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