Let $\alpha$ have the minimal polynomial $x^2 + x+1 =0$ over $\mathbb Q$ , I have to find the expression of the element $\frac {\alpha^2 +1}{\alpha^2 -1}$ in the form $a+b\alpha$ for some $a,b \in \mathbb Q $ .
We have $\alpha^2 + \alpha + 1 = 0 $ , also
$\frac {\alpha^2 +1}{\alpha^2 -1}$ = ${\alpha \over \alpha - 1} - \frac {1}{\alpha + 1}$ , I can't move further, Any help!
Thank you ..
On
The gcd of $x^2 + x + 1$ and $x^2 - 1$ is $1$ and you have $$1 = \frac{1}{3} (x-1) (x^2 - 1) - \frac{1}{3}(x-2) (x^2 + x + 1)$$ Plugging in $\alpha$ makes the second summand zero, so you have $$1 = \frac{1}{3} (\alpha - 1) (\alpha^2 - 1) $$ In other words $$ \frac{1}{\alpha^2 - 1} = \frac{1}{3}(\alpha -1)$$ It follows that $$ \frac{\alpha^2+1}{\alpha^2 -1} = \frac{1}{3}(\alpha^2 +1)(\alpha -1)$$ which you can expand and simplify using the relation $\alpha^2 + \alpha + 1 = 0$.
You could write $\alpha=-\frac12(1+i\sqrt3)$ and use the usual arithmetic of complex numbers.
Alternatively $$\frac{\alpha^2+1}{\alpha^2-1}=\frac{-\alpha}{-\alpha-2}.$$ You need to find real $x$ and $y$ with $$\alpha=(x+y\alpha)(\alpha+2).$$ You can expand this out, using $\alpha^2=-\alpha-1$ to get two linear equations for $x$ and $y$.