Examples of Splitting Fields

Let us now give some examples of splitting fields. Throughout this section let F be a field, f F[X], E_f = a splitting field of f over F.

Example 1: F = Q, f = X² + aX + b (a,b Q).

the zeros of f in C are given by the quadratic formula

(-a + )/2,    (-a - )/2,

where denotes one of the square roots of a² - 4b in C. Therefore, since a Q, we see that

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 E_f = Q and deg(E_f /Q) = 1.

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,

Example 2: F = Q, f = Xⁿ - 1.

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 Q(). we see that

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 Q() over Q. The best way to do this is to determine Irr_Q(,X). For by Theorem 3 of the section on algebraic numbers,

deg(Q()/Q) = deg(Irr_Q(,X)).

Set

_n(X) = (X - ⁱ ).

Then _n(X) is called the nth cyclotomic polynomial. It first appears that _n(X) has complex coefficients. But actually the coefficients of _n(X) are rational and, in fact, are even integers. Note the zeros of _n(X) are just the (n) distinct nth roots of 1. Moreover, if is an nth root of 1, let d be the order of as an element of the group X_n of the nth roots of 1. Then d is the smallest positive integer such that ^d = 1 and is a primitive dth root of 1. Moreover, since the order of X_n is n and the order of an element divides the order of the group, we see that d|n. Thus, every nth root of 1 is a primitive dth root of unity for some uniquely determined d which divides n. Therefore,

Xⁿ - 1 = (X - )

=

= _d(X).

Let us use formula (4) to compute _n(X) for the first few n. For n = 1, (4) reads

₁(X) = X - 1.

For n = 2, formula (4) reads

₁(X) · ₂(X) = X² - 1.

Therefore, by (5),

₂(X) = X + 1.

For n = 3, (4) reads

X³ - 1 = ₁ · ₃(X),

and thus by (5),

₃(X) = (X³ - 1)/(X - 1) = X² + X + 1

For n = 4, (4) reads

X⁴ - 1 = ₁(X) · ₂(X) · ₄(X),

and thus by (5) and (6),

₄(X) = (X⁴ - 1)/[(X - 1)(X + 1)]

From these few computation, it becomes clear that if we have computed _d(X) for all d < n, then we may use (4) to compute _n(X) from

_n(X) = (Xⁿ - 1)/_d(X).

We observe that ₁(X) has rational coefficients. Moreover, if _d(X) is assumed to have rational coefficients for all d < n, then (9) implies that _n(X) has rational coefficients, since _n(X) is computed from _d(X) (d < n) using only rational operations. Thus, by induction,

_n(X) Q[X]   (n > 1).

A more careful analysis of the algorithm for computing _n shows that _n Z[X]. It is clear that _n(X) is a monic polynomial having as a zero. Thus if we knew that _n(X) were irreducible in Q[X], we could deduce that

_n(X) = Irr_Q(,X),

deg(Q()/Q) = deg(Irr_Q(,X))

= deg(_n(X))

= (n),

where (n) denotes Euler's -function. The proof of the irreducibility of _n(X) is included in an appendix, but for the moment, let us record

Theorem 1: If is a primitive nth root of 1, then deg(Q()/Q) = (n).

Example 3: F = Q, f = X² - 2.

Let denote the real cube root of 2. Then the zeros of X³ - 2 in C are all the cube roots of 2, given by

,  ,   ²,

where = (-1 + i)/2 is a primitive cube root of 1. Moreover,

E_f = Q(, , ²)

= Q(, ).

Since X³ - 2 is irreducible in Q[X], by Eisenstein's criterion,

deg(Q()/Q) = 3.

Moreover, is a zero of X² + X + 1 Q()[X], Irr_Q()(,X)|X² + X + 1. and therefore,

deg(Q(,)/Q()) < 2.

Thus, by Theorem 3 in the section on algebraic extensions,

deg(Q(, )/Q) = deg(Q(, )/Q()) · deg(Q()/Q)

Note, however, that X² + X + 1 is irreducible over Q by Example 1. Therefore,

deg(Q()/Q) = 2.

But by Theorem 3 of algebraic extensions and (14),

deg(Q()/Q)|deg(Q(, )/Q)

2|deg(Q(, )/Q)

Similarly,

deg(Q()/Q)|deg(Q(, )/Q)

3|deg(Q(, )/Q)

by (15). By (15) and (16), we see that

6|deg(Q(, )/Q)

deg(Q(, )/Q) > 6.

Finally, by (13) and (17),

deg(Q(, )/Q) = 6.

Example 4: Let F = Q, f = Xⁿ - a (n > 1, a Q).

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/n2, ..., a^1/nn-1.

Thus,

E_f = Q(a^1/n, a^1/n, ..., a^1/nn-1)

= 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

E_f Q() Q

and

deg(E_f /Q) = deg(E_f /Q()) · deg(Q()/Q)

= (n)deg(E_f /Q())

by Theorem 1. Therefore, in order to compute deg(E_f /Q), we must compute deg(E_f /Q()). And, in turn, to compute deg(E_f /Q()), we must compute

deg(Irr_Q()(a^1/n,X)).

Since Xⁿ - a is a polynomial in Q()[X] having a^1/n as a zero, we see that

Irr_Q()(a^1/n,X)|Xⁿ - a,

and therefore,

deg(E_f /Q()) = deg(Irr_Q()(a^1/n,X))

Thus, finally, by (19) we see that

deg(E_f /Q) < n(n).

In special cases, we can compute deg(E_f /Q) exactly. For example,

Theorem 2: Let p be prime and suppose that f = X^p - a is irreducible in Q[X]. Then

Proof: Since p is prime, we have

(p) = p - 1,

By equations (19) and (21), p - 1|deg(E_f /Q). Since X^p - a is irreducible in Q[X],

Therefore, since

we see that

Thus, by (22), we have p(p - 1)|deg(E_f /Q). However, by (20), we see that deg(E_f /Q) < p(p - 1), and hence the proof of the theorem is complete.

Example 5: Let F be a subfield of C which contains the nth roots of 1 for some n > 1. and let f = Xⁿ - 1 (a F). If a^1/n is one nth root of a in C, then the zeros of f are

a^1/n, a^1/n, ..., a^1/nn-1,

where is a primitive nth root of 1. Therefore, since F,

E_f = F(a^1/n, a^1/n, ..., a^1/nn-1)

The extension F(a^1/n)/F is a typical example of what is called a Kummer extension.