Let $\mathbb{R} \supset \mathbb{Q}$ and $\alpha = 2 + \sqrt[3]{4}$. Find the minimal polynomial of $\alpha$.

Question

Problem: Let $\mathbb{R} \supset \mathbb{Q}$ and $\alpha = 2 + \sqrt[3]{4}$. Find the minimal polynomial of $\alpha$.

My attempt: We have $(\alpha - 2)^3 = 4 \Leftrightarrow \alpha^3 - 6 \alpha^2 + 12 \alpha - 8$. Consider polynomial $f(x) = x^3 - 6x^2 + 12x - 8 \in \mathbb{Q}[x]$ and $f(\alpha) = 0$. If $\frac{p}{q} \in \mathbb{Q}$ with $(p,q)=1$ is a root of $f(x)$ then $p \mid 8$ and $q \mid 1$. So maybe $1$ and $8$ are roots of $f(x)$. $f(1) = -1, f(8) = 216$. So $f(x)$ is irreducible on $\mathbb{Q}$, implies $f(x)$ is the minimal polynomial of $\alpha$.

Does the proof right?

Answer 1

First, you forgot to subtract $4$ from the constant term and the polynomial should be $x^3-6x^2+12x-12$.

You also can't just test $1$ and $8$. You also need to check that $\pm p$ is not a root for each combination of sign and factor of $8$, which is actually true. Thus we have the minimal polynomial, but we can only do this because it is cubic (and therefore if it was reducible it must have a rational linear factor).

Answer 2

There is no need to test polynomial $(x-2)^3-4$ for irreducibility by hand:

The polynomial $x^3-4$ is irreducible over $\mathbb Q$ because it has no roots in $\mathbb Q$. Therefore, $(x-2)^3-4$ is irreducible because $x \mapsto x-2$ induces an automorphism of $\mathbb Q[x]$.

Concretely, if $(x-2)^3-4=f(x)g(x)$, then $x^3-4=f(x+2)g(x+2)$ and this transformation preserves degrees.