## 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.

- Page 357:2,3,6,17,23
- 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(
^{3}√2). Let z denote the cube root of 2, so z is a root of x^{3}-2. Also, let ω be a root of the polynomial x^{2} + x +1.
- Show that ω
^{3}=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.)
- Show that the other roots of x
^{3}-2 are given by zω and zω^{2}.
- Now we are going to find the inverse of a + bz +cz
^{2}, where at least one of a, b or c is non-zero. Step 1: show that the quantity (a + bz +cz^{2})(a + bzω +cz^{2}ω^{2})(a + bzω^{2} +cz^{2}ω) is a rational number.
- 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 +Cz
^{2} 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 +cz^{2}, and multiply numerator and denominator by p(zω)p(zω^{2}). 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).