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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.

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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.

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    Fantastic, Lubin. Thank you very much for this very helpful method. Could you clarify a question, though? We have $A,$ which is the matrix that represents the linear transformation $T(x) = \alpha x,$ correct? It follows that $T^2(x) = \alpha^2 x,$ hence the matrix of $\alpha^2$ is given by $A^2.$ Furthermore, I would imagine that the matrix of $\alpha^n$ is given by $A^n.$ Is this correct? I am very curious about this method and where it comes from, so if you happen to know of any literature, I would enjoy reading it. Once again, thanks a lot! – 2017-01-22
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    Sorry, but I’m probably the worst possible person for you to ask about history of mathematics. As I said, this is called the regular representation of the algebra $\Bbb Q(\alpha)$. My understanding is that it has been known for very long, and for all I know, it might predate the method that we now teach by means of polynomials and gcds. I’ll have to leave the history to others, sorry. – 2017-01-23
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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 $x^3-9x+3$.

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    $$(240x+247)\,(x^3-9x+3)\,-\,(80x^2+29x-766)\,(3x^2+2x+1)\,=\, 1507$$ – 2017-01-13
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    Yeah, @BillDubuque, the Norm of $1+2a + 3a^2$ is $1507=11\cdot137$, not a pleasing number to show up in your computations. It would have been much friendlier for the person setting the problem to have arranged a unit as the number to be inverted. – 2017-01-15
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    @BilDubuque To add context to the question. This came from a graduate algebra qualifying exam at our university. The poster took the exam yesterday and was essentially looking for an efficient method of solving the problem (given that the exam is 6 questions and 3 hours). – 2017-01-15
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    Thank you for clarifying, Ken, and thank you for your efforts! – 2017-01-22