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Examples of Splitting FieldsLet us now give some examples of splitting fields. Throughout this section let F be a field, Example 1: the zeros of f in C are given by the quadratic formula
(a + )/2, (a  )/2,
where denotes one of the square roots of
E_{f} = Q().
Let us analyze the situation further: We consider two cases. Case 1: f is reducible in Q[X]. In this case, the zeros of f are contained in Q and Case 2: f is irreducible in Q[X]. Note that by (1), E_{f} can be obtained from Q by adjoining either one of those zeros of f. Therefore, by Theorem 3 of the section on algebraic numbers,
deg(E_{f} /Q) = deg(f ) = 2.
Example 2: The zeros of f in C are just the nth roots of 1. Let be a primitive nth root of 1. Then the zeros of f are
1,,^{2},...,^{n1}.
Since all these roots are contained in
E_{f} = Q().
The field E_{f} is called the nth cyclotomic field. The term "cyclotomic" means "circle dividing" and the reason for this terminology is that the field E_{f} was first considered in connection with the problem of constructing a regular polygon of n sides using only a ruler and compass. Let us determine the degree of Set (3)
_{n}(X) = (X  ^{ i} ).
Then
X^{n}  1 = (X  )
(4)
=
= _{d}(X).
Let us use formula (4) to compute
_{1}(X) = X  1.
For
_{1}(X) · _{2}(X) = X^{2}  1.
Therefore, by (5), (6)
_{2}(X) = X + 1.
For
X^{3}  1 = _{1} · _{3}(X),
and thus by (5), (7)
_{3}(X) = (X^{3}  1)/(X  1) = X^{2} + X + 1
For
X^{4}  1 = _{1}(X) · _{2}(X) · _{4}(X),
and thus by (5) and (6),
_{4}(X) = (X^{4}  1)/[(X  1)(X + 1)]
(8)
= X^{2} + 1.
From these few computation, it becomes clear that if we have computed
_{n}(X) = (X^{n}  1)/_{d}(X).
We observe that
_{n}(X) Q[X] (n > 1).
A more careful analysis of the algorithm for computing _{n} shows that
_{n}(X) = Irr_{Q}(,X),
= deg(_{n}(X))
= (n),
where (n) denotes Euler's function. The proof of the irreducibility of
Theorem 1: If is a primitive nth root of 1, then Example 3: Let denote the real cube root of 2. Then the zeros of
, , ^{2},
where
E_{f} = Q(, , ^{2})
(11)
= Q(, ).
Since
Moreover, is a zero of Thus, by Theorem 3 in the section on algebraic extensions, (13)
< 2 · 3 = 6
Note, however, that But by Theorem 3 of algebraic extensions and (14), (15)Similarly, (16)by (15). By (15) and (16), we see that (17)Finally, by (13) and (17), (18)Example 4: Let Let a^{1/n} be any nth root of a in C and let be a primitive nth root of 1. The zeros of f are just the nth roots of a and therefore are
a^{1/n}, a^{1/n}, a^{1/n}^{2}, ..., a^{1/n}^{n1}.
Thus,
E_{f} = Q(a^{1/n}, a^{1/n}, ..., a^{1/n}^{n1})
= Q(a^{1/n}, ).
Calculating the degree of E_{f} over Q is a rather intricate problem, and we will not consider a complete solution at this time. However, let us say what we can without undue expenditure of effort. Note that and (19)
by Theorem 1. Therefore, in order to compute
deg(Irr_{Q()}(a^{1/n},X)).
Since
Irr_{Q()}(a^{1/n},X)X^{n}  a,
and therefore,
< deg(X^{n}  a)
= n.
Thus, finally, by (19) we see that (20)
In special cases, we can compute
Theorem 2: Let p be prime and suppose that
deg(E_{f} /Q) = p(p  1).
Proof: Since p is prime, we have (21)
(p) = p  1,
(22)
(p  1, p) = 1.
By equations (19) and (21), Therefore, since
= p · deg(E_{f} /Q(a^{1/p})),
we see that
pdeg(E_{f} /Q).
Thus, by (22), we have Example 5: Let F be a subfield of C which contains the nth roots of 1 for some
a^{1/n}, a^{1/n}, ..., a^{1/n}^{n1},
where is a primitive nth root of 1. Therefore, since
E_{f} = F(a^{1/n}, a^{1/n}, ..., a^{1/n}^{n1})
= F(a^{1/n}).
The extension 
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