+
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.
1. Find the splitting field over Q for the polynomial x4 + 4.
Solution: We have the factorization x4 + 4 = (x2 + 2x +2)(x2 - 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 x4 = -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 Z2 for the polynomial (x3 + x + 1)(x2 + 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 Z2 must be 6.
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3. Find the degree [F:Q], where F is the splitting field of the polynomial x3 - 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
x2 +x +1 and
F = Q (,
),
we have [F:Q] = 6.
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4. Determine the splitting field over Q for x4 + 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 x4-2 over Q.
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5. Determine the splitting field over Q for x4 + x2 + 1.
Solution:
Be careful here--this polynomial is not irreducible.
In fact, x6 - 1 factors in two ways, and provides an important clue.
Note that
x6 - 1
= (x3)2 - 1
= (x3 - 1)(x3 + 1)
= (x-1)(x2+x+1)(x+1)(x2-x+1) and
x6 - 1
= (x2)3 - 1
= (x2 - 1)(x4 +x2 +1).
Thus x4+x2+1 = (x2+x+1)(x2-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 x2-x+1 gives all 4 roots,
and so the splitting field
Q()
has degree 2 over Q.
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