**1.**
Find the splitting field over **Q** for the polynomial
x^{4} + 4.

*Solution:*
We have the factorization
x^{4} + 4 = (x^{2} + 2x +2)(x^{2} - 2x +2),
where the factors are irreducible by
Eisenstein's criterion
(using p = 2).
The roots are ± 1 ± i,
so the splitting field is **Q**(i), which has degree 2 over **Q**.

An alternate solution is to solve x^{4} = -4,
To find one root, use DeMoivre's theorem to get
1/ +
1/ i,
and then multiply by to get 1+i.
The other roots are found by multiplying by the powers of i,
because it is a primitive 4th root of unity.

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**2.**
Find the degree of the splitting field over **Z**_{2}
for the polynomial (x^{3} + x + 1)(x^{2} + x + 1).

*Solution:*
The two polynomials are irreducible (you can check that they have no roots).
Therefore the splitting field must have subfields
of degree 3 and of degree 2,
so the degree of the splitting field over **Z**_{2} must be 6.

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**3.**
Find the degree [F:**Q**],
where F is the splitting field of the polynomial x^{3} - 11
over the field **Q** of rational numbers.

*Solution:*
Letting
be the cube root of 11,
the roots of the polynomial are
,
,
and ^{2}
,
where
is a primitive cube root of unity.
Since is not real,
it cannot belong to **Q**().
Since is a root of
x^{2} +x +1 and
F = **Q** (,
),
we have [F:**Q**] = 6.

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**4.**
Determine the splitting field over **Q** for x^{4} + 2.

*Solution:*
To get the splitting field F, we need to adjoin the 4th roots of -2,
which have the form
^{i}
,
where is the 4th root of 2,
is a primitive 8th root of unity,
and i = 1,3,5,7.
To construct these roots we only need to adjoin
and i.

To show this, using the polar form
cos
+ i sin
of the complex numbers,
we can see that

= / 2
+ / 2 i,
^{3}
= - / 2
+ / 2 i,

^{5}
= - / 2
- / 2 i, and
^{7}
= / 2
- / 2 i.

Thus
=
+ ^{7}
must belong to F, and then the cube of this element,
which is 4 ,
must also belong to F.
Therefore
belongs to F (which is somewhat surprising)
and the square of this element is ,
so it follows that
belongs to F, and therefore i belongs to F.
The splitting field is thus
**Q** ( , i),
which has degree 8 over **Q**.

*Note:* This is the same field as in Example 8.3.2 of the text,
which computes the Galois group of x^{4}-2 over **Q**.

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**5.**
Determine the splitting field over **Q** for
x^{4} + x^{2} + 1.

*Solution:*
Be careful here--this polynomial is not irreducible.
In fact, x^{6} - 1 factors in two ways, and provides an important clue.
Note that

x^{6} - 1
= (x^{3})^{2} - 1
= (x^{3} - 1)(x^{3} + 1)
= (x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1) and

x^{6} - 1
= (x^{2})^{3} - 1
= (x^{2} - 1)(x^{4} +x^{2} +1).

Thus x^{4}+x^{2}+1 = (x^{2}+x+1)(x^{2}-x+1),
and the roots of the first factor are the primitive 3rd roots of unity,
while the roots of the second factor are the primitive 6th roots of unity.
Adjoining a root
of x^{2}-x+1 gives all 4 roots,
and so the splitting field
**Q**()
has degree 2 over **Q**.

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