Minimum polynomial of $\sqrt[3]{9} + \sqrt[3]{3} - 1$ over $\mathbb{Q}$

Question

I am stuck trying to find the minimum polynomial of $\sqrt[3]{9} + \sqrt[3]{3} - 1$ over $\mathbb{Q}$. I have tried using this method and this method but none of them seem to be working in this case. Can anyone give me a hint?

Answer 1

Let's work in $\mathbb Q( \sqrt[3]{3})$. Since $x^3-3$ is irreducible over $\mathbb Q$, a basis for $\mathbb Q( \sqrt[3]{3})$ over $\mathbb Q$ is $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$.

Let $\alpha = \sqrt[3]{9} + \sqrt[3]{3} - 1$. The matrix of the $\mathbb Q$-linear map $x \mapsto \alpha x$ in the basis $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ is $$ A= \begin{pmatrix} -1 & \hphantom-3 & \hphantom-3 \\ \hphantom-1 & -1 & \hphantom-3 \\ \hphantom-1 & \hphantom-1 & -1 \end{pmatrix} $$ The set of rational polynomials that kill $A$ and the set of rational polynomials that kill $\alpha$ are exactly the same. In particular, the minimum polynomial of $A$ is the minimum polynomial of $\alpha$.

Now, the characteristic polynomial of $A$ is $\chi(x)=x^3+3x^2-6x-20$. Since it has no rational roots, $\chi$ is irreducible. Since the characteristic and minimum polynomials of $A$ share the same irreducible factors, $\chi$ is the minimum polynomial of $A$, hence of $\alpha$.

Answer 2

Let $\sqrt[3]9+\sqrt[3]3-1=x.$

Now, by using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and since $3\neq9,$ it's equivalent to $$9+3-(x+1)^3+3\sqrt[3]{27}(x+1)=0$$ or $$x^3+3x^2-6x-20=0,$$ which has no integer roots and from here has no rational roots.

Answer 3

First let us find a polynomial that vanished on that number. Notice that $$ \alpha = \sqrt[3]{9} + \sqrt[3]{3} - 1 \; \Rightarrow \; (\alpha + 1)^3 = (\sqrt[3]{9} + \sqrt[3]{3})^3 = 9 + 9 \sqrt[3]{3} + 9 \sqrt[3]{9} + 3 = 12 + 9(\alpha + 1) $$ where the last equation follows trivially as $\alpha + 1 = \sqrt[3]{9} + \sqrt[3]{3}$. So we know that $(x + 1)^3 - 9(x + 1) - 12$ is divided by the minimum polynomial of $\alpha$. Next step is to observe that this polynomial is irreducible over $\mathbb{Q}$ since it has no rational roots, which concludes the proof.