Splitting Fields
We can construct the complex numbers C by adjoining to R a zero of the polynomial X^{2} + 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 = X ^{n} + a _{n1}X ^{n1} + ... + a _{0}, a _{i} 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 wellknown 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 f_{1} is an irreducible factor of f in F[X], then every zero of f_{1} 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
F F[X]/I,
a a + 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 E_{1} of F in which f(X) has a zero _{1}. By Proposition 1, f(X) is divisible by X  _{1} in E_{1}[X]. Therefore. there exists a monic polynomial g_{1}(X) E_{1}[X] such that f(X) = (X  _{1})g_{1}(X). Applying the same reasoning to g_{1}(X) that we just applied to f(X), we see that there exists an extension E_{2} of E_{1}, _{2} E_{2} and g_{2}(X) such that g_{1}(X) = (X  _{2})g_{2}(X). Therefore, f(X) = (X  _{1})(X  _{2})g_{2}(X). Proceeding this way, we can construct an extension E_{n} of F such that
f = g _{n}(X) (X  _{i}), g _{n}(X) E _{n}[X].
But since n = deg f(X), deg g_{n}(X) = 0, so that g_{n}(X) E_{n}. However, because f is monic, g_{n}(X) = 1. Therefore, we may set E = E_{n}.
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 f_{1}(X), f_{2}(X),..., f_{r}(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 f_{1}(X) splits into a product of linear factors.
