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

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

Since there are only three non-zero elements, $\displaystyle \alpha$ is indeed the generator of that cyclic group. Next, for the field of 9 elements, you need to extend $\displaystyle \mathbb{F}_3$ by $\displaystyle g(x)$. 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