By Wilkins D.R.

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12 that there exist polynomials u and v with coefficients in K such that 1 = ug + vf . Then h = ugh + vf h. But f divides ugh + vf h, since f divides gh. It follows that f divides h, as required. 15 Let K be a field, and let (f ) be the ideal of K[x] generated by an irreducible polynomial f with coefficients in K. Then K[x]/(f ) is a field. Proof Let I = (f ). Then the quotient ring K[x]/I is commutative and has a multiplicative identity element I +1. Let g ∈ K[x]. Suppose that I +g = I. Now the only factors of f are constant polynomials and constant multiples of f , since f is irreducible.

Thus the problem is closely related to that of expressing the roots of a given polynomial in terms of its coefficients by means of algebraic formulae which involve only finite addition, subtraction, multiplication, division and the successive extraction of pth roots for appropriate prime numbers p. 5 Splitting Fields Definition Let L: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. The polynomial f is said to split over L if f is a constant polynomial or if there exist elements α1 , α2 , .

Therefore any proper normal subgroup of A5 is trivial, and therefore A5 is simple. 18 Solvable Groups The concept of a solvable group was introduced into mathematics by Evariste Galois, in order to state and prove his fundamental general theorems concerning the solvability of polynomial equations. We now investigate the basic properties of such solvable groups. Definition A group G is said to be solvable (or soluble) if there exists a finite sequence G0 , G1 , . . , Gn of subgroups of G, where G0 = {1} and Gn = G, such that Gi−1 is normal in Gi and Gi /Gi−1 is Abelian for i = 1, 2, .