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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Dummit and Foote Exercise 2, Section 13.2, page 529 reads as follows:

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

So we have [TEX] \mathbb F_2 = \{ 0. 1 \}[/TEX] and f(x) as above.

The elements of [TEX] \mathbb F_2 ( \alpha ) [/TEX], then, are as follows: (see attachment)

[TEX]0 + 0. \alpha = 0 [/TEX],

[TEX]1 + 0. \alpha = 1 [/TEX],

[TEX] 0 + 1. \alpha = \alpha [/TEX],

and [TEX] 1 + 1. \alpha = 1 + \alpha [/TEX]


The multiplication table can then be composed using the following:

[TEX] {\alpha}^2 + \alpha - 1 = 0 [/TEX]

That is [TEX] {\alpha}^2 = 1 - \alpha [/TEX] ... ... ... ... ... (1)

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

[TEX] 0 \times 0 = 0 \ , \ 0 \times 1 = 0 \ , \ 0 \times \alpha = 0 \ , \ 0 \times {1 + \alpha} = 0 [/TEX]

[TEX] 1 \times 0 = 0 \ , \ 1 \times 1 = 1 \ , \ 1 \times \alpha = \alpha \ , \ 1 \times (1 + \alpha) = (1 + \alpha) [/TEX]

[TEX] \alpha \times 0 = 0 \ , \ \alpha \times 1 = \alpha \ , \ \alpha \times \alpha = (1 - \alpha) \ , \ \alpha \times (1 + \alpha) = 1 [/TEX]

[TEX] (1 + \alpha) \times 0 = 0 \ , \ (1 + \alpha) \times 1 = (1 + \alpha) \ , (\ 1 + \alpha) \times \alpha = 1 \ , \ (1 + \alpha) \times (1 + \alpha) = (2 + \alpha) [/TEX]


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 [TEX] \alpha [/TEX] we find:

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

and

[TEX] {\alpha}^3 = \alpha (1 - \alpha) = \alpha - {\alpha}^2 = \alpha - (1 - \alpha) = 2\alpha - 1 [/TEX] (?? not an element of the group)

Can anyone help?

Peter

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

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 now


Peter


(PS sorry for delay ... Commitments to my day job intervened unfortunately :). )
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks for the help, Deveno

Just working on this now


Peter


(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][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]α[/FONT][FONT=MathJax_Main])[/FONT] is isomorphic to [FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]2[/FONT] that you are not contesting the fact that \(\displaystyle \mathbb{F}_2 ( \alpha ) = \{ a_0 + a_1 \alpha \ | \ a_0, a_1 \in \mathbb{F}_2 \) - but are just expressing the fact that \(\displaystyle \mathbb{F}_2 ( \alpha ) \cong \mathbb{F}_2 \times \mathbb{F}_2 \) (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 \(\displaystyle \mathbb{F}_2 \) has characteristic 2 to upgrade my multiplication table for \(\displaystyle \mathbb{F}_2 ( \alpha ) \) - 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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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 $F_p$).