On Galois groups of prime degree polynomials with complex roots
Let f be an irreducible polynomial of prime degree p≥5 over Q, with precisely k pairs of complex roots. Using a result of Jens Hochsmann (1999), show that if p≥4k+1 then Gal(f/Q) is isomorphic to Ap or Sp. This improves the algorithm for computing the Galois group of an irreducible polynomial of p...
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| Цитувати: | On Galois groups of prime degree polynomials with complex roots / Oz Ben-Shimol // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 2. — С. 99–107. — Бібліогр.: 19 назв. — англ. |
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irk-123456789-1546102019-06-16T01:28:12Z On Galois groups of prime degree polynomials with complex roots Oz Ben-Shimol Let f be an irreducible polynomial of prime degree p≥5 over Q, with precisely k pairs of complex roots. Using a result of Jens Hochsmann (1999), show that if p≥4k+1 then Gal(f/Q) is isomorphic to Ap or Sp. This improves the algorithm for computing the Galois group of an irreducible polynomial of prime degree, introduced by A. Bialostocki and T. Shaska. If such a polynomial f is solvable by radicals then its Galois group is a Frobenius group of degree p. Conversely, any Frobenius group of degree p and of even order, can be realized as the Galois group of an irreducible polynomial of degree p over Q having complex roots. 2009 Article On Galois groups of prime degree polynomials with complex roots / Oz Ben-Shimol // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 2. — С. 99–107. — Бібліогр.: 19 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:20B35; 12F12. http://dspace.nbuv.gov.ua/handle/123456789/154610 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let f be an irreducible polynomial of prime degree p≥5 over Q, with precisely k pairs of complex roots. Using a result of Jens Hochsmann (1999), show that if p≥4k+1 then Gal(f/Q) is isomorphic to Ap or Sp. This improves the algorithm for computing the Galois group of an irreducible polynomial of prime degree, introduced by A. Bialostocki and T. Shaska.
If such a polynomial f is solvable by radicals then its Galois group is a Frobenius group of degree p. Conversely, any Frobenius group of degree p and of even order, can be realized as the Galois group of an irreducible polynomial of degree p over Q having complex roots. |
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Oz Ben-Shimol |
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Oz Ben-Shimol On Galois groups of prime degree polynomials with complex roots Algebra and Discrete Mathematics |
| author_facet |
Oz Ben-Shimol |
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Oz Ben-Shimol |
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On Galois groups of prime degree polynomials with complex roots |
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On Galois groups of prime degree polynomials with complex roots |
| title_full |
On Galois groups of prime degree polynomials with complex roots |
| title_fullStr |
On Galois groups of prime degree polynomials with complex roots |
| title_full_unstemmed |
On Galois groups of prime degree polynomials with complex roots |
| title_sort |
on galois groups of prime degree polynomials with complex roots |
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Інститут прикладної математики і механіки НАН України |
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2009 |
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http://dspace.nbuv.gov.ua/handle/123456789/154610 |
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On Galois groups of prime degree polynomials with complex roots / Oz Ben-Shimol // Algebra and Discrete Mathematics. — 2009. — Vol. 8, № 2. — С. 99–107. — Бібліогр.: 19 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT ozbenshimol ongaloisgroupsofprimedegreepolynomialswithcomplexroots |
| first_indexed |
2025-07-14T06:13:37Z |
| last_indexed |
2025-07-14T06:13:37Z |
| _version_ |
1837601776316973056 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2009). pp. 99 – 107
c⃝ Journal “Algebra and Discrete Mathematics”
On Galois groups of prime degree polynomials
with complex roots
Oz Ben-Shimol
Communicated by V. I. Sushchansky
Abstract. Let f be an irreducible polynomial of prime degree
p ≥ 5 over ℚ, with precisely k pairs of complex roots. Using a
result of Jens Höchsmann (1999), we show that if p ≥ 4k + 1 then
Gal(f/ℚ) is isomorphic to Ap or Sp. This improves the algorithm
for computing the Galois group of an irreducible polynomial of
prime degree, introduced by A. Bialostocki and T. Shaska.
If such a polynomial f is solvable by radicals then its Galois
group is a Frobenius group of degree p. Conversely, any Frobenius
group of degree p and of even order, can be realized as the Ga-
lois group of an irreducible polynomial of degree p over ℚ having
complex roots.
1. Introduction
A classical theorem in Galois theory says that an irreducible polynomial
f of prime degree p ≥ 5 over ℚ which has precisely one pair of complex
(i.e., non-real) roots, has the symmetric 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 irreducible polynomial of
prime degree p with precisely k pairs of complex roots. What is its Ga-
lois group Gal(f/ℚ)?. If one tries to imitate the proof of the classical
theorem (i.e., the case k = 1), one would find, constructively, the sub-
group of Sp which is generated by the p-cycle (1 2 ... p) and an involution
(a1 a2) ⋅ ⋅ ⋅ (a2k−1 a2k). My unsuccessful attempts (so far) to solve the
problem in this way indicated that the difference between the degree p
2000 Mathematics Subject Classification: 20B35; 12F12.
100 On Galois groups of prime degree polynomials
and the number 2k of the complex roots, need not be "large" in order to
obtain the alternating group Ap at least (i.e., Gal(f/ℚ) is isomorphic to
Ap or Sp).
More general observations on such 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 which does not contain the
alternating group An, and let m be its minimal degree. Find the infimum
for m in terms of n.
If f is an irreducible polynomial of prime degree p with k > 0 pairs
of complex 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) necessarily
contains the alternating group Ap when 2k ≤ B(p). Thus, as B(p) ap-
proaches the infimum, the difference p− 2k gets smaller, as required.
Returning to the group-theoretic problem stated above (for degree n,
not necessarily 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 Bochert [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 Bochert estimates can be found also in Dixon & Mortimor [7],
Theorem 3.3D and Theorem 5.4A, respectively. More recently, Liebeck
and Saxl [11], using the classification of finite simple groups, have proved
B(n) = n/3.
Finally, Höchsmann [8], using a concept suggested by W.Knapp which
refines 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, which in the worst case meet Liebeck and
Saxl’s bounds. Since the group we are dealing with is of prime degree,
and we have information about its 2-minimal degree, Hochmann’s result
serves us better than that of Liebeck and Saxl.
The paper of A.Bialostocki and T.Shaska [2] focuses on the practical
aspects of this theoretical problem, in the process of computing the Galois
group of prime degree polynomials over ℚ: 1. The existing techniques,
which 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 process.
2. Polynomials in general have plenty of complex roots. 3. Checking
whether a polynomial has complex roots is very efficient since numerical
methods can be used. Therefore, checking first if the polynomial has
complex roots, and then use a "good" bound for the difference between
the polynomial’s degree and the number of its complex roots, makes the
computation of its Galois group much easier. However, they make a use
of estimate due to Jordan (summarized in Wielandt [19, page 42]), which
O. Ben-Shimol 101
is not sharp at all (as the authors point out in their paper). In fact,
Jordan’s bound holds for any primitive group of any finite degree - not
necessarily 2-transitive of prime degree. In the present paper, we improve
their algorithm and discuss some theoretical aspects of the subject.
2. Galois groups of prime degree polynomials with com-
plex roots
A Frobenius group is a transitive permutation group which is not regular,
but in which only the identity has more then one fixed point. In other
words, a Frobenius group G is a transitive permutation group on a set
Ω in which G� ∕= 1 for some � ∈ Ω, but G� ∩ G� = 1 for all �, � ∈ Ω,
� ∕= �. It can be shown that the set of elements fixing no letters of Ω,
together with the identity, form a normal subgroup K called the Frobenius
kernel of G. Frobenius groups are characterized as non-trivial semi-direct
products G = K⋊H such that no element of H ∖{1} commutes with any
element of K∖{1}. Basic examples of Frobenius groups are the subgroups
of AGL1(F ) - the group of the 1-dimensional affine transformations of a
field F , i.e. the group consisting of the permutations of the form t�,� :
� 7→ �� + �, � ∈ F ∗, �, � ∈ 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
particular, if F = Fp - the field of p elements (p prime), then AGL1(p) :=
AGL1(Fp) ∼= Fp ⋊ U , where U is a subgroup of F∗
p (so U is a cyclic of
order n, where n ∕= 1 and n divides p− 1), is a Frobenius group of degree
p. The structure of a Frobenius group of degree p ≥ 5 is described in the
following theorem.
Theorem 2.1. (Galois) Let G be a transitive permutation group of prime
degree p ≥ 5, and of order > p. Then the following statements are
equivalent:
i. G has a unique p-Sylow subgroup.
ii. G is a solvable group.
iii. G is isomorphic to a subgroup of AGL1(p).
iv. G is a Frobenius group of degree p.
Proof. See Huppert [9].
Let G ∼= Fp ⋊ U , U cyclic of order n, n ∕= 1, n∣p− 1, be a Frobenius
group of degree p. Then it is customary 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
Xp − a ∈ ℚ[X], where a ∈ ℚ∗ ∖ (ℚ∗)p. For constructive realization of
102 On Galois groups of prime degree polynomials
Frobenius groups of degree p, see A.A.Bruen, C.Jensen and N.Yui [4].
If f is an irreducible polynomial of degree p ≥ 5 over ℚ, then its
Galois group G = Gal(f/ℚ), as a permutation group acting on the p-set
consisting of the p roots of f , is a transitive group of order p (if and only
if G contains a p-cycle). Complex conjugation is a ℚ-automorphism of
ℂ and, therefore, induces a ℚ-automorphism of the splitting field of f .
This leaves the real roots of f fixed, while transposing the complex roots.
Therefore, if f has a pair of complex roots, then ∣G∣ > p. Furthermore, if,
in addition, f has more then one real root, then the complex conjugation
has more then one fixed point. In particular, G is not a Frobenius group
of degree p. By Theorem 1, G is not solvable, thus, f is not solvable by
radicals. So we have
Corollary 2.2. Let f be an irreducible polynomial of prime degree p ≥ 5
over ℚ, which has a pair of complex roots. If f is solvable by radicals
then Gal(f/ℚ) is a Frobenius group of degree p, and f has exactly one
real root.
Let f be an irreducible polynomial of prime degree p ≥ 5 and with
k > 0 pairs of complex roots. By Corollary 2.2, if p > 2k + 1 then
G = Gal(f/ℚ) is not solvable. Our purpose is to show that if p ≥ 4k+1
then G contains the alternating group (i.e., G isomorphic to Ap or to Sp).
Theorem 2.3. (Burnside) A non-solvable transitive permutation group
of prime degree is 2-transitive.
Therefore, a transitive permutation group of prime degree is either
2-transitive or a Frobenius group (see Theorem 2.1).
Proof. See, Burnside [5], or Dixon & Mortimor [7, Corollary 3.5B].
Recall that the minimal degree m(G) of a permutation group G acting
on a set Ω is the minimum of the supports of the non-identity elements:
m(G) := min{∣ supp(x)∣ : x ∈ G, x ∕= 1}. Hence, 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 define 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 classification of the finite simple groups, J.
Höchsmann [8] has proved
O. Ben-Shimol 103
Theorem 2.4. (Höchsmann) Let G be a 2-transitive group of degree n
which does not contain the alternating group, and let r be a prime divisor
of ∣G∣. Then
i. mr(G) ≥ r−1
r ⋅ n or
ii. G ≥ PSL(2, 2m), r = 2m − 1 ≥ 7 is a Mersenne prime and mr(G) =
r = n− 2 or
iii. G = PSp(2m, 2), n = 2m−1 ⋅ (2m − 1) with m > 2, r = 2 and
mr(G) = 2m−1
−1
2m−1 ⋅ n ≥ 3
7 ⋅ n.
In any case mr(G) ≥ r−1
r+1 ⋅ n.
An immediate consequence (in fact, a special case) of this theorem is
Corollary 2.5. Let G be a 2-transitive group of prime degree p which
does not contain the alternating group. Then m2(G) ≥ p
2 .
Theorem 2.6. Let f be an irreducible polynomial of prime degree p ≥ 5
over ℚ. Suppose that f has precisely k > 0 pairs of complex roots. If
p ≥ 4k + 1 then G = Gal(f/ℚ) is isomorphic to Ap or to Sp. Clearly, if
k is odd then G ∼= Sp.
Proof. Complex conjugation has support 2k, hence m2(G) ≤ 2k. By
Corollary 2.2, G is not solvable (f has more than one real root). By The-
orem 2.4, G is 2-transitive and, by Corollary 2.5, G necessarily contains
the alternating group.
Therefore, the algorithm given in [2] for computing the Galois group
of an irreducible prime degree polynomial, can be improved:
Input: An irreducible polynomial f(x) ∈ ℚ[x] of prime degree p.
Output: The Galois group Gal(f/ℚ).
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 Δ(f) is a complete square then
Gal(f/ℚ) = Ap;
else
Gal(f/ℚ) = Sp;
endif;
endif;
else
104 On Galois groups of prime degree polynomials
ReductionMethod(f(x));
endif
end;
Remark 2.7. Δ(f) denotes the discriminant of f(x). It is well known
that if f is a polynomial of degree n with coefficients in a field K, char(K)∕=
2, then Δ(f) is a perfect square in K if and only if Gal(f/K) is isomor-
phic to a subgroup of An. See e.g., Stewart [18, Theorem 22.7].
Remark 2.8. A short discussion on the reduction modulo p method, can
be found in [2] and in Cox [6, page 401].
Remark 2.9. Corollary 1 in [2] can also be improved: (replace their r
with our k - the number of pairs of the complex roots of a given irreducible
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.
3. Non-real realization of Fpn
As stated in Corollary 2.2, an irreducible solvable polynomial of prime
degree p ≥ 5 over ℚ, which has complex roots, has a Frobenius group of
degree p (and of even order, of course) 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 irreducible polynomial of degree p over ℚ having complex roots.
Theorem 3.1. (Dirichlet) Let k,ℎ be integers such that k > 0 and
(ℎ, k) = 1. Then there are infinitely many primes in the arithmetic
progression nk + ℎ, n = 0, 1, 2, . . ..
Proof. See e.g., Serre [15] or Apostol [1].
Lemma 3.2. Let l be a positive integer, and let � be a primitive l-th
root of unity. Then 1, �, . . . , �'(l)−1 form a ℤ-basis for the ring of integers
of ℚ(�).
Proof. See e.g., Neukirch [12, Chapter I, Proposition 10.2].
Lemma 3.3. (Galois) Let f be a polynomial of prime degree over ℚ.
Then, f is solvable by radicals if and only if any two distinct roots of f
generate its splitting field.
Proof. See Cox [6, Theorem 14.1.1].
O. Ben-Shimol 105
Theorem 3.4. (Scholz) A splitting embedding problem has a proper
solution over number fields. (That is, let K be a number field and let
M/K be a Galois extension with Galois group H. Suppose that H acts
on an abelian group A. Then, there exist a Galois extension L/K which
contains M/K such that Gal(L/K) ∼= A⋊H).
Proof. See Scholz [14].
Theorem 3.5. Let Fpn be a Frobenius group of degree p and of even
order. Then Fpn occurs as Galois group of an irreducible polynomial f
of degree p over ℚ having complex roots. Furthermore, the splitting field
of f is ℚ(a, ib) for every complex root a+ ib of f .
Proof. We describe an explicit non-real Cn-extensions over ℚ. By Theo-
rem 3.1, there exist a prime q such 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 since n is even. Thus, such a prime
q does exist. Let m be a primitive root modulo q (that is, a generator of
F∗
q). Consider the sum
�n = �q + �m
n
q + �m
2n
q + . . .+ �m
( q−1
n −1)n
q , (1)
where �q is a primitive q-th root of unity. Then Gal(ℚ(�q)/ℚ) is cyclic of
order q − 1 and generated by the automorphism � : �q 7→ �mq . We shall
see that ℚ(�n)/ℚ is a non-real Cn-extension, and then we shall apply
Theorem 3.4.
ℚ(�n)/ℚ is a Cn-extension: By the Fundamental Theorem of Galois
Theory, it is enough to prove ℚ(�n) = ℚ(�q)
�n
. The inclusion ℚ(�n) ⊆
ℚ(�q)
�n
is because �n moves cyclically the summands of (1) (in fact,
�n is the image of �q under the trace map Trℚ(�q)/ℚ(�q)�
n , hence �n is
an element of ℚ(�q)
�n
). Suppose that ℚ(�n) ⫋ ℚ(�n)
�n . There exist a
proper divisor d of n such that ℚ(�n) = ℚ(�q)
�d
. In particular, �d(�n) =
�n, or
q−1
n
−1∑
j=0
�m
jn+d
q −
q−1
n
−1∑
j=0
�m
jn
q = 0. (2)
We shall see in a moment that the summands in (2) are distinct in
pairs. Taking it as a fact, there are 2(q − 1)/n (≤ q − 1) summands,
and dividing each of them by �q gives us a linear dependence among the
1, �q, �
2
q , . . . , �
q−2
q in contradiction to Lemma 1. Now, if �m
jn+d
q = �m
in
q
for some i, j = 0, 1, . . . , q−1
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 <
106 On Galois groups of prime degree polynomials
( q−1
n − 1)n+ n = q − 1, a contradiction. Therefore, all the summands in
(2) are distinct in pairs.
�n is not real: No summand in (1) is a complex conjugate of the other.
Indeed, if �m
jn
q = �−min
q for some i, j = 0, 1, . . . , q−1
n − 1, j ≥ i, then
m(j−i)n ≡ −1( mod q), so m2(j−i)n ≡ 1( mod q). Therefore, the odd
number (q−1)/n divides 2(j−i), thus divides j−i. But j−i < (q−1)/n.
We conclude that no summand in (1) is a complex conjugate of the other.
Finally, if �n was real, then 1
�q
(�n −�n) = 0 and by the same considera-
tions above, we get a contradiction to Lemma 3.2.
Now by Theorem 3.4, we can embed the non-real Cn-extension ℚ(�n)/ℚ
in a Fpn-extension L/ℚ (say). Let ℚ(�)/ℚ be an intermediate extension
of degree p which corresponds to (the isomorphic copy of) U ∼= Cn. No
non-trivial subgroup of U is normal in Fpn, hence L/ℚ is the splitting
field of the minimal polynomial f of the primitive element �. f is the
required polynomial.
If a+ ib is a complex root of f then L = ℚ(a+ ib, a− ib) = ℚ(a, ib)
by Lemma 3.3 and Corollary 2.2.
Remark 3.6. Any Frobenius group can be realized as Galois group over
ℚ (the realizations are not necessarily non-real). I.R.Šafarevič [13] proved
that any solvable group appears as Galois group over number fields, and
J.Sonn [16,17] proved that any non-solvable Frobenius group appears as
Galois group over ℚ.
Acknowledgment
The author is grateful to Moshe Roitman, Jack Sonn, Tanush Shaska and
John Dixon for useful discussions.
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[19] H.Wielant, Finite permutation groups. Academic Press, New York-London, 1964.
Contact information
O. Ben-Shimol Department of Mathematics,
University of Haifa,
Mount Carmel 31905, Haifa, Israel
E-Mail: obenshim@math.haifa.ac.il
Received by the editors: 06.09.2008
and in final form 15.04.2009.
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