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

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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 Φ n ( x ) = Φ 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 Φ pk ( x ) = Φ 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 0 , where ( p, k 0 ) = 1. Then, we know x pk - 1 = Y d | pk Φ d ( x ) , so it follows that Φ pk ( x ) = x pk - 1 Q d | pk d<pk Φ d ( x ) . Now, 1

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Y d | pk d<pk Φ d ( x ) = Y d | k Φ d ( x ) Y d | pk 0 d<pk 0 Φ p m +1 d ( x ) = ( x k - 1) Y d | pk 0 d<pk 0 Φ p m +1 d ( x ) = [( x p ) k 0 - 1] Y d | pk 0 d<pk 0 Φ p m d ( x p ) = Y d | k 0 Φ d ( x p ) Y d | pk 0 d<pk 0 Φ p m d ( x p ) = Y d | k d<k Φ d ( x p ) , by the induction hypothesis. Therefore, we get Φ pk ( x ) = x pk - 1 Q d | pk d<pk Φ d ( x ) = x pk - 1 Q d | k d<k Φ d ( x p ) = Φ 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: Φ n ( x ) = Φ p 1 ...p t ( x p m 1 - 1 1 ...p m t - 1 t ) . Alternative Solution. We know deg(Φ n ( x )) = ϕ ( 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(Φ p 1 ...p t ( x )) = ϕ ( p 1 . . . p t ) = ( p 1 - 1) . . . ( p t - 1) , 2
i.e. deg(Φ 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(Φ n ( x )) . Furthermore, it is easy to see, because Φ 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 ζ be a primitive n th root of unity. We claim that ζ p m 1 - 1 1 ...p m t - 1 t is a primitive p 1 . . . p t th root of unity. This is because ( ζ p m 1 - 1 1 ...p m t - 1 t ) p 1 ...p t = ζ p m 1 1 ...p m t t = ζ n = 1 , and for any k N , where k < p 1 . . . p t , we have ( ζ 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 ζ p m 1 - 1 1 ...p m t - 1 t is a root of Φ p 1 ...p t ( x ). However, this simply means that ζ is a root of Φ p 1 ...p t ( x p m 1 - 1 1 ...p m t - 1 t ). Hence, Φ p 1 ...p t ( x p m 1 - 1 1 ...p m t - 1 t ) has ζ has a root, is monic, and has the same degree as Irr( ζ, Q , x ) = Φ n ( x ), and therefore we must have Φ n ( x ) = Φ 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 σ : E E be an F -linear embedding of E into itself. Show that σ is an automorphism. Solution. Let σ : E E be the F -linear embedding. Then, σ ( F ) = F σ ( 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 σ to the coefficients of f , and ob- tain σ ( f ) = f , since σ ( F ) = F . Then, f ( σ ( a )) = σ ( f )( σ ( a )) = σ ( f ( a )) = 0. So, σ sends a to another root of f . Since σ is a homomorphism between fields, it must necessarily be injective. Hence, σ permutes the roots of f . In particular, there is a root b of f such that σ ( b ) = a . Hence σ is surjective. Since σ is injective, it must be bijective, and hence it is an automorphism.

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

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