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solutions02 - Problem Set#2 Solutions Chapter III 2.2 1 If...

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Problem Set #2 Solutions Chapter III. 2.2. 1. If F E K and E is stable, then G E G = G ( K : F ) . Proof. Let θ ∈ G E , φ G , and a E . Since E is stable, φ ( a ) E , and so θφ ( a ) = φ ( a ). Thus φ - 1 θφ ( a ) = φ - 1 φ ( a ) = a , so φ - 1 θφ ∈ G E . 2. If H G , then F H is stable. Proof. Let φ G = G ( K : F ) and θ H . Since H is a normal subgroup of G we have φ - 1 θφ H , and so φ - 1 θφ ( a ) = a whenever a ∈ F H . But then θφ ( a ) = φ ( a ), so φ ( a ) ∈ F H (since θ H was abitrary), and hence F H is stable (since a ∈ F H was abitrary). 3.1. Suppose F is a field of characteristic p > 0 . 1. ( a + b ) p = a p + b p and ( a - b ) p = a p - b p for all a, b F . Proof. We have ( a ± b ) p = p i =0 p i a i ( ± b ) p - i . Furthermore, p | p ! i !( p - i )! = p i whenever i = 0 , p, since p is prime and 0 < i, p - i < p (that is, p divides the numerator but not the denominator). But char( F ) = p so p · 1 F = 0, and hence all the middle terms of ( a ± b ) p vanish. If p is odd, ( - 1) p = - 1; otherwise p = 2, in which case ( - 1) p = ( - 1) 2 = 1 = - 1. Thus either way ( a ± b ) p = a p + ( ± b ) p = a p ± b p . 2. Let K = F ( t ) for an indeterminate t and let f ( x ) = x p - t K [ x ] . Then f ( x ) is irreducible with just one root with multiplicity p in any splitting field. Proof. First note that F [ t ] is a PID, t F is a nonzero nonunit in F [ t ] and hence prime, and f ( x ) is primitive in ( F [ t ])[ x ]. Thus by Eisensteins criterion, f ( x ) is irreducible in ( F [ t ])[ x ]. But F [ t ] is also a UFD (since PID’s are always UFD’s, by prop II.5.10), so by proposition II.5.15
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