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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
Ef = 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), Ef 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(Ef /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,...,n-1.
Since all these roots are contained in
Ef = Q().
The field Ef is called the nth cyclotomic field. The term "cyclotomic" means "circle dividing" and the reason for this terminology is that the field Ef 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
Xn - 1 = (X - )
(4)
=
= d(X).
Let us use formula (4) to compute
1(X) = X - 1.
For
1(X) · 2(X) = X2 - 1.
Therefore, by (5), (6)
2(X) = X + 1.
For
X3 - 1 = 1 · 3(X),
and thus by (5), (7)
3(X) = (X3 - 1)/(X - 1) = X2 + X + 1
For
X4 - 1 = 1(X) · 2(X) · 4(X),
and thus by (5) and (6),
4(X) = (X4 - 1)/[(X - 1)(X + 1)]
(8)
= X2 + 1.
From these few computation, it becomes clear that if we have computed
n(X) = (Xn - 1)/d(X).
We observe that
n(X) Q[X] (n > 1).
A more careful analysis of the algorithm for computing n shows that
= 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
Ef = 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 a1/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
a1/n, a1/n, a1/n2, ..., a1/nn-1.
Thus,
Ef = Q(a1/n, a1/n, ..., a1/nn-1)
= Q(a1/n, ).
Calculating the degree of Ef 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
Since and therefore,
< deg(Xn - 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(Ef /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(Ef /Q(a1/p)),
we see that
p|deg(Ef /Q).
Thus, by (22), we have Example 5: Let F be a subfield of C which contains the nth roots of 1 for some
a1/n, a1/n, ..., a1/nn-1,
where is a primitive nth root of 1. Therefore, since
Ef = F(a1/n, a1/n, ..., a1/nn-1)
= F(a1/n).
The extension |
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