Solutions for Field Theory Problem Set 5
A. Let β =
√
√
2
2
+
i.
2
2
Let K = Q(β). Find all subfields of K. Justify your answer carefully.
SOLUTION. All subfields of K must automatically contain Q. Thus, this problem concerns
the intermediate fields for the extension
√ K/Q. In a previous problem set, it was proved
that [K : Q] = 4 and that K = Q( 2, i). Furthermore, it was pointed out in class that
Gal(K/Q) ∼
= U(Z/8Z) and that this group is isomorphic to the Klein 4-group. Thus, every
element in Gal(K/Q) has order 2, except for the identity element. Therefore, Gal(K/Q) has
exactly three elements of order 2 and hence exactly three subgroups of order 2. Therefore,
by the Fundamental Theorem of Galois Theory, it follows that K contains exactly three
subfields L such that [K : L] = 2. Those are the subfields L of K such that
[L : Q] =
[K : Q]
4
=
= 2.
[K : L]
2
The three subfields
√
L1 = Q( 2),
L2 = Q(i),
√
√
L3 = Q( 2i) = Q( −2)
are distinct (see below) and hence must be all of the subfields of K of degree 2 over Q. In
addition, Gal(K/Q) obviously has exactly one subgroup of order 1 and exactly one subgroup
of order 4. The corresponding subfields of K are K itself and Q.
One could prove directly that the three fields L1 , L2 , and L3 are distinct. But we can
also use the following lemma which was proved in the solution for problem B in problem set
3. We will give the proof again.
√
√
Lemma: Suppose that d1 and d2 are nonzero integers. Then Q( d1 ) = Q( d2 ) if and
only if d1 d2 is the square of an integer.
Proof. One direction is straightforward. Suppose that d1 d2 = n2 , where
n ∈ Z. We are
√
assuming
that d1 and d2 are nonzero.
Any subfield of C containing d1 will√also contain
√
√
n( d1 )−1√and hence will contain d2 . Similarly, any subfield of C containing d2 will also
contain d1 . All subfields√of C contain
√ Q. Thus, assuming that d1 d2 is the square of an
integer, it follows that Q( d1 ) = Q( d2 ).
√
√
Now assume that Q( d1 ) = Q( d2 ). This field coincides with Q if and only if x2 − d1
and x2 − d2 are reducible over Q. This means that those polynomials have a root in Q. By
the rational root test, any rational root of those polynomials will have denominator dividing
1. Hence those roots will be integers and both d1 and d2 will be squares of integers. Thus,
their product will be a square of an integer.
√
√
We must now consider the case where Q( d1 ) = Q( d2 ) and this field is an extension
of Q of degree 2. Let K denote that field. Then K is the splitting field for the polynomial
x2 − d1 over Q and [K : Q] = 2. Hence Aut(K/Q)
is a group of order 2. Let σ denote √
its
√
2
nontrivial element. For d = d1√or d = d√
,
σ(
d)
must
be
a
root
of
x
−
d
different
from
d
2
itself. Hence we must have σ( d) = − d. Thus, we have
p
p
p
p
σ( d1 ) = − d1 ,
σ( d2 ) = − d2 .
√ √
Let η = d1 d2 . Then η ∈ K. Furthermore,
p p
p
p
p
p
p p
σ(η) = σ( d1 d2 ) = σ( d1 )σ( d2 ) = (− d1 )(− d2 ) = d1 d2 = η .
We will show that η ∈ Q. To see this, assume to the contrary that η 6∈ Q. Then Q(η) is a
subfield of K and [Q(η) : Q] > 1. It would follow that [Q(η) : Q] = 2 and that Q(η) = K.
Since σ is determined by σ(η) and σ(η) = η, it follows that σ is the identity map. This
contradicts the fact that σ is the nontrivial element in Aut(K/Q). It follows that η ∈ Q.
Note that η 2 = d1 d2 . Thus η is a root of the polynomial x2 − d1 d2 . By the rational root
test,
of η divides 1. Thus, η is actually an integer. Hence, assuming that
√
√ the denominator
Q( d1 ) = Q( d2 ), it follows that d1 d2 is the square of an integer. We have proved the
lemma.
The lemma tells us that L1 , L2 , and L3 are distinct because 2 · (−1), 2 · (−2), and
(−1) · (−2) are not squares in Z.
B. Let f (x) = x4 + x3 + x2 + x + 1. Let ω be a root of the polynomial f (x) in C. Let
K = Q(ω). Find all subfields of K. Justify your answer carefully.
SOLUTION. We proved in class that [K : Q] = 4 and that Gal(K/Q) is a cyclic group
of order 4. Thus, Gal(K/Q) has exactly three subgroups, namely the trivial subgroup, a
subgroup of order 2, and Gal(K/Q) itself. Therefore, by the Fundamental Theorem of Galois
Theory, there are exactly three distinct intermediate fields for the extension K/Q. Every
subfield of K will be one of those three fields.
√
We also showed in class that K contains L = Q( 5). Some other subfields of K are K
itself and Q. These subfields of K, namely the fields Q, L, and K, are clearly distinct and
hence must be the three distinct subfields of K.
C. Let K = Q(ω), where ω = cos( 2π
) + sin( 2π
)i. Prove that K contains a unique subfield
17
17
L such that [L : Q] = 8. Prove that L is a Galois extension of Q. Find an element β ∈ L
such that L = Q(β).
SOLUTION. We proved in class that there is a group isomorphism
Gal(K/Q) ∼
= U(Z 17Z) .
However, one can check easily that U(Z 17Z) is a cyclic group of order 16. (To see this,
verify that 3 + 17Z is an element of U(Z 17Z) of order 16.)
A cyclic group of order 16 will have a unique subgroup H of order 2. Thus, there will be
a unique intermediate field L for K/Q such that [K : L] = 2. Thus,
[L : Q] =
16
[K : Q]
=
= 8.
[K : L]
2
Thus, it is clear that L is the unique intermediate field with [L : Q] = 8. As proven in class,
any subfield of K contains Q and hence is an intermediate field for the extension K/Q.
Note that Gal(K/Q) is abelian and hence every subgroup of Gal(K/Q) is a normal
subgroup of Gal(K/Q). In particular, Gal(K/L) is a normal subgroup of Gal(K/Q). As
discussed in class, it follows that L/Q is a Galois extension.
Finally, consider β = ω + ω −1 = 2cos(2π/17). Then β ∈ K. Let M = Q(β). Thus, M
is a subfield of K. Also, β ∈ R and hence M ⊂ R. Since K 6⊂ R, we have M 6= K. Hence
[K : M] ≥ 2. Now ω is a root of the following polynomial
f (x) = (x − ω)(x − ω −1 ) = x2 − (ω + ω −1 )x + ωω −1 = x2 − βx + 1
Note that f (x) ∈ M[x] and that K = M(ω). It follows that [K : M] ≤ 2. Since we also
have [K : M] ≥ 2, it follows that [K : M] = 2. Therefore, M is a subfield of K and
[M : Q] = [K : Q] [K : M] = 16/2 = 8. Since L is the unique subfield of K which
has
degree 8 over Q, we must have M = L. Thus, we have L = Q(β) = Q cos(2π/17) .
D. Suppose that K is a finite Galois extension of Q and that Gal(K/Q) ∼
= S4 . Prove that
there exists a polynomial g(x) ∈ Q[x] such that g(x) has degree 4 and K is the splitting
field for g(x) over Q.
SOLUTION. We are given that K is a finite, Galois extension of Q. Let G = Gal(K/Q).
By assumption, we have G ∼
= S4 . Now S4 contains a subgroup of order 6, namely
{ g ∈ S4 g(4) = 4 } ,
a subgroup of S4 which is isomorphic to S3 . It follows that G has a subgroup H such that
H∼
= S3 . We let L denote K H . Note that
[L : Q] = [G : H] =
|G|
24
=
=4 .
|H|
6
Thus, by the Primitive Element Theorem, L = Q(β) for some β ∈ C. Since [L : Q] = 4, it
follows that β has degree 4 over Q. Let m(x) be the minimal polynomial for β over Q.
We have m(x) ∈ Q[x] and deg m(x) = 4. Let M denote the splitting field for m(x)
over Q. Since K is a Galois extension of Q and β ∈ K, it follows that all of the complex
roots of m(x) are also in K. We proved this in class. Furthermore, it follows that M ⊂ K.
Also, L ⊆ M. Thus, we have L ⊆ M ⊆ K.
We have Gal(K/L) = H ∼
= S3 . Now [K : L] = |Gal(K/L)| = 6. Therefore, we have
[K : L] = 6. By the degree formula, we have [K : M][M : L] = [K : L] = 6. Thus, [K : M]
divides 6.
Since M is a Galois extension of Q, it follows that Gal(K/M) is a normal subgroup
of G = Gal(K/Q). But |Gal(K/M)| = [K : M] divides 6. Recall that S4 has normal
subgroups only of orders 24, 12, 4, and 1. Since G ∼
= S4 , the same statement is true for G.
Since Gal(K/M) is a normal subgroup of G and its order divides 6, the only possibility is
that |Gal(K/M)| = 1. Hence Gal(K/M) = {idG }. This means that M = K. That is, the
splitting field over Q for m(x) is indeed the field K.
E. Suppose that √
K is a finite Galois extension of Q and that Gal(K/Q) ∼
= S5 . Show that
5
K cannot contain 2.
√
5
2 ∈ K. We know that the minimal polynomial
SOLUTION.
Assume
to
the
contrary
that
√
for 5 2 over Q is m(x) = x5 − 2. As explained in class, since K is a Galois extension of Q
which contains one root of m(x), it follows that K contains all of the roots of m(x).
Let F be the splitting field for x5 − 2 over Q. The above remark shows that F ⊆ K. We
also know that [F : Q] = 20. This was proved in a previous problem set. Furthermore, since
F/Q is a Galois extension, it follows that there is a surjective group homomorphism
r : Gal(K/Q) −→ Gal(F/Q)
and that Ker(r) = Gal(K/F ) is a normal subgroup of Gal(K/Q) and has order equal to
[K : F ] = [K : Q] [F : Q] = 120/20 = 6.
We obtain a contradiction because Gal(K/Q) ∼
= S5 and S5 has no normal subgroups of
order 6.
F. Suppose that r ∈ Q. Let β = cos(rπ). Prove that β is algebraic over Q. Let K = Q(β).
Prove that Q(β) is a Galois extension of Q and that Gal(K/Q) is an abelian group.
Let n be the denominator of that rational number r/2. Then r = 2k/n, where k is an integer.
Let ω = cos(2π/n) + sin(2π/n)i. Thus, ω is a root of the polynomial xn − 1. Therefore,
Q(ω) is a finite extension of Q. Let M = Q(ω). In fact, M contains all of the roots of xn − 1
and hence M is the splitting field over Q for xn − 1. Thus, M is a finite, Galois extension
of Q. Note that M contains
ω k + ω −k = cos(2kπ/n) + sin(2kπ/n)i + cos(2kπ/n) − sin(2kπ/n)i
= 2cos(2kπ/n) = 2cos(rπ) = 2β
and therefore β ∈ M.
Since M is a finite extension of Q and β ∈ M, it follows that β is algebraic over Q.
Thus, Q ⊆ K ⊆ M. Recall that Gal(M/Q) is abelian. (See problem C in problem set 3.) It
follows that every subgroup of Gal(M/Q) is a normal subgroup. In particular, Gal(M/K)
is a normal subgroup of Gal(M/Q). This implies that K is a Galois extension of Q.
Gal(M/K). Since Gal(M/Q) is an
Furthermore, we have Gal(K/Q) ∼
= Gal(M/Q)
abelian group, the quotient group Gal(M/Q) Gal(M/K) must also be abelian. Hence
Gal(K/Q) is indeed an abelian group.
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