HW 07 - MAT 322: Algebra with Galois Theory Professor Claus...

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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 7 Rik Sengupta November 20, 2009 1. Let n = p m 1 1 ...p m t t be the prime factorization. Prove that n ( x ) = p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t ) . Solution. We will first show that if p | k , then pk ( x ) = k ( x p ), using induction on k . Obviously the base case is trivial, because if k = 1, then there is nothing to show. So suppose k > 1, and that we have shown the result for all 1 m < k . Now, write k = p m k , where ( p,k ) = 1. Then, we know x pk- 1 = Y d | pk d ( x ) , so it follows that pk ( x ) = x pk- 1 Q d | pk d<pk d ( x ) . Now, 1 Y d | pk d<pk d ( x ) = Y d | k d ( x ) Y d | pk d<pk p m +1 d ( x ) = ( x k- 1) Y d | pk d<pk p m +1 d ( x ) = [( x p ) k- 1] Y d | pk d<pk p m d ( x p ) = Y d | k d ( x p ) Y d | pk d<pk p m d ( x p ) = Y d | k d<k d ( x p ) , by the induction hypothesis. Therefore, we get pk ( x ) = x pk- 1 Q d | pk d<pk d ( x ) = x pk- 1 Q d | k d<k d ( x p ) = k ( x p ) . Now, if we let k = p 1 ...p t , applying this result m i- 1 times to each p i shows our desired result: n ( x ) = p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t ) . Alternative Solution. We know deg( n ( x )) = ( n ) = n (1- 1 p 1 ) ... (1- 1 p t ) = ( p 1- 1) ... ( p t- 1) p m 1- 1 1 ...p m t- 1 t . and deg( p 1 ...p t ( x )) = ( p 1 ...p t ) = ( p 1- 1) ... ( p t- 1) , 2 i.e. deg( p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t )) = ( p 1- 1) ... ( p t- 1) p m 1- 1 1 ...p m t- 1 t = deg( n ( x )) . Furthermore, it is easy to see, because n is always a monic polynomial, and the arguments in both these cases have a leading coefficient of 1, both the polynomials are monic. Now, let be a primitive n th root of unity. We claim that p m 1- 1 1 ...p m t- 1 t is a primitive p 1 ...p t th root of unity. This is because ( p m 1- 1 1 ...p m t- 1 t ) p 1 ...p t = p m 1 1 ...p m t t = n = 1 , and for any k N , where k < p 1 ...p t , we have ( p m 1- 1 1 ...p m t- 1 t ) k 6 = 1, as p m 1- 1 1 ...p m t- 1 t k < n . Hence, from the definition of the cyclotomic polynomials, it follows that p m 1- 1 1 ...p m t- 1 t is a root of p 1 ...p t ( x ). However, this simply means that is a root of p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t ). Hence, p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t ) has has a root, is monic, and has the same degree as Irr( , Q ,x ) = n ( x ), and therefore we must have n ( x ) = p 1 ...p t ( x p m 1- 1 1 ...p m t- 1 t ) . 2. Let E be an algebraic extension of F , and let : E E be an F-linear embedding of E into itself. Show that is an automorphism. Solution. Let : E E be the F-linear embedding. Then, ( F ) = F ( E ). Now, given any a E , since E/F is algebraic, there must be some f F [ x ] such that f ( a ) = 0. We apply to the coefficients of f , and ob- tain ( f ) = f , since ( F ) = F . Then, f ( ( a )) =...
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HW 07 - MAT 322: Algebra with Galois Theory Professor Claus...

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