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Math Help - How to use Euclidian Algorithm to find multiplicative inverse of a polynomial??

  1. #1
    Member elninio's Avatar
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    How to use Euclidian Algorithm to find multiplicative inverse of a polynomial??

    For example, finding the multiplicative inverse of  [x] in Z_5[x]/<x^2+x+1>

    or even something like

    [a+bx] in R[x]/<x^2+1>

    I cant seem to figure out this concept.

    Thank you!
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  2. #2
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    Quote Originally Posted by elninio View Post
    For example, finding the multiplicative inverse of  [x] in Z_5[x]/<x^2+x+1>

    or even something like

    [a+bx] in R[x]/<x^2+1>

    I cant seem to figure out this concept.

    Thank you!

    Divide x^2+1 by bx+a with residue:

    x^2+1=(bx+a)(dx+c)+r , with r=0\,\,\,or\,\,\,\deg(r)<1\Longrightarrow r\in\mathbb{R}\setminus{0}.

    As x^2+1 has no real roots, it can't be r=0\Longrightarrow 0\ne r\in\mathbb{R} , and thus (bx+a)\left(\frac{d}{r}x+\frac{c}{r}\right)=1\!\!\  !\pmod{x^2+1}\Longrightarrow (bx+a)^{-1}=\frac{d}{r}x+\frac{c}{r}\,\,\,in\,\,\,\mathbb{R  }[x]\slash<x^2+1>

    Tonio
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  3. #3
    Senior Member Shanks's Avatar
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    Tonio, I think, the inverse of bx+a should be -(dx+c)/r.
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    Quote Originally Posted by Shanks View Post
    Tonio, I think, the inverse of bx+a should be -(dx+c)/r.

    Yes, of course: forgot the sign. Thanx

    Tonio
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  5. #5
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    Can we not write it in terms of a and b like

    (bx+a)^{-1}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}x?
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