Irreducible implies minimal polynomial?

Assume we're looking for the minimal polynomial of $\alpha$, $\mu$ say, and suppose we have another polynomial $f$ satisfying $f(\alpha)=0$.

We can write $f=g\mu+r$, where $g$ and $r$ are polynomials and $\mathrm{deg}(r)<\mathrm{deg}(\mu)$ (Euclidean division).

Clearly, $r(\alpha)=0$, but then $r=0$ since $\mu$ was chosen such that it is the non-zero polynomial of minimal degree with this property. Hence, $\mu$ divides $f$.

Therefore we know that any "candidate" for the minimal polynomial is a multiple of the minimal polynomial and if our candidate is irreducible, then it is the minimal polynomial and we're done.

Note that this works over any field, not just $\mathbb{Q}$.

Suppose there is a polynomial $p(x)$ of degree $\lt 3$, with rational coefficients, which is not a constant times our polynomial $q(x)$ of degree $3$, such that $p(\sqrt[3]{2})=0$. If the gcd $d(x)$ of $p(x)$ and $q(x)$ has degree $\ge 1$, we have contradicted irreducibility.

If the gcd is $1$, there are polynomials $A(x)$ and $B(x)$, with rational coefficients, such that $A(x)p(x)+B(x)q(x)=1$. This is impossible, for it would imply that $\sqrt[3]{2}$ is a root of the constant polynomial $1$.

The same argument works in general.