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

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    Super Member Bernhard's Avatar
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    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
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    Last edited by Bernhard; September 28th 2013 at 12:23 AM.
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    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.
    Thanks from Bernhard
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529

    Thanks SlipEternal - again, most helpful

    Peter
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