Suppose $\xi\in\mathbb{C}$ is an algebraic number, and suppose $m(x)\in\mathbb{Q}[x]$ is the minimal polynomial of $\xi$. If $\xi$ is a root of some monic polynomial $g(x)\in\mathbb{Z}[x]$, how can we show that $m(x)\in\mathbb{Z}[x]$? any help is appreciated.
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By the division algorithm, there are $v(x), r(x)\in\mathbb{Q}[x]$ such that $g(x)=m(x)v(x)+r(x)$ with either $r(x)=0$ or $0\le\deg r<\deg m$. Since $m(\xi)=g(\xi)=0$, we have $$r(\xi)=g(\xi)-m(\xi)v(\xi)=0.$$ Note that if $r(x)\neq 0$, we can find some $\ell\in\mathbb{Z}$ such that $\frac{1}{\ell}\cdot r(x)\in\mathbb{Q}[x]$ is monic with $\deg\frac{1}{\ell}r<\deg m$ which has $\xi$ has a root, contradiction (since $m$ is the minimal polynomial of $\xi$). Thus, $g(x)=m(x)v(x)\in\mathbb{Q}[x]$. Thus, by Gauss' Lemma, there exist $m_0(x), v_0(x)\in\mathbb{Z}[x]$ such that $g(x)=m_0(x)v_0(x)$ and $m_0(x)=\alpha m(x)$ for some $\alpha\in\mathbb{Q}_{>0}$. WLOG, let the leading coefficient of $m_0(x)$ be positive (otherwise we can just consider $g(x)=(-m_0(x))(-v_0(x))$). Since $g$ is monic, $m_0$ is as well, so comparing coefficients of $m_0(x)$ and $\alpha m(x)$ yields $\alpha=1\implies m=m_0\in\mathbb{Z}[x]$, as desired.
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