Abstract Algebra Theory and Applications - Computer Science ...

19.2 SPLITTING FIELDS 335Lemma 19.18 Let φ : E → F be an isomorphism of fields. Let K be anextension field of E and α ∈ K be algebraic over E with minimal polynomialp(x). Suppose that L is an extension field of F such that β is root of thepolynomial in F [x] obtained from p(x) under the image of φ. Then φ extendsto a unique isomorphism ψ : E(α) → F (β) such that ψ(α) = β and ψ agreeswith φ on E.Proof. If p(x) has degree n, then by Theorem 19.5 we can write anyelement in E(α) as a linear combination of 1, α, . . . , α n−1 . Therefore, theisomorphism that we are seeking must beψ(a 0 + a 1 α + · · · + a n−1 α n−1 ) = φ(a 0 ) + φ(a 1 )β + · · · + φ(a n−1 )β n−1 ,wherea 0 + a 1 α + · · · + a n−1 α n−1is an element in E(α). The fact that ψ is an isomorphism could be checked bydirect computation; however, it is easier to observe that ψ is a compositionof maps that we already know to be isomorphisms.We can extend φ to be an isomorphism from E[x] to F [x], which we willalso denote by φ, by lettingφ(a 0 + a 1 x + · · · + a n x n ) = φ(a 0 ) + φ(a 1 )x + · · · + φ(a n )x n .This extension agrees with the original isomorphism φ : E → F , sinceconstant polynomials get mapped to constant polynomials. By assumption,φ(p(x)) = q(x); hence, φ maps 〈p(x)〉 onto 〈q(x)〉. Consequently, we havean isomorphism φ : E[x]/〈 p(x)〉 → F [x]/〈 q(x)〉. By Theorem 19.4, we haveisomorphisms σ : E[x]/〈p(x)〉 → F (α) and τ : F [x]/〈q(x)〉 → F (β), definedby evaluation at α and β, respectively. Therefore, ψ = τ −1 φσ is the requiredisomorphism.ψE(α)−→ F (β)⏐⏐↓σ↓τE[x]/〈p(x)〉⏐↓Eφ−→φ−→F [x]/〈q(x)〉⏐↓FWe leave the proof of uniqueness as a exercise.□