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Math Help - annihilating polynomial

  1. #1
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    annihilating polynomial

    I'm trying to understand the following worked problem:




    So, \alpha^4 = \alpha^2 . \alpha^2 = (1+x^2)(1+x^2)=1+2x^2+x^4. And did they subtract m(x)=x^4+x^3+x^2+x+1 from \alpha^4 to get x+x^2+x^3 because the result had degree 4? Does this mean every time the degree gets larger than or equal to 4, we must subtract m(x) from it (to have everything in mod m(x))?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: annihilating polynomial

    More exactly is...

    \alpha^{4} = \alpha^{2}\ \alpha^{2}= (1+x^{2})\ (1+x^{2})= 1 + x^{4}= x + x^{2} + x^{3}\ \text{mod}\ m(x)

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor

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    Re: annihilating polynomial

    it IS kind of confusing because when the author says " \alpha = x+1" what he really means is:

    \alpha is the coset of x+1 in K. and in K, multiples of m(x) are set equal to 0.

    so in K, (remembering "x" in K is really x + m(x)), x^4 = x^3 + x^2 + x + 1

    (more precisely, the coset x^4 + (m(x)) is the same coset as x^3 + x^2 + x + 1, since

    x^4 - (x^3 + x^2 + x + 1) = x^4 + (-1)(x^3 + x^2 + x + 1) = x^4 + x^3 + x^2 + x + 1

     = m(x) \in (m(x)) = 0 + (m(x)), the identity of \mathbb{Z}_2[x]/(m(x))

    with this (unfortunately standard) abuse of notation:

    (\alpha)^4 = (x+1)^4 = x^4 + 1 (since 4 is a power of 2, the x^3, x^2 and x terms are all 0)

    = x^3 + x^2 + x + 1 + 1 = x^3 + x^2 + x.
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