. Then Gal(Q(ζq )/Q) is y li of order
q − 1 and generated by the automorphism σ : ζq 7→ ζqm . We shall see that Q(αn )/Q
is a non-real Cn -extension, and then we shall apply Theorem 6.
Q(αn )/Q is a Cn -extension: By the Fundamental Theorem of Galois Theory, it
n
n
is enough to prove Q(αn ) = Q(ζq )σ . The in lusion Q(αn ) ⊆ Q(ζq )σ is be ause σ n
moves y li ly the summands of (1) (in fa t, αn is the image of ζq under the tra e
n
map TrQ(ζq )/Q(ζq )σn , hen e αn is an element of Q(ζq )σ ). Suppose that Q(αn ) $
d
Q(ζn )σn . There exist a proper divisor d of n su h that Q(αn ) = Q(ζq )σ . In
6
OZ BEN-SHIMOL
parti ular, σd (αn ) = αn , or
q−1
n −1
(2)
X
j=0
jn+d
ζqm
q−1
n −1
−
X
ζqm
jn
= 0.
j=0
We shall see in a moment that the summands in (2) are distin t in pairs. Taking
it as a fa t, there are 2(q − 1)/n (≤ q − 1) summands, and dividing ea h of them
by ζq gives us a linear dependen e among the 1, ζq , ζq2 , . . . , ζqq−2 in ontradi tion to
jn+d
in
= ζqm for some i, j = 0, 1, . . . , q−1
Lemma 1. Now, if ζqm
n − 1, j ≥ i, then
m(j−i)n+d ≡ 1( mod q). m is primitive modulo q so q − 1 divides (j − i)n + d. But,
(j − i)n + d < ( q−1
n − 1)n + n = q − 1, a ontradi tion. Therefore, all the summands
in (2) are distin t in pairs.
αn is not real: No summand in (1) is a omplex onjugate of the other. Indeed, if
jn
in
(j−i)n
ζqm = ζq−m for some i, j = 0, 1, . . . , q−1
≡ −1( mod q),
n − 1, j ≥ i, then m
2(j−i)n
≡ 1( mod q). Therefore, the odd number (q − 1)/n divides 2(j − i),
so m
thus divides j − i. But j − i < (q − 1)/n. We on lude that no summand in (1) is a
omplex onjugate of the other. Finally, if αn was real, then ζ1q (αn − αn ) = 0 and
by the same onsiderations above, we get a ontradi tion to Lemma 1.
Now by Theorem 6, we an embed the non-real Cn -extension Q(αn )/Q in a Fpn extension L/Q (say). Let Q(β)/Q be an intermediate extension of degree p whi h
orresponds to (the isomorphi opy of) U ∼
= Cn . No non-trivial subgroup of U is
normal in Fpn , hen e L/Q is the splitting eld of the minimal polynomial f of the
primitive element β . f is the required polynomial.
If a + ib is a omplex root of f then L = Q(a + ib, a − ib) = Q(a, ib) by Lemma
2 and Corollary 1.
Any Frobenius group an be realized as Galois group over Q (the
realizations are not ne essarily non-real). I.R.afarevi£ [13℄ proved that any solvable
group appears as Galois group over number elds, and J.Sonn [16,17℄ proved that
any non-solvable Frobenius group appears as Galois group over Q.
Remark 4.
4.
A knowledgment
The author is grateful to Moshe Roitman, Ja k Sonn, Tanush Shaska and John
Dixon for useful dis ussions.
Referen es
[1℄ T.M.Apostol, Introdu tion to Analyti Number Theory, Undergraduate Texts in Mathemati s, Springer-Verlag, 1976.
[2℄ A.Bialosto ki & T.Shaska, Computing the Galois group of prime degree polynomials with
nonreal roots, Le t. Notes in Computing, 13, 243-255, (2005).
[3℄ A.Bo hert, Über die Klasse der Transitiven Substitutionengruppen II. Math. Ann. 49, 133144, (1897).
[4℄ A.A.Bruen, C.U.Jensen & N.Yui, Polynomials with Frobenius Groups of Prime Degree as
Galois Groups II, Journal of Number Thoery 24, 305-359 (1986).
[5℄ W.Burnside, Theory of Groups of Finite Order, Cambridge: at the university press, 1897.
[6℄ D.A.Cox, Galois Theory, Wiley-Inters ien e, 2004.
[7℄ J.D.Dixon & B.Mortimor, Permutation Groups, Springer-Verlag, 1996.
[8℄ J.Hö hsmann, On minimal p-degrees in 2-transitive permutation groups, Ar hiv der Mathematik, 72, 405-417, (1999).
ON GALOIS GROUPS OF PRIME DEGREE POLYNOMIALS WITH COMPLEX ROOTS
7
[9℄ B.Huppert, Endli he Gruppen I, Grundlehren der mathematis hen Wissens haften 134,
Springer-Verlag, 1967.
[10℄ C.Jordan, Theoremes sur les groupes primitifs, J. Math. Pure Appl. 16, 383-408 (1871).
[11℄ M.W.Liebe k and J.Saxl, Minimal degrees of primitive permutation groups, with an appliation to monodromy groups of overs of Riemann surfa es, Pro . London Math. So . (3)
63, 266-314 (1991).
[12℄ J.Neukir h, Algebrai Number Theory, A Series of Comprehensive Studies in Mathemati s
322, Springer-Verlage, 1999.
[13℄ I.R.afarevi£, Constru tion of elds of algebrai numbers with given solvable Galois groups,
Amer. Math. So . Transl. Ser 2 4, 185-237, (1956).
[14℄ S.S holz, Über die Bildung algebraisher Zahlkörper mit auösbarer Galoiss he Gruppe,
Math.Z.30, 332-356, (1929).
[15℄ J.P.Serre, A Course in Arithmeti , Springer-Verlag, New York, 1987.
[16℄ J.Sonn, Frobenius Galois groups over quadrati elds, Israel J. Math. 31, 91-96 (1978).
[17℄ J.Sonn, SL(2, 5) and Frobenius Galois groups over Q, Can. J.Math 32 (2), 281-293 (1980).
[18℄ I.Stewart, Galois Theory, Third Edition, Chapman & Hall/CRC, 2004.
[19℄ H.Wielant, Finite permutation groups. A ademi Press, New York-London, 1964.
Oz Ben-Shimol
Department of Mathemati s
University of Haifa
Mount Carmel 31905, Haifa, Israel
E-mail Address: obenshimmath.haifa.a .il