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**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...

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