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Let F be an algebraic extension of the field K. Then F is said to be a normal extension of K if every irreducible polynomial in K[x] that contains a root in F is a product of linear factors in F [x]. With this definition, the following theorem and its corollary can be proved from previous results. 6 are satisfied. 6 The following are equivalent for an extension field F of K: (1) F is the splitting field over K of a separable polynomial; (2) K = F G for some finite group G of automorphisms of F ; (3) F is a finite, normal, separable extension of K.

7). 1 1 −1 0 Solution: Let a = and b = . Then it is easy to check that 0 1 0 1 1 1 −1 0 1 1 a has order 4 and b has order 2. Since aba = = 0 1 0 1 0 1 −1 −1 1 1 −1 0 = b, we have the identity ba = a−1 b. = 0 1 0 1 0 1 Finally, each element has the form ai b, so the group is isomorphic to D4 .    1 0 0  12. Let G be the subgroup of GL3 (Z2 ) defined by the set  a 1 0  such   b c 1 that a, b, c ∈ Z2 . Show that G is isomorphic to a known group of order 8. Solution: The computation in Review Problem 9 shows the following.

This theorem states that for every positive integer n, the Galois group of the nth cyclotomic polynomial Φn (x) over Q is isomorphic to Z× n . 3 shows more generally that if K is a field of characteristic zero that contains all nth roots of unity, a ∈ K, and F is the splitting field of xn − a over K, then Gal(F/K) is a cyclic group whose order is a divisor of n. We have finally reached our goal, stated in the following two theorems. 6. Let f (x) be a polynomial over a field K of characteristic zero.