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

[Math Amateur]

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 notFor 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

[Note; The above has also been posted on MHF]

 

[Deveno]

OK, I think you are a bit confused.

If the base field is $F_2 = \{0,1\} = \Bbb Z_2$ with multiplication and addition modulo 2, and our irreducible polynomial is of degree 2 (so that $[F_2(\alpha):F] = 2$), then viewed as a vector space, $F_2(\alpha)$ is isomorphic to $F_2 \times F_2$:

$0 = 0 + 0\alpha \leftrightarrow (0,0)$
$1 = 1 + 0\alpha \leftrightarrow (1,0)$
$\alpha = 0 + 1\cdot\alpha \leftrightarrow (0,1)$
$\alpha + 1 = 1 + 1\cdot\alpha \leftrightarrow (1,1)$

This clearly has just 4 elements, and thus just THREE non-zero elements. I claim $\alpha$ generates the non-zero elements:

$\alpha^2 = 1 - \alpha$ (since $\alpha^2 + \alpha - 1 = 0$). But in a field $F$ of characteristic 2, $-u = u$ for ALL $u \in F$, since:

$u+u = 1\cdot u + 1\cdot u = (1 + 1)u = 0u = 0$,

Thus $1 - \alpha = 1 + \alpha = \alpha + 1$.

Finally:

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

$= (\alpha + \alpha) + 1 = 0 + 1 = 1$, which shows that $\alpha$ is of order 3.

 

[Math Amateur]

OK, I think you are a bit confused.

If the base field is $F_2 = \{0,1\} = \Bbb Z_2$ with multiplication and addition modulo 2, and our irreducible polynomial is of degree 2 (so that $[F_2(\alpha):F] = 2$), then viewed as a vector space, $F_2(\alpha)$ is isomorphic to $F_2 \times F_2$:

$0 = 0 + 0\alpha \leftrightarrow (0,0)$
$1 = 1 + 0\alpha \leftrightarrow (1,0)$
$\alpha = 0 + 1\cdot\alpha \leftrightarrow (0,1)$
$\alpha + 1 = 1 + 1\cdot\alpha \leftrightarrow (1,1)$

This clearly has just 4 elements, and thus just THREE non-zero elements. I claim $\alpha$ generates the non-zero elements:

$\alpha^2 = 1 - \alpha$ (since $\alpha^2 + \alpha - 1 = 0$). But in a field $F$ of characteristic 2, $-u = u$ for ALL $u \in F$, since:

$u+u = 1\cdot u + 1\cdot u = (1 + 1)u = 0u = 0$,

Thus $1 - \alpha = 1 + \alpha = \alpha + 1$.

Finally:

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

$= (\alpha + \alpha) + 1 = 0 + 1 = 1$, which shows that $\alpha$ is of order 3.

Thanks for the help, Deveno

Just working on this nowPeter(PS sorry for delay ... Commitments to my day job intervened unfortunately :-). )

 

[Math Amateur]

Thanks for the help, Deveno

Just working on this nowPeter(PS sorry for delay ... Commitments to my day job intervened unfortunately :-). )

I have worked through the post.. It was very helpful.

I am assuming that where you write [FONT=MathJax_Math]F[FONT=MathJax_Main]2[FONT=MathJax_Main]([FONT=MathJax_Math]α[FONT=MathJax_Main]) is isomorphic to [FONT=MathJax_Math]F[FONT=MathJax_Main]2[FONT=MathJax_Main]×[FONT=MathJax_Math]F[FONT=MathJax_Main]2 that you are not contesting the fact that $$ \mathbb{F}_2 ( \alpha ) = \{ a_0 + a_1 \alpha \ | \ a_0, a_1 \in \mathbb{F}_2 $$ - but are just expressing the fact that $$ \mathbb{F}_2 ( \alpha ) \cong \mathbb{F}_2 \times \mathbb{F}_2 $$ (I do not think you are contesting this - but just checking with you)

I have used the points you have made regarding the fact that $$ \mathbb{F}_2 $$ has characteristic 2 to upgrade my multiplication table for $$ \mathbb{F}_2 ( \alpha ) $$ - I have attached the multiplication table as I had it before - labelled previous post - and the new multiplication table - labelled new post.

Can someone please confirm that the new multiplication table is correct.

Peter

 

[Deveno]

Yes, it is correct. Note we have two ways to view this finite field:

1) As a vector space over $F_2$, this makes it easy to add elements.

2) As $\{0\} \cup \langle \alpha \rangle$, this makes it easy to multiply elements.

To pass between the two ways of looking at a finite field requires finding a generator, or primitive element, so we can create a "discrete log table". This is not always an easy task.

Now do the same thing with $F_3$, which will be a lot more interesting (and give you a better idea of how bad things can get when $p$ is large in the field $F_p$).