Let E. Then is a zero of f if and only if X - divides f in E[X]; that is, f = (X - )g for some g 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 F[X], where F is an arbitrary field. Throughout this section, let

f = Xn + an-1Xn-1 + ... + a0,   ai 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 E. Then is a zero of f if and only if X - divides f in E[X]; that is, f = (X - )g for some g E[X].

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

f = (X - )g + r,    g,r E[X],

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

f() = ,

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

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

Proof: Let 1,...,k be distinct zeros of f in E. Then, in E[X], f is divisible by (X-1)(X - 2)···(X - k). 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 in E, then f is divisible by X - 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

FF[X]/I,
aa + I

is an isomorphism. Thus, let us identify F with the subring {a + I | a 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 = X + I E. Then

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

so that 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) = (X - i),   i E.

Proof: By Theorem 4, there exists an extension E1 of F in which f(X) has a zero 1. By Proposition 1, f(X) is divisible by X - 1 in E1[X]. Therefore. there exists a monic polynomial g1(X) E1[X] such that f(X) = (X - 1)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, 2 E2 and g2(X) such that g1(X) = (X - 2)g2(X). Therefore, f(X) = (X - 1)(X - 2)g2(X). Proceeding this way, we can construct an extension En of F such that

f = gn(X)(X - i),   gn(X) En[X].

But since n = deg f(X), deg gn(X) = 0, so that gn(X) 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) = (X - i),    i E.

The subfield F(1,...,n) of E gotten by adjoining all zeros of f(X) to F is called the splitting field of f. It is clear that F(1,...,n) 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.