solutions01

# solutions01 - Problem Set #1 Solutions Chapter II. 1.2. Let...

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

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.

Unformatted text preview: Problem Set #1 Solutions Chapter II. 1.2. Let A denote the algebraic numbers, i.e., the set of a ∈ C which are algebraic over Q . Then [ A : Q ] = ∞ . Proof. Suppose to the contrary that [ A : Q ] = n . Let a = 2 1 /n +1 be a root of f ( x ) = x n +1- 2 ∈ Q [ x ]; then a is an element of A . We claim f ( x ) is irreducible: f ( x ) ∈ Z [ x ] is primitive, Z is a PID, and letting p = 2 we have p | (- 2), p 2 6 | (- 2), and p 6 | 1 (note we cannot do this over Q since 2 is a unit, and hence not a prime in Q ). Moreover, since Z is UFD and Q is its fraction field, f ( x ) is also irreducible when viewed as an element in Q [ x ] by proposition II.5.15. Thus [ Q ( a ) : Q ] = deg( f ( x )) = n + 1 by proposition 1.3. But A is an extension field of Q ( a ), and so by proposition 1.1, [ A : Q ] = [ A : Q ( a )][ Q ( a ) : Q ] = [ A : Q ( a )] · ( n + 1) > n, a contradiction. 1.4 Suppose K is an algebraic extension of F . Then if F is a finite field, K is finite or countable....
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 / 2

solutions01 - Problem Set #1 Solutions Chapter II. 1.2. Let...

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

View Full Document
Ask a homework question - tutors are online