Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529

• Sep 28th 2013, 12:03 AM
Bernhard
Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529
Dummit and Foote Exercise 2, Section 13.2, page 529 reads as follows:

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2. Let $g(x) = x^2 + x -1$ and let $h(x) = x^3 - x + 1$. 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 $F = \mathbb F_2$ or $\mathbb F_3$. 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 $f(x) = x^2 + x -1$ and $F = \mathbb F_2$

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

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

$0 + 0. \alpha = 0$,

$1 + 0. \alpha = 1$,

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

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

The multiplication table can then be composed using the following:

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

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

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

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

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

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

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

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 $\alpha$ we find:

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

and

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

Can anyone help?

Peter
• Sep 28th 2013, 09:30 AM
SlipEternal
Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529
In $\mathbb{F}_2$, $1=-1$ and $2=0$. In $\mathbb{F}_3$, you have $3=0$ and $2=-1$. So, your multiplication table for the extension of $\mathbb{F}_2$ by $g(x)$ is misleading since $2+\alpha = 0+\alpha = \alpha$. So, you have:

$\alpha^2 = 1-\alpha = 1+\alpha$
$\alpha^3 = \alpha(1+\alpha) = 1$

Since there are only three non-zero elements, $\alpha$ is indeed the generator of that cyclic group. Next, for the field of 9 elements, you need to extend $\mathbb{F}_3$ by $g(x)$. There will be 8 nonzero elements of that field extension, and they will, indeed, form a cyclic group. Just make sure you remember how addition and multiplication work in the finite fields. It works just like modular arithmetic.
• Sep 30th 2013, 08:00 PM
Bernhard
Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529
Thanks SlipEternal - again, most helpful

Peter