Abstract Algebra Theory and Applications - Computer Science ...

19.1 EXTENSION FIELDS 327Example 6. Let f(x) = x 2 − 2 and g(x) = x 4 − 4x 2 + 1. These polynomialsare the minimal polynomials of √ 2 and √ 2 + √ 3, respectively. Proposition 19.4 Let E be a field extension of F and α ∈ E be algebraicover F . Then F (α) ∼ = F [x]/〈p(x)〉, where p(x) is the minimal polynomial ofα over F .Proof. Let φ α : F [x] → E be the evaluation homomorphism. The kernelof this map is the minimal polynomial p(x) of α. By the First IsomorphismTheorem for rings, the image of φ α in E is isomorphic to F (α) since itcontains both F and α.□Theorem 19.5 Let E = F (α) be a simple extension of F , where α ∈ Eis algebraic over F . Suppose that the degree of α over F is n. Then everyelement β ∈ E can be expressed uniquely in the formfor b i ∈ F .β = b 0 + b 1 α + · · · + b n−1 α n−1Proof. Since φ α (F [x]) = F (α), every element in E = F (α) must be of theform φ α (f(x)) = f(α), where f(α) is a polynomial in α with coefficients inF . Letp(x) = x n + a n−1 x n−1 + · · · + a 0be the minimal polynomial of α. Then p(α) = 0; hence,Similarly,α n+1 = αα nα n = −a n−1 α n−1 − · · · − a 0 .= −a n−1 α n − a n−2 α n−1 − · · · − a 0 α= −a n−1 (−a n−1 α n−1 − · · · − a 0 ) − a n−2 α n−1 − · · · − a 0 α.Continuing in this manner, we can express every monomial α m , m ≥ n, as alinear combination of powers of α that are less than n. Hence, any β ∈ F (α)can be written asβ = b 0 + b 1 α + · · · + b n−1 α n−1 .To show uniqueness, suppose thatβ = b 0 + b 1 α + · · · + b n−1 α n−1 = c 0 + c 1 α + · · · + c n−1 α n−1