This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 &gt; 1, and that we have shown the result for all 1 m &lt; 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&lt;pk d ( x ) . Now, 1 Y d  pk d&lt;pk d ( x ) = Y d  k d ( x ) Y d  pk d&lt;pk p m +1 d ( x ) = ( x k 1) Y d  pk d&lt;pk p m +1 d ( x ) = [( x p ) k 1] Y d  pk d&lt;pk p m d ( x p ) = Y d  k d ( x p ) Y d  pk d&lt;pk p m d ( x p ) = Y d  k d&lt;k d ( x p ) , by the induction hypothesis. Therefore, we get pk ( x ) = x pk 1 Q d  pk d&lt;pk d ( x ) = x pk 1 Q d  k d&lt;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 &lt; 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 &lt; 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 Flinear embedding of E into itself. Show that is an automorphism. Solution. Let : E E be the Flinear 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 )) =...
View
Full
Document
This note was uploaded on 05/07/2010 for the course MAT 322 at Princeton.
 '09
 CLAUSSORENSEN
 Algebra

Click to edit the document details