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Math Help - Inverses over a field F

  1. #1
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    Inverses over a field F

    I'm given the field Z_2[t]/(t^3 + t + 1) for which I need to calculate the inverse of each element.

    I assume the way to do this is similar to the Extended Euclidean Algorithm and this is what I have for computing the inverse of t^2 + 1. I'm a little bit skeptical with my reasoning however and this is partially due to my lack of familiarity of working with polynomials over Z_p[x].

    1) First use the Division algorithm to obtain t^3 + t + 1 = t*(t^2 + 1) + 1

    From the above, this gives us -1*(t^3 + t + 1) + 1 = -t*(t^2 + 1).

    On the other hand, as we are working over Z_2[t], it follows that we must have (t^3 + t + 1) + 1 = t*(t^2 + 1) since -1 \equiv 1. Hence the inverse of t^2 + 1 is just t.

    Is this correct?
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  2. #2
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    Quote Originally Posted by jamix View Post
    I'm given the field Z_2[t]/(t^3 + t + 1) for which I need to calculate the inverse of each element.

    I assume the way to do this is similar to the Extended Euclidean Algorithm and this is what I have for computing the inverse of t^2 + 1. I'm a little bit skeptical with my reasoning however and this is partially due to my lack of familiarity of working with polynomials over Z_p[x].

    1) First use the Division algorithm to obtain t^3 + t + 1 = t*(t^2 + 1) + 1

    From the above, this gives us -1*(t^3 + t + 1) + 1 = -t*(t^2 + 1).

    On the other hand, as we are working over Z_2[t], it follows that we must have (t^3 + t + 1) + 1 = t*(t^2 + 1) since -1 \equiv 1. Hence the inverse of t^2 + 1 is just t.

    Is this correct?
    there's a much neater way to solve this problem: let F=\mathbb{Z}_2[t]/(t^3 + t + 1) and for simplicity let's show the image of t in F by t. then t^3=t+1 and so t^6=t^2+1 \neq 1. that means t, as an

    element of the multiplicative group F^{\times}=F-\{0 \}, has order 7. thus F^{\times}=<t>=\{t^k: \ 0 \leq t \leq 6 \}. clearly the inverse of t^k is t^{7-k}.


    Note: (some extra explanations!) i probably don't have to mention that since t^3=t^2+1, we'll have t^4=t^2+t, \ t^5=t^2+t+1, \ t^6=t^2+1 and t^7=1. so, for example, the inverse of t^2+t=t^4
    is t^{7-4}=t^3=t+1, etc. this method can be used in any finite field because we know that the multiplicative group of any finite finite field is cyclic.
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