HW11_selected_solns

# HW11_selected_solns - College of the Holy Cross, Fall 2009...

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: College of the Holy Cross, Fall 2009 Math 351 – Homework 11 selected solutions Chapter 21, #14. By calculating ( √- 3 + √ 2) 4 and ( √- 3 + √ 2) 2 , one sees that ( √- 3 + √ 2) satisfies the polynomial x 4 + 2 x 2 + 25. There are two ways one can show this is the minimal polynomial for ( √- 3 + √ 2): the first is to show it’s irreducible over Q , and the second is to show that ( √- 3 + √ 2) cannot be a root of a polynomial of lower degree. In this case, it’s very hard to show that x 4 + 2 x 2 + 25 is irreducible over Q . However, using the method of Example 6 on p.374, one can show that Q ( √- 3 + √ 2) = Q ( √- 3 , √ 2), and the latter field has degree 4 over Q . Thus [ Q ( √- 3+ √ 2) : Q ] = 4, and it follows that the minimal polynomial of √- 3+ √ 2 over Q has degree 4. Thus the minimal polynomial is in fact x 4 + 2 x 2 + 25. Chapter 21, #21. Assume that both αβ and α + β are algebraic over Q . Then Q ( αβ,α + β ) is an algebraic extension of...
View Full Document

## This document was uploaded on 11/03/2010.

Ask a homework question - tutors are online