Abstract Algebra Theory and Applications - Computer Science ...

EXERCISES 383[K : R] must be odd. Consequently, K = R(β) with β having a minimalpolynomial f(x) of odd degree. Therefore, K = R.We now know that G(L/R) must be a 2-group. It follows that G(L/C)is a 2-group. We have assumed that L ≠ C; therefore, |G(L/C)| ≥ 2. By thefirst Sylow Theorem and the Fundamental Theorem of Galois Theory, thereexists a subgroup G of G(L/C) of index 2 and a field E fixed elementwiseby G. Then [E : C] = 2 and there exists an element γ ∈ E with minimalpolynomial x 2 +bx+c in C[x]. This polynomial has roots (−b± √ b 2 − 4c )/2that are in C, since b 2 − 4c is in C. This is impossible; hence, L = C. □Although our proof was strictly algebraic, we were forced to rely onresults from calculus. It is necessary to assume the completeness axiomfrom analysis to show that every polynomial of odd degree has a real rootand that every positive real number has a square root. It seems that thereis no possible way to avoid this difficulty and formulate a purely algebraicargument. It is somewhat amazing that there are several elegant proofs ofthe Fundamental Theorem of Algebra that use complex analysis. It is alsointeresting to note that we can obtain a proof of such an important theoremfrom two very different fields of mathematics.Exercises1. Compute each of the following Galois groups. Which of these field extensionsare normal field extensions? If the extension is not normal, find a normalextension of Q in which the extension field is contained.(a) G(Q( √ 30 )/Q)(c) G(Q( √ 2, √ 3, √ 5 )/Q)(e) G(Q( √ 6, i)/Q)(b) G(Q( 4√ 5 )/Q)(d) G(Q( √ 2, 3√ 2, i)/Q)2. Determine the separability of each of the following polynomials.(a) x 3 + 2x 2 − x − 2 over Q(c) x 4 + x 2 + 1 over Z 3(b) x 4 + 2x 2 + 1 over Q(d) x 3 + x 2 + 1 over Z 23. Give the order and describe a generator of the Galois group of GF(729)over GF(9).4. Determine the Galois groups of each of the following polynomials in Q[x];hence, determine the solvability by radicals of each of the polynomials.