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Thread: Polinomio in K[x]

  1. #1
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    Polinomio in K[x]

    Hi I have problems with exercise


    Let $\displaystyle p(x) = x^3+x^2-2x-1$ and let $\displaystyle k = \mathbb{Q}$ [$\displaystyle x$] / $\displaystyle (p)$ Show that, in $\displaystyle K$,


    $\displaystyle p(x^2-2) = p(x^3-3x) = 0$


    Hence the three roots of $\displaystyle p $in $\displaystyle K$ are $\displaystyle x + (p)$, $\displaystyle x^2-2+(p)$ and $\displaystyle x^3-3x + (p)$


    I do not know how to start

    Thanks
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  2. #2
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    Re: Polinomio in K[x]

    Let $\displaystyle x+(p)=u\in \mathbb{Q}[x]/(p) $

    $\displaystyle p(u)=0 $ is obvious

    Show that

    $\displaystyle p\left(u^2-2\right)=p(u)\left(u^3-u^2-2u+1\right)=0$

    For the third root it is easier to first note that

    $\displaystyle u^3=-u^2+2u+1$ and therefore $\displaystyle u^3-3u=-u^2-u+1$

    and then show that $\displaystyle p\left(-u^2-u+1\right)=0$
    Thanks from cristianoceli
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  3. #3
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    Re: Polinomio in K[x]

    Thanks
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