MA 3419: Galois theory
Selected answers/solutions to the assignment due December 1, 2015
1. (a) Note that this polynomial is irreducible (Eisenstein for p = 2). Also, we note that
f(100) > 0, f(1) = −1 < 0, f(−1) = 5 > 0, f(−100) < 0, so this polynomial has at least three
real roots, and f 0 (x) = 5x4 − 4, so this polynomial has two extrema, which means that it
cannot have more than three roots. Thus, two of the roots are complex conjugate, and hence
the Galois group contains a transposition (induced by the complex conjugation). We know
from class that a transitive subgroup of S5 (transitivity follows from irreducibility) containing
a transposition must coincide with S5 .
(b) This polynomial is irreducible (Eisenstein for p = 2 again), and f(100) > 0,
f(1) = −1 < 0, f(−100) > 0, so there are at least two real roots. Also, f 0 (x) = 4x3 − 4, so
the only extremal point is at x = 1. This means that this polynomial cannot have more than
two roots. Thus, two of the root are complex conjugate, and the Galois group contains a
transposition.
Let a1 , a2 , a3 , a4 be the roots of this polynomial; consider, as we discussed in class in the
beginning of this semester, the quantities x1 , x2 , x3 determined by
2a1 = x1 + x2 + x3 ,
2a2 = x1 − x2 − x3 ,
2a3 = −x1 + x2 − x3 ,
2a4 = −x1 − x2 + x3 .
We have
x21 + x22 + x23 = 0,
x1 x2 x3 = −4,
x21 x22
+ x21 x23 + x22 x23 = −8.
Thus, x21 , x22 , x23 are roots of the polynomial
t3 − 8t − 16 = 0.
This polynomial is irreducible
over rational numbers because it has no rational roots. Also, the
p
extremal points are ± 8/3, and by direct inspection, the values at both of these are negative,
so this polynomial has just one real root, and complex conjugation is an automorphism. This
immediately implies that the Galois group is S3 . The field Q(x21 , x22 , x23 ) is a subfield of the
splitting field, and it is a normal extension of Q being a splitting field itself, so its Galois
group is a quotient of the Galois group of the original equation. Thus, the Galois group is
a subgroup of S4 admitting S3 as a quotient; it must have 6, 12, or 24 elements. A group
of 6 elements must be isomorphic to S3 and as such cannot be transitive on four elements.
A group of 12 elements is necessarily isomorphic to A4 and cannot contain a transposition.
Hence, the group is isomorphic to S4 .
2. Without loss of generality, the cycle is (123). By transitivity, each element 1,2,3,4,5
is involved in at least one 3-cycle, so there is another 3-cycle involving at least one other
element. Without loss of generality, that cycle is (124) (in which case there is also a cycle
involving 5) or (145). In the former case, conjugating (124) with (123), we also obtain the
cycles (234) and (314), and by taking their inverses we obtain all 3-cycles in A4 . Since
A4 is generated by 3-cycles, we get the whole of A4 and then subsequently the whole of
A5 by conjugating the cycle involving 5. In the other case, if our group contains (123)
and (145), conjugating these cycles by each other and their inverses (132) and (154), we
obtain the cycles (234) = (145)(123)(154), (235) = (154)(123)(145), (245) = (123)(145)(132),
(345) = (132)(145)(123), and further conjugating we obtain (124) = (132)(234)(123),
(125) = (253)(132)(235), (134) = (234)(123)(243), (135) = (235)(123)(253), all 3-cycles in
A5 .
3. (a) The multiplicative group of a finite field is cyclic; any generator of that group can
be taken as a primitive element.
∼ F2 [τ], where τ2 + τ + 1 = 0. The Galois group takes τ to 1 + τ. Thus,
(b) We have F4 =
τ, 1 + τ can be taken as a normal basis.
∼ F2 [λ], where λ3 + λ + 1 = 0. Note that over F2 we have
We have F8 =
(λ2 )3 + λ2 + 1 = λ6 + λ2 + 1 = (λ + 1)2 + λ2 + 1 = 0,
so λ 7→ λ2 is an automorphism, and its square which sends λ to λ4 = λ2 + λ is another one.
It easily follows that 1 + λ, 1 + λ2 , 1 + λ + λ2 can be taken as a normal basis.
∼ F2 [β], where β4 + β + 1 = 0. Note that over F2 we have
We have F16 =
(β2 )4 + β2 + 1 = β8 + β2 + 1 = (β + 1)2 + β2 + 1 = 0,
so β 7→ β2 is an automorphism, its square which sends β to β4 = β + 1 is another one, and
its cube that sends β to β8 = β2 + 1 is another one. We see that the orbits of 1, β, β2 do
not contain β3 , so cannot form a normal basis. Let us try β3 . Its orbit is β3 , β6 = β3 + β2 ,
(β + 1)3 = β3 + β2 + β + 1, (β2 + 1)3 = β3 + β, and these are easily seen to form a basis.
4. (a) Note that [L : K] = p2 , since we adjoin two p-th roots. (The polynomial tp − x is
√
irreducible over K, and the polynomial tp − y is irreducible over K( p x)). However, it is clear
that for each element of a ∈ L, we have ap ∈ K, so K(a) generates an extension of degree at
most p.
√
√
(b) Each element p x + a p y, where a ∈ k, generates a nontrivial subfield, and these fields
are clearly distinct.
√ √ √
√ √ √
√ √ √
√ √ √
5. The elements 1 + 2 +
√ 3√+ 6, 1 − 2 + 3 − 6, 1 + 2 − 3 − 6, 1 − 2 − 3 + 6
form a normal basis of Q( 2, 3) over Q (they
clearly are from the same orbit, and the

1 1
1
1
1 −1 1 −1

determinant of the matrix 
1 1 −1 −1 is equal to 16, so these elements form a basis).
1 −1 −1 1
√ √
√
√
√
√
√ √
√
4
4
4
4
The elements
(1+i)(1+
2+ 2+ 4 8), (1+i)(1+i
2−√ 2−i 4 8), (1+i)(1−
2+
2− √
8),
√
√
√
√
√
√
√
4
4
4
4
4
4
(1 + i)(1 −√
i 2−
2
+
i
8),
(1
−
i)(1
+
2
+
2
+
8),
(1
−
i)(1
+
i
2
−
2
−
i
8),
√
√
√
√
√
4
4
4
4
(1 − i)(1 − 2 + 2 − 8), (1 − i)(1 − i 2 − 2 + i 8) form a normal basis of the splitting
field of x4 −2 over Q (they clearly
are
the same orbit
of the
group, and
they
a
√
√ from
√
√
√ Galois
√
√
√ form
√
4
4
4
4
4
4
spanning set, since (1+i)(1+
2+ 8)+(1−i)(1+
2+
8) = 2(1+
2+√ 2+ √8),
√
√ 2+√
√
√ 2+
√
√
4
4
4
4
4
similarly,
(1
+
i)(1
−
2
+
2
−
8)
+
(1
−
i)(1
−
2
+
2
−
8)
=
2(1
−
2
+ 2√
− 4 8),
√
√
√
√
√
4
4
4
so 1 + 2 and 2 + 8 can be obtained, and for similar reasons 1 − 2 and 2 − 4 8 can
be obtained, etc.
6. Note that the number of primitive roots is equal of the degree of the cyclotomic field,
and they form a single orbit of the Galois group, so they form a normal basis if and only if
they are linearly independent. The values of n < 16 are
1 and 2: the splitting field is Q, answer is trivially “yes”,
3: the primitive roots are ω and ω2 = −1 − ω, so the answer is yes,
4: the primitive roots are i and −i, clearly not a basis,
5: the primitive roots are ξ5 , ξ25 , ξ35 , ξ45 = −1 − ξ5 − ξ25 − ξ35 , so the answer is yes,
6: the primitive roots are −ω and −ω2 , so the answer is yes as for 3,
7: the cyclotomic polynomial has degree 6, the primitive roots are ξk7 , k 6 6, so the answer
is yes,
8: among the primitive roots we have, as in the case of 4, opposite elements, clearly not
a basis,
9: the primitive roots are third roots of the primitive 3rd root of unity, thus the cyclotomic
polynomial is x6 + x3 + 1; the primitive roots themselves are ξ9 , ξ29 , ξ49 , ξ59 , ξ79 , ξ89 , and we see
that ξ9 + ξ49 + ξ79 = ξ9 (1 + ξ39 + ξ69 ) = 0, so the primitive roots do not form a basis,
10: the cyclotomic polynomial has degree 4, the primitive roots are −ξk5 , 1 6 k 6 4, so
the answer is yes as for 5,
11: the cyclotomic polynomial has degree 10, the primitive roots are ξk11 , k 6 10, so the
answer is yes,
12: among the primitive roots we have, as in the case of 4, opposite elements, clearly not
a basis,
13: the cyclotomic polynomial has degree 12, the primitive roots are ξk13 , k 6 6, so the
answer is yes,
14: the cyclotomic polynomial has degree 6, the primitive roots are −ξk7 , 1 6 k 6 6, so
the answer is yes as for 7,
15: the cyclotomic polynomial has degree 8, the primitive roots are ωk ξl5 , k = 1, 2,
1 6 l 6 4, so the answer is easily seen to be yes.