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Let alpha element of E. Then alpha is a zero of f if and only if X - alpha divides f in E[X]; that is, f = (X - alpha)g for some g element of E[X]. If f has a leading coefficient 1, then we say that f is monic. M is a maximal ideal of R if and only if R/M is a field. An ideal M of R is said to be a maximal ideal if (a) M is a proper ideal and (b) if A is a proper ideal containing M, then M = A. The set of real numbers. The complex numbers.

Splitting Fields

We can construct the complex numbers C by adjoining to R a zero of the polynomial X2 + 1 in R[X]. Let us now take up the general problem of adjoining to F a zero of some polynomial f element of F[X], where F is an arbitrary field. Throughout this section, let

f = Xn + an-1Xn-1 + ... + a0,   ai element of F.

Let us first obtain some information about what zeros f can have in an extension E of F. (The case E = F is not excluded.) Our first result is a generalization of a well-known example from high school algebra.

Proposition 1: Let alpha element of E. Then alpha is a zero of f if and only if X - alpha divides f in E[X]; that is, f = (X - alpha)g for some g element of E[X].

Proof: By the division algorithm in E[X], we may write

f = (X - alpha)g + r,    g,r element of E[X],

where r = 0 or 0 < deg(r) < 1. In either case, r is a constant polynomial, say r = beta element of E, so that f = (X - alpha) g + beta. Evaluating f at alpha, we see that

f(alpha) = beta,

and therefore f = (X - alpha)g + f(alpha). Thus, f(alpha) = 0 if and only if X - alpha divides f in E[X].

Corollary 2: f has at most n zeros in E.

Proof: Let alpha1,...,alphak be distinct zeros of f in E. Then, in E[X], f is divisible by (X-alpha1)(X - alpha2)···(X - alphak). Therefore, n > k.

Corollary 3: Suppose that n > 1 and f is irreducible in E[X]. Then f has no zeros in E.

Proof: For if f has a zero alpha in E, then f is divisible by X - alpha in E[X], so that f is reducible in E[X] (since n > 1).

Let us now prove the main result of this section.

Theorem 4: There exists and extension E of F which contains a zero of f.

Proof: If f1 is an irreducible factor of f in F[X], then every zero of f1 is also a zero of f. Therefore, without loss of generality, let us assume that f is irreducible in F[X]. Then I = f · F[X] is a maximal ideal of F[X], so that by Theorem 8 of the section on prime ideals, F[X]/I is a field. Since f is irreducible, f is not constant. Therefore, the mapping

amapsa + I

is an isomorphism. Thus, let us identify F with the subring {a + I | a element of F} of F[X]/I, so that we can view F[X]/I as an extension of F. Let E = F[X]/I and set alpha = X + I element of E. Then

f(alpha) = f(X) + I
= 0 + I    [since f(X) = f element of I],

so that alpha is a zero of f in E.

A an easy consequence or Theorem 4 we deduce

Corollary 5: There exists and extension E of F such that, in E[X]. f(X) can be written in the form

f(X) = product over i(X - alphai),   alphai element of E.

Proof: By Theorem 4, there exists an extension E1 of F in which f(X) has a zero alpha1. By Proposition 1, f(X) is divisible by X - alpha1 in E1[X]. Therefore. there exists a monic polynomial g1(X) element ofE1[X] such that f(X) = (X - alpha1)g1(X). Applying the same reasoning to g1(X) that we just applied to f(X), we see that there exists an extension E2 of E1, alpha2 element of E2 and g2(X) such that g1(X) = (X - alpha2)g2(X). Therefore, f(X) = (X - alpha1)(X - alpha2)g2(X). Proceeding this way, we can construct an extension En of F such that

f = gn(X)product over i(X - alphai),   gn(X) element of En[X].

But since n = deg f(X), deg gn(X) = 0, so that gn(X) element of En. However, because f is monic, gn(X) = 1. Therefore, we may set E = En.

Let E denote an extension of F such that

f(X) = product over i(X - alphai),    alphaielement of E.

The subfield F(alpha1,...,alphan) of E gotten by adjoining all zeros of f(X) to F is called the splitting field of f. It is clear that F(alpha1,...,alphan) is the smallest subfield of E in which f(X) splits into a product of linear factors, whence the name "splitting field." An arbitrary field in which f(X) factors into linear factors is called a root field. If no proper subfield of the root field exists, it is the splitting field.

A simple argument utilizing Corollary 5 allows us to show

Corollary 6: Let f1(X), f2(X),..., fr(X), be monic polynomials belonging to F[X] such that deg f(X) > 1 (1 < i < r). Then there exists an extension E of F such that in E[X], each polynomial f1(X) splits into a product of linear factors.