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21.2: Splitting Fields

( \newcommand{\kernel}{\mathrm{null}\,}\)

Let F be a field and p(x) be a nonconstant polynomial in F[x]. We already know that we can find a field extension of F that contains a root of p(x). However, we would like to know whether an extension E of F containing all of the roots of p(x) exists. In other words, can we find a field extension of F such that p(x) factors into a product of linear polynomials? What is the “smallest” extension containing all the roots of p(x)?

Let F be a field and p(x)=a0+a1x++anxn be a nonconstant polynomial in F[x]. An extension field E of F is a splitting field of p(x) if there exist elements α1,,αn in E such that E=F(α1,,αn) and

p(x)=(xα1)(xα2)(xαn).

A polynomial p(x)F[x] splits in E if it is the product of linear factors in E[x].

Example 21.29.

Let p(x)=x4+2x28 be in Q[x]. Then

Solution

p(x) has irreducible factors x22 and x2+4. Therefore, the field Q(2,i) is a splitting field for p(x).

Example 21.30.

Let p(x)=x33 be in Q[x]. Then p(x) has a root in the field Q(33).

Solution

However, this field is not a splitting field for p(x) since the complex cube roots of 3,

33±(63)5i2,

are not in Q(33).

Theorem 21.31.

Let p(x)F[x] be a nonconstant polynomial. Then there exists a splitting field E for p(x).

Proof

We will use mathematical induction on the degree of p(x). If degp(x)=1, then p(x) is a linear polynomial and E=F. Assume that the theorem is true for all polynomials of degree k with 1k<n and let degp(x)=n. We can assume that p(x) is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 21.5, there exists a field K such that p(x) has a zero α1 in K. Hence, p(x)=(xα1)q(x), where q(x)K[x]. Since degq(x)=n1, there exists a splitting field EK of q(x) that contains the zeros α2,,αn of p(x) by our induction hypothesis. Consequently,

E=K(α2,,αn)=F(α1,,αn)

is a splitting field of p(x).

The question of uniqueness now arises for splitting fields. This question is answered in the affirmative. Given two splitting fields K and L of a polynomial p(x)F[x], there exists a field isomorphism ϕ:KL that preserves F. In order to prove this result, we must first prove a lemma.

Theorem 21.32.

Let ϕ:EF be an isomorphism of fields. Let K be an extension field of E and αK be algebraic over E with minimal polynomial p(x). Suppose that L is an extension field of F such that β is root of the polynomial in F[x] obtained from p(x) under the image of ϕ. Then ϕ extends to a unique isomorphism ¯ϕ:E(α)F(β) such that ¯ϕ(α)=β and ¯ϕ agrees with ϕ on E.

Proof

If p(x) has degree n, then by Theorem 21.13 we can write any element in E(α) as a linear combination of 1,α,,αn1. Therefore, the isomorphism that we are seeking must be

¯ϕ(a0+a1α++an1αn1)=ϕ(a0)+ϕ(a1)β++ϕ(an1)βn1,

where

a0+a1α++an1αn1

is an element in E(α). The fact that ¯ϕ is an isomorphism could be checked by direct computation; however, it is easier to observe that ¯ϕ is a composition of maps that we already know to be isomorphisms.

We can extend ϕ to be an isomorphism from E[x] to F[x], which we will also denote by ϕ, by letting

ϕ(a0+a1x++anxn)=ϕ(a0)+ϕ(a1)x++ϕ(an)xn.

This extension agrees with the original isomorphism ϕ:EF, since constant polynomials get mapped to constant polynomials. By assumption, ϕ(p(x))=q(x); hence, ϕ maps p(x) onto q(x). Consequently, we have an isomorphism ψ:E[x]/p(x)F[x]/q(x). By Proposition 21.12, we have isomorphisms σ:E[x]/p(x)E(α) and τ:F[x]/q(x)F(β), defined by evaluation at α and β, respectively. Therefore, ¯ϕ=τψσ1 is the required isomorphism (see Figure 21.33).

clipboard_efac96b6fb89d79f1f30f7e405036f7a9.png

Figure 21.33.

We leave the proof of uniqueness as a exercise.

 

Theorem 21.34.

Let ϕ:EF be an isomorphism of fields and let p(x) be a nonconstant polynomial in E[x] and q(x) the corresponding polynomial in F[x] under the isomorphism. If K is a splitting field of p(x) and L is a splitting field of q(x), then ϕ extends to an isomorphism ψ:KL.

Proof

We will use mathematical induction on the degree of p(x). We can assume that p(x) is irreducible over E. Therefore, q(x) is also irreducible over F. If degp(x)=1, then by the definition of a splitting field, K=E and L=F and there is nothing to prove.

Assume that the theorem holds for all polynomials of degree less than n. Since K is a splitting field of p(x), all of the roots of p(x) are in K. Choose one of these roots, say α, such that EE(α)K. Similarly, we can find a root β of q(x) in L such that FF(β)L. By Lemma 21.32, there exists an isomorphism ¯ϕ:E(α)F(β) such that ¯ϕ(α)=β and ¯ϕ agrees with ϕ on E (see Figure 21.35).

clipboard_ed62e8e883f02c4e2e9f57520d6deda9f.png

Figure 21.35.

Now write p(x)=(xα)f(x) and q(x)=(xβ)g(x), where the degrees of f(x) and g(x) are less than the degrees of p(x) and q(x), respectively. The field extension K is a splitting field for f(x) over E(α), and L is a splitting field for g(x) over F(β). By our induction hypothesis there exists an isomorphism ψ:KL such that ψ agrees with ¯ϕ on E(α). Hence, there exists an isomorphism ψ:KL such that ψ agrees with ϕ on E.

 

Corollary 21.36.

Let p(x) be a polynomial in F[x]. Then there exists a splitting field K of p(x) that is unique up to isomorphism.


This page titled 21.2: Splitting Fields is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Thomas W. Judson (Abstract Algebra: Theory and Applications) via source content that was edited to the style and standards of the LibreTexts platform.

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