Problema 2.15.

Problema 2.15. Sea $E=\mathbb{Q}(\alpha)$ donde $\alpha^3+\alpha^2+\alpha+2 = 0$. Expresa a $(\alpha^2+\alpha+1)(\alpha^2+\alpha)$ y a $(\alpha-1)^{-1}$ en la forma $a \alpha^2+ b\alpha+c$ con $a,b,c \in \mathbb{C}$.

Solución:

Notemos que $\alpha^3 = -\alpha^2 - \alpha - 2$

$\alpha^4 = - \alpha^3 - \alpha^2 -2 \alpha = -(- \alpha^2 - \alpha - 2) - \alpha^2 - 2 \alpha = 2 - \alpha$

Ahora veamos que

$(a^2 + \alpha + 1)(\alpha^2 + \alpha) = \alpha^4 + 2 \alpha^3 + 2\alpha^2 + \alpha$
$\hphantom{(a^2 + \alpha + 1)(\alpha^2 + \alpha)}= (2 - \alpha) - 2(- \alpha^2 - \alpha - 2) + 2 \alpha^2 + \alpha$
$\hphantom{(a^2 + \alpha + 1)(\alpha^2 + \alpha)}= 2 - \alpha + 2\alpha^2 + 2\alpha + 4 + 2 \alpha^2 + \alpha$
$\hphantom{(a^2 + \alpha + 1)(\alpha^2 + \alpha)} = 4 \alpha^2 + 2 \alpha + 6$

Veamos que

$\left(\frac{-1}{5}\alpha^2 + \frac{-2}{5}\alpha + \frac{-3}{5}\right)(\alpha - 1) = \frac{-1}{5}\alpha^3 + \frac{-2 + 1}{5}\alpha^2 + \frac{-3 + 2}{5}\alpha + \frac{3}{5}$
$\hphantom{\left(\frac{-1}{5}\alpha^2 + \frac{-2}{4}\alpha + \frac{-3}{4}\right)(\alpha - 1)}= \frac{-1}{5}\alpha^3 + \frac{-1}{5}\alpha^2 + \frac{-1}{5}\alpha - \frac{2}{5} + \frac{5}{5}$
$\hphantom{\left(\frac{-1}{5}\alpha^2 + \frac{-2}{4}\alpha + \frac{-3}{4}\right)(\alpha - 1)}= \frac{1}{5}\left(\alpha^3 + \alpha^2 + \alpha + 2 \right) +1$
$\hphantom{\left(\frac{-1}{5}\alpha^2 + \frac{-2}{4}\alpha + \frac{-3}{4}\right)(\alpha - 1)}= \frac{-1}{5}(0) + 1 = 1$

Por lo que $(\alpha - 1)^{-1}$ = $\frac{-1}{5}\alpha^2 + \frac{-2}{5}\alpha + \frac{-3}{5} \blacksquare$

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License