Abstract Algebra Theory and Applications - Computer Science ...

336 CHAPTER 19 FIELDSTheorem 19.19 Let φ : E → F be an isomorphism of fields and let p(x)be a nonconstant polynomial in E[x] and q(x) the corresponding polynomialin F [x] under the isomorphism. If K is a splitting field of p(x) and L is asplitting field of q(x), then φ extends to an isomorphism ψ : K → L.Proof. We will use mathematical induction on the degree of p(x). We canassume that p(x) is irreducible over E. Therefore, q(x) is also irreducibleover F . If deg p(x) = 1, then by the definition of a splitting field, K = Eand L = F and there is nothing to prove.Assume that the theorem holds for all polynomials of degree less thann. Since K is a splitting field of E, all of the roots of p(x) are in K. Chooseone of these roots, say α, such that E ⊂ E(α) ⊂ K. Similarly, we can find aroot β of q(x) in L such that F ⊂ F (β) ⊂ L. By Lemma 19.18, there existsan isomorphism φ : E(α) → F (β) such that φ(α) = β and φ agrees with φon E.ψK−→ L⏐ ⏐↓ ↓E(α)⏐↓Eφ−→φ−→F (β)⏐↓Now write p(x) = (x − α)f(x) and q(x) = (x − β)g(x), where the degreesof 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 asplitting field for g(x) over F (β). By our induction hypothesis there existsan isomorphism ψ : K → L such that ψ agrees with φ on E(α). Hence,there exists an isomorphism ψ : K → L such that ψ agrees with φ on E. □Corollary 19.20 Let p(x) be a polynomial in F [x]. Then there exists asplitting field K of p(x) that is unique up to isomorphism.F19.3 Geometric ConstructionsIn ancient Greece, three classic problems were posed. These problems are geometricin nature and involve straightedge-and-compass constructions fromwhat is now high school geometry; that is, we are allowed to use only astraightedge and compass to solve them. The problems can be stated asfollows.