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Thread: splitting over Z mod 2

  1. #1
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    splitting over Z mod 2

    let $\displaystyle \alpha$ be a zero of $\displaystyle f(x)=x^3+x^2+1$ over $\displaystyle \mathbb{Z}_{2}$, Show that $\displaystyle f(x)$ splits over $\displaystyle \mathbb{Z}_{2}(\alpha)$

    I don't understand this question because$\displaystyle f(x)$ is irreducible in $\displaystyle \mathbb{Z}_{2}$ ($\displaystyle f(0)=1$ and $\displaystyle f(1)=1$) so how can $\displaystyle \alpha$ be its zero.
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  2. #2
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    The zero does not lie in Z_2. It lies in an extension of Z_2 containing alpha.

    To help solve problem note that alpha^2 is also a root in Z_2(alpha).
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  3. #3
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    Quote Originally Posted by frankdent1 View Post
    let $\displaystyle \alpha$ be a zero of $\displaystyle f(x)=x^3+x^2+1$ over $\displaystyle \mathbb{Z}_{2}$, Show that $\displaystyle f(x)$ splits over $\displaystyle \mathbb{Z}_{2}(\alpha)$

    I don't understand this question because$\displaystyle f(x)$ is irreducible in $\displaystyle \mathbb{Z}_{2}$ ($\displaystyle f(0)=1$ and $\displaystyle f(1)=1$) so how can $\displaystyle \alpha$ be its zero.
    as whipflip15 suggested: show that if $\displaystyle \alpha$ is a zero of $\displaystyle x^3 + x^2 + 1 \in \mathbb{Z}_2[x],$ then: $\displaystyle x^3+x^2+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4).$
    Last edited by NonCommAlg; Apr 13th 2009 at 09:48 PM.
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