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


    p(x^2-2) = p(x^3-3x) = 0


    Hence the three roots of p in K are x + (p), x^2-2+(p) and 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 x+(p)=u\in \mathbb{Q}[x]/(p)

    p(u)=0 is obvious

    Show that

    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

    u^3=-u^2+2u+1 and therefore u^3-3u=-u^2-u+1

    and then show that 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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