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Moreover the polynomial σ∗ (m) is irreducible over K2 , since σ: K1 → K2 induces an isomorphism between the polynomial rings K1 [x] and K2 [x]. Choose a root β of σ∗ (m). 52 Let g and h be polynomials with coefficients in K1 . Now g(α) = h(α) if and only if m divides g − h. Similarly σ∗ (g)(β) = σ∗ (h)(β) if and only if σ∗ (m) divides σ∗ (g) − σ∗ (h). Therefore σ∗ (g)(β) = σ∗ (h)(β) if and only if g(α) = h(α), and thus there is a well-defined isomorphism ϕ: K1 (α) → K2 (β) which sends g(α) to σ∗ (g)(β) for any polynomial g with coefficients in K.

10). But then z = f (α) since m(α) = 0. Suppose that z = h(α) for some polynomial h ∈ K[x], where either h = 0 or deg h < deg m. Then m divides h−f , since α is a zero of h−f . But if h−f were non-zero then its degree would be less than that of m, and thus h − f would not be divisible by m. We therefore conclude that h = f . Thus any element z of K(α) can be expressed in the form z = f (α) for some uniquely determined polynomial f ∈ K[x] satisfying either f = 0 or deg f < deg m. Thus if n = deg m then 1, α, α2 .

Then S and T are constructible points in the plane, and the point R at which the line ST intersects the line P Q is the midpoint of the line segment P Q. Thus this midpoint is a constructible point. 10 If any three vertices of a parallelogram in the plane are constructible, then so is the fourth vertex. Proof Let the vertices of the parallelogram listed in anticlockwise (or in clockwise) order be A, B, C and D, where A, B and D are constructible points. We must show that C is also constructible.

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Abstract Algebra (Course 311) by Wilkins D.R.


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