A GALOIS THEORY EXAMPLE
BRIAN OSSERMAN
Let F be the splitting field of x4 − 2 over Q. This is normal because it is a splitting field, and
separable because we are in characteristic 0. We compute the Galois group and all intermediate
subfields. First let α be the positive, real 4th root of 2, and next observe that the roots of x4 − 2
are precisely α, −α, iα, −iα. It follows that F = Q(α, i), and we also easily see that x4 − 2 is
irreducible over Q, since no root is in Q, and no product of two roots is in Q. Thus, [Q(α) : Q] = 4.
Furthermore, because Q(α) ⊆ R, it does not contain i, so since i is a root of quadratic polynomial
(over Q and also over Q(α)), the field Q(α, i) must have degee 2 over Q(α). We conclude that
[F : Q] = 8, so Gal(F/Q) has order 8.
Now, every element of Gal(F/Q) must permute the roots of x4 − 2. Furthermore, since F is the
splitting field of this polynomial, it is by definition generated by its roots, so we have that Gal(F/Q)
is a subgroup of the group of permutations of the roots of x4 − 2. If we label the roots in the order
α, −α, iα, −iα, we can consider Gal(F/Q) ⊆ S4 . Complex conjugation holds α and −α fixed, and
permutes iα and −iα, so under our labeling it corresponds to the transposition g = (3, 4). Next,
we observe that α and i are independent generators, in the sense that the relations they satisfy
are generated by α4 − 2 = 0 and i2 + 1 = 0. Thus, we can map α to any other root of x2 − 2
and i to itself or −i, and we will obtain a unique automorphism of F . In particular, there is an
automorphism of F which holds i fixed, and maps α to iα. Under our labeling, this corresponds
to the permutation h = (1, 3, 2, 4). Now, it is easily verified that ghg = h−1 , so g, h generate the
dihedral group of order 8, and this must be the Galois group.
The nontrivial subgroups are then the cyclic subgroups generated by h, by h2 , by g, by gh, by
2
gh , and by gh3 , as well as the subgroups (each isomorphic to Z/2Z × Z/2Z) generated by gh, h2
and by g, h2 . The lattice of subgroups is as follows:
G = hg, hi
g, h2
r
rrr
r
r
rrr
r
MMM
MMM
MMM
M
hhi
gh, h2
KK
II
y
KK
II
rrr
yy
r
K
II
y
r
K
r
KK
II
r
yy
r
y
K
r
I
K
y
r
y
2 r
3
2
hghi
h
gh
hgi TTT
gh
i
TTTT
ii
q
i
i
q
i
TTTT LLLLL
q
q
iiii
TTTT LL
qqqiiiiiii
q
TTTTLLL
q
i
TTTL
qq ii
iqiii
hei
The subgroup generated by h has order 4, so the corresponding fixed field must have degree
8/4 = 2 over Q, and must contain Q(i), so it is equal to Q(i). Next, h2 sends α to −α and iα to
−iα, so it fixes α2 (the positive square root
√ of 2), and since the fixed field must have degree 8/2 = 4
over Q, we see that the fixed field is Q(i, 2). Now, g fixes α, and the fixed field has degree 8/2 = 4
over Q, so it must be Q(α).
Next, gh has order 2, and sends α to −iα while sending i to −i. The fixed field has degree 8/2 = 4
over Q, and contains α − iα. One checks that this is not the root of any quadratic polynomial, so
1
it generates a quartic extension which is necessarily the fixed field. Similarly, the fixed field of the
group generated by gh3 is generated by α + iα. The fixed field of the group generated by gh2 is
generated by iα.
We then have that the fixed field of the group generated
by g, h2 is the intersection of the fixed
√
2
The intersection of these is
fields
√ for g and for h , which were Q(α) and Q(i, 2), respectively.
2 is the intersection of Q(α − iα) with
Q( √
2). Similarly, the fixed field of group
generated
by
gh,
h
√
√
√
2 =
Q(i, 2), and noting that (α − iα)√
2(1 − 2i − 1) = 2i 2, we see that i 2 is in both fields, so
the desired fixed field must be Q(i 2). We thus obtain the lattice of subfields as follows (where we
have mirrored everything from top to bottom in comparing to the lattice of subgroups):
F = Q(α, i) W
Q(α)
OOO WWWW
OOO WWWWWW
WWWWW
OOO
WWWWW
OO
W
√
ii q
iiiiqqqqq
i
i
i
ii
qq
iiii
qqq
iiii
Q(iα)
GG
GG
GG
GG
G
√
Q( 2)
2)
Q(i,
Q(i)
MMM
MMM
MMM
MMM
Q
Q(α − iα)
NNN
NNN
NNN
NN
r
rrr
r
r
r
rrr
oo
ooo
o
o
ooo
ooo
2
Q(α + iα)
qq
qqq
q
q
qqq
√ q
Q(i 2)