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Thread: Another field contruction

  1. #1
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    Another field contruction

    Construct a field of order 125.

    I don't know, this chapter is just not coming to me for some reasons, I can't really understand how to construct a field. So I pick a field, then pick a poly that is irreducible and has order of 125, how do I do that?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Construct a field of order 125.

    I don't know, this chapter is just not coming to me for some reasons, I can't really understand how to construct a field. So I pick a field, then pick a poly that is irreducible and has order of 125, how do I do that?
    Again remember what I said. Let $\displaystyle F$ be a finite field $\displaystyle \mathbb{Z}_p$. Let $\displaystyle p(x) = a_nx^n+...+a_1x+a_0$ be an irreducible polynomial over $\displaystyle F$. Then $\displaystyle F/\left< p(x) \right>$ is a field. Now any element in $\displaystyle F/\left< p(x) \right>$ can be written uniquely as $\displaystyle b_0+b_1x+...+b_{n-1}x^{n-1}$ now for $\displaystyle b_i$ (for $\displaystyle 0\leq i \leq n-1$) we have $\displaystyle p$ choices for the coefficients. Thus, in total there are $\displaystyle p^n$ such elements in this larger field.

    1)Let $\displaystyle F=\mathbb{Z}_5$
    2)Let $\displaystyle p(x)$ be an order $\displaystyle 3$ irreducible polynomial in $\displaystyle F$.
    3)The factor ring $\displaystyle F/\left< p(x)\right>$ is a field with $\displaystyle 5^3=125$ elements.

    So the thing remaining now is for you to find an irreducible degree polynomial.
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  3. #3
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    $\displaystyle (x-2)^3$, it has zero in there. would that be it?
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    Quote Originally Posted by tttcomrader View Post
    $\displaystyle (x-2)^3$, it has zero in there. would that be it?
    I am not sure what you are asking. Do you follow what I did? You need to find and irreducible polynomial of degree 3 over Z_5.
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