It was an exercise in Isaac's character theory:
Let $\alpha$ be an algebraic integer (so $\alpha$ satisfies monic irreducible polynomial $g(x)$ over $\mathbb{Z}$).
Let $\alpha$ satisfies a monic irreducible polynomial $f(x)$ over $\mathbb{Q}$.
Show that $f(x)\in\mathbb{Z}[x]$.
My question is about conclusion. This $f(x)$ is actually $g(x)$ (which is in $\mathbb{Z}[x]$), which I want to conclude. This is done as follows.
Since $\alpha$ is algebraic integer, it satisfies a monic irreducible polynomial $g(x)$ over $\mathbb{Z}$; by Gauss lemma, the polynomial $g(x)$ is irreducible over $\mathbb{Q}$; then finally, monic irreducible polynomial over a field satisfied by an element (of some extension field) should be unique, hence $f(x)=g(x)$ [and so $f(x)\in\mathbb{Z}[x]$].
Q. Is this conclusion right?