Dummit and Foote Exercise 2, Section 13.2, page 529 reads as follows:

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2. Let $\displaystyle g(x) = x^2 + x -1 $ and let $\displaystyle h(x) = x^3 - x + 1 $. Obtain fields of 4, 8, 9 and 27 elements by adjoining a root of f(x) to a field F where f(x) = g(x) or h(x) and $\displaystyle F = \mathbb F_2 $ or $\displaystyle \mathbb F_3 $. Write down the multiplication tables for for the fields with 4 and 9 elements and show that the non-zero elements form a cyclic group.

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In my first attempt at this exercise I took $\displaystyle f(x) = x^2 + x -1 $ and $\displaystyle F = \mathbb F_2 $

So we have $\displaystyle \mathbb F_2 = \{ 0. 1 \}$ and f(x) as above.

The elements of $\displaystyle \mathbb F_2 ( \alpha ) $, then, are as follows:(see attachment)

$\displaystyle 0 + 0. \alpha = 0 $,

$\displaystyle 1 + 0. \alpha = 1 $,

$\displaystyle 0 + 1. \alpha = \alpha $,

and $\displaystyle 1 + 1. \alpha = 1 + \alpha $

The multiplication table can then be composed using the following:

$\displaystyle {\alpha}^2 + \alpha - 1 = 0 $

That is $\displaystyle {\alpha}^2 = 1 - \alpha $ ... ... ... ... ... (1)

So the multiplication table, composed using (1) is as follows:(see attachment)

$\displaystyle 0 \times 0 = 0 \ , \ 0 \times 1 = 0 \ , \ 0 \times \alpha = 0 \ , \ 0 \times {1 + \alpha} = 0 $

$\displaystyle 1 \times 0 = 0 \ , \ 1 \times 1 = 1 \ , \ 1 \times \alpha = \alpha \ , \ 1 \times (1 + \alpha) = (1 + \alpha) $

$\displaystyle \alpha \times 0 = 0 \ , \ \alpha \times 1 = \alpha \ , \ \alpha \times \alpha = (1 - \alpha) \ , \ \alpha \times (1 + \alpha) = 1 $

$\displaystyle (1 + \alpha) \times 0 = 0 \ , \ (1 + \alpha) \times 1 = (1 + \alpha) \ , (\ 1 + \alpha) \times \alpha = 1 \ , \ (1 + \alpha) \times (1 + \alpha) = (2 + \alpha) $

So far so good (i think?) but when I test whether the 9 non-zero elements in the multiplication table form a cyclic group they do not

For example if you try a cyclic group for the non-zero elements based on $\displaystyle \alpha $ we find:

$\displaystyle {\alpha}^2 = 1 - \alpha $ (?? should generate another member of the group but does not!)

and

$\displaystyle {\alpha}^3 = \alpha (1 - \alpha) = \alpha - {\alpha}^2 = \alpha - (1 - \alpha) = 2\alpha - 1 $ (?? not an element of the group)

Can anyone help?

Peter