{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ma5c-hw7-soln

# ma5c-hw7-soln - M a 5c HOMEWORK 7 SOLUTION SPRING 09 The...

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

Ma 5 c HOMEWORK 7 SOLUTION SPRING 09 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummit and Foote, unless stated otherwise. Page 582, 17 . c) Gal ( K/F ) = { 1 , σ } , with σ ( a + b D ) = a - b D . Therefore N K/F ( a + b D ) = ( a + b D )( a - b D ) = a 2 - b 2 D . d) Since [ F ( α ) : F ] = d , d | n follows from the tower law. α has d conjugates since the minimal polynomial is of degree d . Suppose H 0 is the subgroup of G corresponding to F ( α ), then H H 0 of index k = n d . Therefore, α is fixed by k embeddings of K , and each conjugate α i appears k times in Q σ ( α ). Furthermore, Q α i = ( - 1) d a o . Therefore N K/F ( α ) = Q σ ( α i ) = Q ( α i ) k = ( - 1) n ( a 0 ) k . 2.a) The minimal polynomial for 1 - ζ is Φ p r (1 - x ) = (1 - x ) p r - 1 (1 - x ) p r - 1 - 1 = 1 + t + · · · + t p - 1 , where t = (1 - x ) p r - 1 . N (1 - ζ ) = Φ p r (1 - 0) = p . b)Suppose n = p r m , where p doesn’t divide m , q | m for some other prime q . then Φ n ( x ) = Φ m ( x p r ) Φ m ( x p r - 1 ) , since the numberator and denomina- tor each has the same number of terms, Φ n (1 - x ) has constant term 1. Therefore N (1 - ζ ) = 1.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

ma5c-hw7-soln - M a 5c HOMEWORK 7 SOLUTION SPRING 09 The...

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

View Full Document
Ask a homework question - tutors are online