Expressing $(\alpha-1)^{-1}$ where $\alpha^3-\alpha^2+\alpha+2=0$ Let $E=\Bbb{Q}[\alpha]$ where $\alpha^3-\alpha^2+\alpha+2=0$
I would like to express $(\alpha^2+\alpha+1)(\alpha^2-1)$ and $(\alpha-1)^{-1}$ on the form $a+b\alpha+c\alpha^2$ where $(a,b,c)\in\Bbb{Q}^3.$
For the first one I get $-4\alpha-2$, but for the inverser I am not sure how can I do that, I would like to say that it's $\alpha-1$ but it seems 'silly'.
 A: Let's try to represent $(\alpha - 1)^{-1}$ as $a + b\alpha + c\alpha^2$. We must have
$$ (\alpha - 1)(a + b\alpha + c\alpha^2) = a\alpha + b\alpha^2 + c\alpha^3 - a - b\alpha - c\alpha^2 \\
= c\alpha^3 + (b - c)\alpha^2 + (a-b)\alpha -a = \\
c(\alpha^3 - \alpha^2+\alpha +2) +b\alpha^2+(a-b-c)\alpha - (2c + a) = 1. $$
Thus, by comparing coefficients, we see that we want
$$ b = 0, a - b - c = 0, 2c + a = -1 $$
which is a linear system of equations whose solution is $b = 0$ and $a = c = -\frac{1}{3}$. That is, $(\alpha - 1)^{-1} = -\frac{1}{3} (1 + \alpha^2)$. Checking ourselves, we indeed have
$$ -\frac{1}{3}(\alpha - 1)(1 + \alpha^2) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha - 1 \right) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha + 2 - 3 \right) \\ 
= -\frac{1}{3}( -3) = 1.$$
A: I'm surprised that no one has mentioned this yet, but the standard way of computing multiplicative inverses in algebraic extensions is via polynomial division.
Recall that $\frac{\mathbb{Q}[x]}{(m(x))} \cong \mathbb{Q}[\alpha]$ via the map sending $x \mapsto \alpha$, where $m$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.  Under this isomorphism, the element $\alpha - 1 \in \mathbb{Q}(\alpha)$ corresponds to the polynomial $x - 1 \in \mathbb{Q}[x]$ modulo $m(x)$.  Since $m(x)$ is irreducible and $x - 1$ is of lower degree (assuming $\alpha \notin \mathbb{Q}$), then $\gcd(m(x),x-1) = 1$.  By Bézout's identity, then there exist polynomials $a(x), b(x) \in \mathbb{Q}[x]$ such that $1 = a(x) m(x) + b(x) (x-1)$.  Moreover, the polynomials $a(x),b(x)$ can be computed by the extended Euclidean algorithm.  Since we are working mod $m(x)$, this yields
$$
1 = a(x) m(x) + b(x) (x-1) \equiv b(x) (x-1) \pmod{m(x)}
$$
so we have found the multiplicative inverse of $x-1$: it is just $b(x)$.  Using the isomorphism above, then $(\alpha - 1)^{-1} = b(\alpha)$.  (For reference, see $\S$13.1 of Dummit and Foote, p. 515.) 
In your example, we have $m(x) = x^3 - x^2 + x + 2$.  Dividing $m(x)$ by $x-1$, we find
$$
m(x) = x^3 - x^2 + x + 2 = (x^2+1)(x-1) + 3 \, .
$$
Rearranging this equation, we obtain
$$
1 = -\frac{1}{3}(x^2+1)(x-1) -\frac{1}{3} m(x) \equiv-\frac{1}{3}(x^2+1)(x-1) \pmod{m(x)}
$$
hence $(\alpha - 1)^{-1} = -\frac{1}{3}(\alpha^2+1)$.
A: Well, the inverse you can do it with a trick
$$
0 = \alpha^3-\alpha^2+\alpha+2
=
\alpha^{2} (\alpha - 1) + \alpha  - 1 + 3 = (\alpha - 1) (\alpha^{2} + 1) + 3,
$$
whence
$$
(\alpha - 1)^{-1} = - \frac{1}{3} (\alpha^{2} + 1).
$$
