solutions03

solutions03 - Problem Set #3 Solutions Chapter III. 4.1 If...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem Set #3 Solutions Chapter III. 4.1 If f ( x ) F [ x ] and K is a splitting field for f ( x ) over F , denote by S the set of distinct roots of f ( x ) in K and let G = G ( K : F ) . i. If f ( x ) is irreducible over F , then G is transitive on S . Since K is a splitting field for f ( x ), f ( x ) = c Q s S ( x- s ) n s in K [ x ] for some n s Z 1 , c F . Thus 1 c f ( x ) is a minimal polynomial over F since it is irreducible and monic with roots S , so all the s S are conjugate over F . If S consists of a single element G is trivially transitive, so let s,s S . By the corollary to proposition 1.9, since s,s K are conjugate there is an isomorphism : F ( s ) F ( s ) , with ( s ) = s and ( b ) = b for all b F . Notice f ( x ) F [ x ] F ( s )[ x ], ( f ( x )) = f ( x ) F [ x ] F ( s )[ x ], and since K is a splitting field for f ( x ) = ( f ( x )), by theorem 1.10 can be extended to an isomorphism : K K . In particular, ( b ) = b for all b F , so G . But ( s ) = s , so G is transitive. ii. If f ( x ) has no repeated roots and G is transitive on S , then f ( x ) is irre- ducible over F . If | S | = 1 then f ( x ) is linear and hence irreducible. Otherwise, if f ( x ) = Q s S ( x- s ) were reducible over F , then for some subset S S we would have h ( x ) = Q s S ( x- s ) F [ x ]. Extend each G to K [ x ] by setting ( x ) = x (this perserves isomorphism by theorem II.3.4). Then for any G , ( h ( x )) = Q s S ( x- ( s )). If ( S ) = S for all G or ( S ) = S \ S for all G , G would not be transitive, so there exists a such that ( S ) 6 = S ,S \ S . Thus since h ( x ) is a product of distinct linear factors, ( h ( x )) 6 = h ( x ). (Note that if h ( x ) were not a product of distinct linear factors, it is easy to construct an example where this would fail by using exercise 3.1.) But h ( x ) F [ x ] and is an F-automorphism, so ( h ( x )) = h ( x ), a contradiction. 4.2. If K is an n th root of unity, 6 = 1 , then 1 + + 2 + + n- 1 = 0 . This follows since n = 1, 6 = 1, and 1 + x + x 2 + + x n- 1 = x n- 1 x- 1 ....
View Full Document

This note was uploaded on 08/06/2008 for the course MATH 220 taught by Professor Morrison during the Spring '08 term at UCSB.

Page1 / 5

solutions03 - Problem Set #3 Solutions Chapter III. 4.1 If...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online