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
Unformatted text preview: Math 5126 Monday, February 18 Fifth Homework Solutions 1. Section 13.1, Exercise 2 on page 519. Show that x 3 2 x 2 is irreducible over Q and let θ be a root. Compute ( 1 + θ )( 1 + θ + θ 2 ) and ( 1 + θ ) / ( 1 + θ + θ 2 ) in Q ( θ ) . If x 3 2 x 2 was not irreducible, then it would have a degree one factor, and then by Gauss’s lemma we could assume that this factor was of the form x a where a ∈ Z and a 2. Thus ± 1 or ± 2 would be a root, which is clearly not the case. From x 3 2 x 2 = 0, we see that θ 3 = 2 θ + 2. Therefore ( 1 + θ )( 1 + θ + θ 2 ) = θ 3 + 2 θ 2 + 2 θ + 1 = 2 θ + 2 + 2 θ 2 + 2 θ + 1 = 2 θ 2 + 4 θ + 3 . Let us write ( 1 + θ ) / ( 1 + θ + θ 2 ) = a + b θ + c θ 2 , where a , b , c ∈ Q . Then we have 1 + θ = c θ 4 + ( b + c ) θ 3 + ( a + b + c ) θ 2 + ( a + b ) θ + a . Using x 3 2 x 2 = 0, we see that θ 3 = 2 θ + 2 and θ 4 = 2 θ 2 + 2 θ . Substituting back in the above, we obtain 1 + θ = ( a + 2 b + 2 c )+( a + 3 b + 4...
View
Full
Document
This note was uploaded on 01/29/2009 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

Click to edit the document details