By Paulo Ribenboim

The exposition of the classical thought of algebraic numbers is apparent and thorough, and there is a huge variety of workouts in addition to labored out numerical examples. A cautious examine of this booklet will supply an outstanding historical past to the educational of newer topics.

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Xn) E Kn such that j(x1, .. , Xn) i 0. Hint: Proceed by recurrence on n. 28 2. Commutative Fields 24. Let V be a vector space of dimension n over K; let W 1 , ... , lt1m be subspaces of V, distinct from V. Show that if K is an infinite field then W1 U · · · U Wm =J V. Hint: Use the previous exercise. 25. Prove the theorem of the primitive element: If L is a separable extension of finite degree over a field K, then there exists an element t E L such that L = K(t). Hint: Consider first the case when K is finite; then, letting K be infinite, consider the sets {x E L I a-i(x) = a-j(x)} where a-i, O"j are distinct K-isomorphisms from L into an algebraic closure of K; conclude using the previous exercise.

Rt, ... 12. Discriminant and Resultant of Polynomials 21 Another expression for the discriminant may be obtained in terms of the K-isomorphisms 0"1, ... , O"n of L: discrLIK(x1, ... , Xn) = [det(CJi(xj))] 2. In order that x 1, ... r1, ... , Xn) I 0. Consider the special basis { 1, t, ... , tn- 1} where t is a primitive element, L = K(t); we obtain discrLIK(1, t, ... , tn- 1) = IJ(ti- tj) 2, i

Prove that the subfield generated by the union of all Abelian extensions of K is an Abelian extension K' of K; it is called the Abelian closure of K. Exercises 31 53. Let A be an integral domain and let R(f, g) be the resultant of polynomials J, g E A[X] as defined in Chapter 2, Section 11. Denote n = deg J, m = deg g. Show: (a) R(g,f) = (-1)mnR(f,g). (b) If hE A[X] then R(f, gh) = R(f, g)· R(f, h). (c) If f(X) = ao TI~= 1 (X- ai) and g(X) = bo TI7'= 1 (X- f3J) then n R(f,g) = a0 IJg(ai) = (-l)mnb 0 j=l i=l = a0 b0 m IJ f(f3J) n m II II (ai- f3J)· i=l j=l 54.