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F
that respects the field
operations, i.e.,
˃
(
xy
)=
˃
(
x
)
˃
(
y
) and
˃
(
x
+
y
)=
˃
(
x
)+
˃
(
y
). The set of all field
automorphism of a field
F
forms a group denoted by Aut(
F
). Given a field extension
K
of
F
,thesubset of Aut(
K
) that leaves
F
unchanged is itself a group, called the
Galois group
of the extension, and is denoted
A
field automorphism
of the field
F
is a bijection
˃
:
F
ₒ
Gal(
K/F
)=
{
˃
∈
Aut(
K
)
|
˃
(
x
)=
x
for all
x
∈
F
}
.
The
splitting field
of a polynomial,
p
, with rational coefficients, denoted split(
p
) is the
smallest subfield of the complex numbers that contains all the roots of the polynomial.
Each automorphism in Gal(split(
p
)
/
Q
) permutes the roots of the polynomial, no two
automorphisms permute the roots in the same way, and these permutations form a
group, so Gal(split(
p
)
/
Q
) can be thought of as a permutation group.
Lemma 7.
If
ʱ
can be computed byaradicalcomputation tree and
K
is the splitting
field of an irreducible polynomial with
ʱ
asone of its roots, then
Gal(
K/
Q
)
does not
contain
S
n
as a subgroupforany
n
≥
5
.
Proof.
If
ʱ
is computable by a radical computation tree, it can be written as an expres-
sion using nested radicals. If
K
is the splitting field of an irreducible polynomial with
such an expression as a root, Gal(
K/
Q
) is a solvable group (Def. 8.1.1 of [19], p. 191
and Theorem 8.3.3 of [19], p. 204). But
S
n
is not solvable for
n
5 (Theorem 8.4.5
of [19], p. 213), and every subgroup of a solvable group is solvable (Proposition 8.1.3
of [19], p. 192). Thus, Gal(
K/
Q
) cannot contain
S
n
(
n
≥
≥
5)asasubgroup.
The next lemma allows us to infer properties of a Galois group from the coefficients
of a
monic
polynomial, that is, a polynomial with integer coefficients whose first co-
efficient is one. The
discriminant
of a monic polynomial is (uptosign) the product of
the squared differences of all pairs of its roots; it can also be computed as a polynomial
function of the coefficients. The lemma is due to Dedekind and proven in [19].
Lemma 8 (Dedekind's theorem).
Let
f
(
x
)
be an irreducible monic polynomialin
Z
[
x
]
and
p
a prime not dividing thediscriminantof
f
.If
f
(
x
)
factors into a product
of irreducibles of degrees
d
0
,d
1
,...d
r
over
Z
/p
Z
,then
Gal(split(
f
)
/
Q
)
contains a
permutation thatisthecomposition of disjointcycles of lengths
d
0
,d
1
,...,d
r
.
Apermutation groupis
transitive
if, for every two elements
x
and
y
of the elements
being permuted, the groupincludes a permutation that maps
x
to
y
.If
K
is the splitting
field of an irreducible polynomial of degree
n
,thenGal(
K/
Q
) (viewed as a permu-
tation group on the roots) is necessarily transitive. The next lemma allows ustouse
Dedekind's theorem to prove that Gal(
K/
Q
) equals
S
n
. It is a standard exercise in
abstract algebra (e.g., [20], Exercise 3, p. 305).
Lemma 9.
If a transitivesubgroup
G
of
S
n
contains a transpositionand an
(
n
−
1)
-
cycle, then
G
=
S
n
.
