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Unformatted text preview: Math 220C Spring 2008 Midterm Solutions 1. Let L be a subfield of the complex numbers C such that [ L : Q ] = 2. Let α ∈ Q such that √ α 6∈ Q and Q ( √ α ) 6 = L . Let M be the smallest subfield of C containing both Q ( √ α ) and L . Show that [ M : Q ] = 4. How many intermediate subfields are there? Solution : Let β ∈ L \ Q . The minimal polynomial p ( x ) of β over Q has degree 2, and since β 6∈ Q ( √ α ), p ( x ) must be irreducible in Q ( √ α )[ x ]. The field M = Q ( √ α,β ) is obtained by adjoining a root of p ( x ) to Q ( √ α ), so it has degree 2 over Q ( √ α ) and hence has degree 4 over Q . This is a Galois extension with Galois group Z 2 × Z 2 ; the intermediate fields are L , Q ( √ α ), and a third field fixed by the diagonal Z 2 in the product Z 2 × Z 2 . (You can identify that third field more explicitly by completing the square on p ( x ) = x 2 + ax + b and introducing β = β + a 2 which has the property that ( β ) 2 = a 2 4 b...
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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
 MORRISON
 Algebra, Complex Numbers

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