MA 3419: Galois theory
Selected answers/solutions to the assignment due December 10, 2015
1. (a) First of all, an extension of a field K has the same characteristic as K, so p = p 0 .
Also, an extension of a field K is a vector space over K, so an extension of Fq has qm elements,
0
where m is the dimension of the extension as a vector space over Fq . Thus, pn = (pn )m , and
n = n 0 m.
n
(b) We know that Fpn is the splitting field of xp − x over Fp , so it is normal. It is
manifestly an extension of a finite degree n. Finally, an extension of finite fields is always
separable.
(c) We have (xy)p = xp yp , and (x + y)p = xp + yp (the latter because we are in charack
teristic p), so x 7→ xp is an automorphism. The k-th power of it is x 7→ xp , which is different
from x 7→ x for k < n, since the multiplicative group of Fpn is cyclic, and therefore contains
k
an element of order pn − 1; for such an element η, we have ηp 6= η for k < n. A Galois
extension has a Galois group of order equal to the degree, so we found all automorphisms.
n
n0
(d) If n is divisible by n 0 , then every root of xp − x is a root of xp − x, since pn − 1 is
0
divisible by pn − 1, therefore there is inclusion between splitting fields.
(e) As before, it is normal and separable and finite. The Galois group is the group of all
n0
elements fixing Fpn 0 , that is the subgroup generated by x 7→ xp . This subgroup is isomorphic
to Z/(n/n 0 )Z. A Galois extension has a Galois group of order equal to the degree, so we
found all automorphisms.
2. The multiplicative group (Z/3Z)× is cyclic generated by 2; the powers of 2 modulo
13 are, in the order of the exponent, 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7. Thus, denoting by ξ the
primitive root e2π/13 of unity of degree 13, we may consider the quantities
A = ξ + ξ4 + ξ3 + ξ12 + ξ9 + ξ10 ,
A 0 = ξ2 + ξ8 + ξ6 + ξ11 + ξ5 + ξ7 .
Clearly, A + A 0 = −1, and AA 0 = −3, so A and A 0 are roots of the quadratic equation
x2 + x − 3 = 0. Next, we consider the quantities
B = ξ + ξ12 ,
B 0 = ξ4 + ξ9 ,
B 00 = ξ3 + ξ10 .
We have B+B 0 +B 00 = A, BB 0 +BB 00 +B 0 B 00 = −1, BB 0 B 00 = 2+A 0 , so B, B 0 , and B 00 are roots
of a cubic equation with coefficients of Q(A, A 0 ). Clearly, ξ + ξ12 = ξ + ξ−1 = 2 cos(2π/13).
3. Note that 720 = 24 · 32 · 5. Note that
∼ Z/24 Z × Z/2Z,
(Z/26 Z)× =
∼ Z/(9 · 2)Z =
∼ Z/9Z × Z/2Z,
(Z/27Z)× =
∼ Z/10Z =
∼ Z/5Z × Z/2Z.
(Z/11Z)× =
Let us consider the cyclotomic field Q(ζ26 ·33 ·11 ). The Galois group of this field is the product
∼ Z/24 Z × Z/9Z × Z/5Z as a
of the abovementioned groups, and as such admits Z/720Z =
quotient (by some subgroup H). Thus Z/720Z is the Galois group of Q(ζ26 ·33 ·11 )H over Q.
Since we are in characteristic zero, for α we can take the primitive element of that extension.
4. (a) Over F2 , we have x5 −x−1 = (x2 −x−1)(x3 −x2 −1), and these two polynomials are
irreducible, so the Galois group over Q contains an element that is a product of a transposition
and a 3-cycle. Over F5 , we have (x + 1)5 − (x + 1) − 1 = x5 − x − 1, so this polynomial cannot
be reducible, since the transformation x 7→ x + 1 acts on its roots transitively; this also shows
that adjoining one root adjoins them all, so the Galois group over Q contains a 5-cycle. The
cube of a product of a transposition and a 3-cycle is a transposition, so we are dealing with
a subgroup containing a transposition and a 5-cycle. Such a subgroup must coincide with all
of S5 , since conjugating a transposition with a 5-cycle, we can get four transpositions that
generate S5 .
(b) The polynomial x(x − 1) · · · (x − (p − 4)) has p − 3 simple roots, each of which is close
to one simple root of
1
p
xp − N3 p3 x(x − 1) · · · (x − (p − 4)) − p = N3 p3
(x
−
p)
−
x(x
−
1)
·
·
·
(x
−
(p
−
4))
.
N3 p3
Also, there is one simple root close to Np, since
x
1
x
p−4
p
x p x
−
−
···
−
−
.
xp −N3 p3 x(x−1) · · · (x−(p−4))−p = (Np)p
Np
Np Np Np
Np
Np
(Np)p
This already gives p − 2 roots. If there were more roots, there would be p of them, and there
will be at least one root different from the roots we found (possibly with multiplicity 2). Then
by Rolle theorem, the derivative of this polynomial would have at least p − 2 different roots,
. . . , the p − 3-rd derivative would have at least 2 different roots. But that derivative is of the
form Ax3 − B, which has just one real root. This implies that the Galois group of the splitting
field of this polynomial contains a transposition (corresponding to the complex conjugation).
Also, by Eisenstein this polynomial is irreducible, so the Galois group is transitive, hence the
Galois group is Sp .