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.
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.