1. ## 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

2. ## 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