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- Jun 22, 2012

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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:

[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:

[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]

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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.

--------------------------------------------------------------------------------------------------------------------

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]

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