## Homework assignments for Math 28b, Spring 2003

### Problem Set 6. Due Thursday, April 3

Read Chapter 20, through p. 350 (example 7). We will be doing Chapter 21, through p. 366, so you may want to look at that as well.

1. Page 357:2,3,6,17,23
2. In this problem, you will work through something like problem 17, but a little more general. The goal is to see explicitly how to find inverses in Q(32). Let z denote the cube root of 2, so z is a root of x3-2. Also, let ω be a root of the polynomial x2 + x +1.
1. Show that ω3=1. You may find it useful to have a formula for ω but it is not necessary. (Note, however, that ω is a complex number.)
2. Show that the other roots of x3-2 are given by zω and zω2.
3. Now we are going to find the inverse of a + bz +cz2, where at least one of a, b or c is non-zero. Step 1: show that the quantity (a + bz +cz2)(a + bzω +cz2ω2)(a + bzω2 +cz2ω) is a rational number.
4. Finish the problem, by using the result from the previous step to give a formula for the inverse. Your formula should look like A + Bz +Cz2 where A,B, and C are rationals. (Ie the ω s have disappeared.)
Discussion of this last problem: The point of the construction is to generalize our method for rationalizing the denominator in expressions of the form 1/(a+b √n). In that case, we multiplied numerator and denominator by a-b √n, and used the difference-of-two-squares identity to eliminate the roots in the denominator. In this case, we start with the polynomial p(z) = a + bz +cz2, and multiply numerator and denominator by p(zω)p(zω2). In the problem, you verify by direct calculation that this works. The real explanation for this procedure lies in Galois theory. A bit of this is described in chapter 32 (see page 551, especially).