**7.**
Determine the group of all automorphisms of a field with 4 elements.

*Solution:*
The automorphism group consists of two elements:
the identity mapping and the
Frobenius automorphism.

Read on only if you need more detail. By Theorem
6.5.2
and Theorem
6.4.5,
up to isomorphism there is only one field with 4 elements.
Since x^{2}+x+1 is irreducible over **Z**_{2},
by Theorem
4.4.6
this field with 4 elements can be constructed as
F = **Z**_{2}[x] / < x^{2}+x+1 >.
Letting a be the coset of x,
we have F = {0, 1, a, 1+a}.
Any automorphism of F must leave 0 and 1 fixed,
so the only possibility for an automorphism other than the identity
is to interchange a and 1+a.
Is this an automorphism?
Since x^{2}+x+1 0,
we have x^{2} -x-1
x+1,
so a^{2} = 1+a and
(1+a)^{2} = 1+2a+a^{2} = a.
Thus the function that fixes 0 and 1
while interchanging a and 1+a
is in fact the Frobenius automorphism of F.

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**8.**
Let F be the splitting field in **C** of x^{4}+1.

(i) Show that [F:**Q**] = 4.

*Solution:*
The polynomial x^{8}-1 factors over **Q** as
x^{8}-1
= (x^{4}-1)(x^{4}+1)
= (x-1)(x+1)(x^{2}+1)(x^{4} +1).
The factor x^{4} +1 is irreducible over **Q** by
Eisenstein's criterion.
The roots of x^{4}+1 are thus the primitive 8th roots of unity,
± / 2
± / 2 i,
and adjoining one of these roots also gives the others,
together with i.
Thus the splitting field is obtained in one step,
by adjoining one root of x^{4}+1,
so its degree over **Q** is 4.

It is clear that the splitting field can also be obtained
by adjoining first and then i,
so it can also be expressed as
**Q**(, i).

(ii) Find automorphisms of F that have fixed fields
**Q**(),
**Q**(i), and
**Q**( i),
respectively.

*Solution:*
These subfields of **Q**(, i)
are the splitting fields of
x^{2}-2, x^{2}+1, and x^{2}+2, respectively.
Any automorphism must take roots to roots,
so if
is an automorphism of
**Q**(, i),
we must have
()
= ± ,
and (i)
= ± i.
These possibilities must in fact define 4 automorphisms of
the splitting field.

If we define _{1}
()
= and
_{1} (i) = -i,
then the subfield fixed by
_{1}
is **Q**().
If we define _{2}
()
= -
and _{2} (i) = i,
then the subfield fixed by
_{2} is **Q**(i).
Finally, for _{3}
= _{2}
_{1} we have
_{3}
()
= -
and (i) = -i,
and thus _{3}
( i)
= i,
so _{3} has
**Q**( i) as its fixed subfield.

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**9.**
Find the Galois group over **Q** of the polynomial x^{4}+4.

*Solution:*
Problem
6.4.1
shows that the splitting field
of the polynomial has degree 2 over **Q**,
and so the Galois group must be cyclic of order 2.

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**10.**
Find the Galois groups of x^{3}-2 over the fields
**Z**_{5} and **Z**_{11}.

*Solution:*
The polynomial is not irreducible over **Z**_{5},
since it factors as
x^{3}-2 = (x+2)(x^{2}-2x-1).
The quadratic factor will have a splitting field of degree 2
over **Z**_{5},
so the Galois group is cyclic of order 2.

A search in **Z**_{11} for roots of x^{3}-2
yields one and only one: x = 7.
Then x^{3}-2 can be factored as
x^{3}-2 = (x-7)(x^{2}+7x+5),
and the second factor must be irreducible.
The splitting field has degree 2 over **Z**_{11},
and can be described as
**Z**_{11}[x] / < x^{2}+7x+5 >.
Thus the Galois group is cyclic of order 2.

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**11.**
Find the Galois group of x^{4}-1 over the field **Z**_{7}.

*Solution:*
We first need to find the the splitting field of
x^{4}-1 over **Z**_{7}.
We have x^{4}-1 = (x-1)(x+1)(x^{2}+1).
A quick check of ±2 and ±3
shows that they are not roots of x^{2}+1 over **Z**_{7},
so x^{2}+1 is irreducible over **Z**_{7}.
To obtain the splitting field we must adjoin a root of x^{2}+1,
so we get a splitting field
**Z**_{7}[x] / < x^{2}+1 >
of degree 2 over **Z**_{7}.

It follows from Corollary
8.1.7
that the Galois group of x^{4}-1 over **Z**_{7}
is cyclic of order 2.

Next problem | Next solution | Table of Contents

**12.**
Find the Galois group of x^{3}-2 over the field **Z**_{7}.

*Solution:*
In this case, x^{3}-2 has no roots in **Z**_{7},
so it is irreducible.
We first adjoin a root a
of x^{3}-2 to **Z**_{7}.
The resulting extension
**Z**_{7}(a)
has degree 3 over **Z**_{7},
so it has 7^{3} = 343 elements,
and each element is a root of the polynomial x^{343}-x.
Let b> be a generator of the multiplicative group of the extension.
Then (b^{114})^{3} = b^{342} = 1,
showing that **Z**_{7}(a)
contains a nontrivial cube root of 1.
It follows that x^{3}-2 has three distinct roots in
**Z**_{7}(a):
a, ab^{114}, and ab^{228},
so therefore **Z**_{7}(a)
is a splitting field for x^{3}-2 over **Z**_{7}.
Since the splitting field has degree 3 over **Z**_{7},
it follows from Corollary
8.1.7
that the Galois group
of the polynomial is cyclic of order 3.

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