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 + a_n-1X^n-1 + ... + a₀,   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 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 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 f₁(X), f₂(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 f₁(X) splits into a product of linear factors.