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If zeta is a primitive nth root of 1, then deg(Q(zeta)/Q) = phi(n). Let f = a0+a1X+...+anXn element of Z[X] be primitive and let p be prime. Assume that p|ai (0 < i < n-1), pdoes not dividean, p2does not dividea0. Then f is irreducible in Z[X]. Let alpha element of E be algebraic over F. alpha the zero of a monic irreducible polynomial p element F[X]. The polynomial p is called the irreducible polynomial of alpha over F denoted IrrF(alpha,X) phi:nmapsphi(n) = the number of integers a (1 < a < n) such that (a,n) = 1. If f has a leading coefficient 1, then we say that f is monic. Let E subset of F subset of G be three fields. Assume that deg(F/E) and deg(G/F) are finite. Then deg(G/E) is finite and deg(G/E) = deg(G/F) · deg(F/E). Further, if {alpha1,...,alphan} is a basis of G over F and {beta1,...,betam} is a basis of F over E, then {alphaibetaj} (1 < i < n, 1 < j < m) is a basis for G over E. Let alpha be algebraic over F, n = deg(IrrF(alpha,X)). Then: (1). deg(F(alpha)/F) = n. (2). {1,alpha,alpha2,...,alphan-1} is a basis for F(alpha) over F. The subfield F(alpha1,...,alphan) of E gotten by adjoining all zeros of f(X) to F is called the splitting field of f. The set of rational numbers. The complex numbers.

Examples of Splitting Fields

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

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

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

(-a + discriminant)/2,    (-a - discriminant)/2,

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

(1)
Ef = Q(discriminant).

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

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: F = Q, f = Xn - 1.

The zeros of f in C are just the nth roots of 1. Let zeta be a primitive nth root of 1. Then the zeros of f are

1,zeta,zeta2,...,zetan-1.

Since all these roots are contained in Q(zeta). we see that

Ef = Q(zeta).

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

(2)
deg(Q(zeta)/Q) = deg(IrrQ(zeta,X)).

Set

(3)
phin(X) = product of relatively prime(X - zeta i ).

Then phin(X) is called the nth cyclotomic polynomial. It first appears that phin(X) has complex coefficients. But actually the coefficients of phin(X) are rational and, in fact, are even integers. Note the zeros of phin(X) are just the phi(n) distinct nth roots of 1. Moreover, if eta is an nth root of 1, let d be the order of eta as an element of the group Xn of the nth roots of 1. Then d is the smallest positive integer such that etad = 1 and eta is a primitive dth root of 1. Moreover, since the order of Xn 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 = product over eta (X - eta)
(4)
= product of primitives
= product over d divides n phid(X).

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

(5)
phi1(X) = X - 1.

For n = 2, formula (4) reads

phi1(X) · phi2(X) = X2 - 1.

Therefore, by (5),

(6)
phi2(X) = X + 1.

For n = 3, (4) reads

X3 - 1 = phi1 · phi3(X),

and thus by (5),

(7)
phi3(X) = (X3 - 1)/(X - 1) = X2 + X + 1

For n = 4, (4) reads

X4 - 1 = phi1(X) · phi2(X) · phi4(X),

and thus by (5) and (6),

phi4(X) = (X4 - 1)/[(X - 1)(X + 1)]
(8)
= X2 + 1.

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

(9)
phin(X) = (Xn - 1)/product of phiphid(X).

We observe that phi1(X) has rational coefficients. Moreover, if phid(X) is assumed to have rational coefficients for all d < n, then (9) implies that phin(X) has rational coefficients, since phin(X) is computed from phid(X) (d < n) using only rational operations. Thus, by induction,

(10)
phin(X) element of Q[X]   (n > 1).

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

phin(X) = IrrQ(zeta,X),
implies deg(Q(zeta)/Q) = deg(IrrQ(zeta,X))
= deg(phin(X))
= phi(n),

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

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

Example 3: F = Q, f = X2 - 2.

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

cube root of 2,  cube root of 2omega,   cube root of 2omega2,

where omega = (-1 + isquare root of 3)/2 is a primitive cube root of 1. Moreover,

Ef = Q(cube root of 2, cube root of 2omega, cube root of 2omega2)
(11)
= Q(cube root of 2, omega).

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

(12)
deg(Q(cube root of 2)/Q) = 3.

Moreover, omega is a zero of X2 + X + 1 element of Q(cube root of 2)[X], IrrQ(cube root of 2)(omega,X)|X2 + X + 1. and therefore,

deg(Q(cube root of 2,omega)/Q(cube root of 2)) < 2.

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

deg(Q(cube root of 2, omega)/Q) = deg(Q(cube root of 2, omega)/Q(cube root of 2)) · deg(Q(cube root of 2)/Q)
(13)
< 2 · 3 = 6

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

(14)
deg(Q(omega)/Q) = 2.

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

deg(Q(omega)/Q)|deg(Q(cube root of 2, omega)/Q)
(15)
implies 2|deg(Q(cube root of 2, omega)/Q)

Similarly,

deg(Q(cube root of 2)/Q)|deg(Q(cube root of 2, omega)/Q)
(16)
implies 3|deg(Q(cube root of 2, omega)/Q)

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

6|deg(Q(cube root of 2, omega)/Q)
(17)
impliesdeg(Q(cube root of 2, omega)/Q) > 6.

Finally, by (13) and (17),

(18)
deg(Q(cube root of 2, omega)/Q) = 6.

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

Let a1/n be any nth root of a in C and let zeta be a primitive nth root of 1. The zeros of f are just the nth roots of a and therefore are

a1/n, a1/nzeta, a1/nzeta2, ..., a1/nzetan-1.

Thus,

Ef = Q(a1/n, a1/nzeta, ..., a1/nzetan-1)
= Q(a1/n, zeta).

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

Ef superset of Q(zeta) superset of Q

and

deg(Ef /Q) = deg(Ef /Q(zeta)) · deg(Q(zeta)/Q)
(19)
= phi(n)deg(Ef /Q(zeta))

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

deg(IrrQ(zeta)(a1/n,X)).

Since Xn - a is a polynomial in Q(zeta)[X] having a1/n as a zero, we see that

IrrQ(zeta)(a1/n,X)|Xn - a,

and therefore,

deg(Ef /Q(zeta)) = deg(IrrQ(zeta)(a1/n,X))
< deg(Xn - a)
= n.

Thus, finally, by (19) we see that

(20)
deg(Ef /Q) < nphi(n).

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

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

deg(Ef /Q) = p(p - 1).

Proof: Since p is prime, we have

(21)
phi(p) = p - 1,
(22)
(p - 1, p) = 1.

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

deg(Q(a1/p)/Q) = p.

Therefore, since

deg(Ef /Q) = deg(Ef /Q(a1/p)) · deg(Q(a1/p)/Q)
= p · deg(Ef /Q(a1/p)),

we see that

p|deg(Ef /Q).

Thus, by (22), we have p(p - 1)|deg(Ef /Q). However, by (20), we see that deg(Ef /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 element of F). If a1/n is one nth root of a in C, then the zeros of f are

a1/n, a1/nzeta, ..., a1/nzetan-1,

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

Ef = F(a1/n, a1/nzeta, ..., a1/nzetan-1)
= F(a1/n).

The extension F(a1/n)/F is a typical example of what is called a Kummer extension.