Irreducibles in $\mathbb{Q}$ and Inverses of Quadratics in $\mathbb{Q}(\alpha)$ Consider $f(x) = x^3 - 9x + 3 \in \mathbb{Q}[x],$ and let $\alpha$ be a root of $f(x).$ Prove that $f(x)$ is irreducible over $\mathbb{Q}.$ Furthermore, in the field $\mathbb{Q}(a),$ find $(3 \alpha^2 + 2 \alpha + 1)^{-1}$ in terms of the basis $1, \alpha, \alpha^2.$
$\textit{Proof.}$ Observe that $f$ is 3-Eisenstein, hence $f$ is irreducible in $\mathbb{Q}[x].$
$\textit{Solution.}$ Because $f$ is irreducible, it follows that $f$ is relatively prime to $g(x) = 3x^2 + 2x + 1.$ By definition of relatively prime, there exist polynomials $u$ and $v$ in $\mathbb{Q}[x]$ such that $$(3x^2 + 2x + 1)u(x) + (x^3 - 9x + 3)v(x) = 1.$$
By the Division (Euclidean) Algorithm, I know that we can explicitly find our $u(x)$ and $v(x),$ but I am not sure how to proceed with the algorithm because it looks a little funky to multiply by these unknown polynomials. Ultimately, if I can find $u(x),$ I know that $u(\alpha) = (3 \alpha^2 + 2 \alpha + 1)^{-1}.$ Can anyone point me in the right direction with this. Thanks very much for your consideration.
 A: Although we all learn that the Euclidean Algorithm for greatest common divisor can be used as a pry-bar for finding the reciprocal of an element of $\Bbb Q(a)$, the method I find less confusing and more direct is to represent $1$, $a$ and $a^2$ as matrices, representing what they do when they multiply the basis $\{1,a,a^2\}$. This is called the regular representation of $\Bbb Q(a)$. The matrix of $a$ is:
$$
A = \begin{pmatrix}
0&0&-3\\1&0&9\\0&1&0
\end{pmatrix}\,,
$$
since multiplying by $a$ has the three results $1\mapsto a$, $a\mapsto a^2$, and $a^2\mapsto a^3=-3+9a$. The matrix of $a^2$ is $A^2$, naturally enough, and we can call the identity matrix $I$. Next you find the inverse of the matrix $I+2A+3A^2$, which is easy by methods I’m sure you know. Express the result as a linear combination of $I$, $A$, and $A^2$, and Bob’s your uncle.
A: I really like Lubin's approach. It is something I haven't seen before. The Euclidean Algorithm is tried and true, but as you will see, it is not fun in this case.
Using the division algorithm, we divide $x^3-9x+3$ by $3x^2 + 2x + 1$ to see that:
$$ x^3 - 9x + 3 = \left( \frac{1}{3}x-\frac{2}{9}\right) (3x^2+2x+1) + \left(-\frac{80}{9}x + \frac{29}{9}\right).$$
Now divide $3x^2 + 2x + 1$ by $\left(-\frac{80}{9}x + \frac{29}{9}\right)$ to see that: 
$$ 3x^2+2x+1 = \left(-\frac{80}{9}x + \frac{29}{9}\right) \left(-\frac{27}{80}x - \frac{2223}{6400} \right) + \frac{13563}{6400}$$
Therefore:
\begin{align*}
1 &= \frac{6400}{13563}(3x^2+2x+1) - \frac{6400}{13563}\left(-\frac{80}{9}x + \frac{29}{9}\right)\left(-\frac{27}{80}x - \frac{2223}{6400} \right)\\
&= \frac{6400}{13563}(3x^2+2x+1) - \frac{6400}{13563}\left((x^3-9x+3) -  \left( \frac{1}{3}x-\frac{2}{9}\right) (3x^2+2x+1) \right)\left(-\frac{27}{80}x - \frac{2223}{6400} \right)_.
\end{align*}
From here it is just some gross simplification, then modding out by $x^3-9x+3$.
