Number Fields

Lemma: A polynomial in $\mathbb{Z}[x]$ is irreducible if and only if it is irreducible over $\mathbb{Q}[x]$.

Proof: Let $m,n$ be the gcd's of the coefficients of $f,g\in \mathbb{Z}[x]$. Then $mn$ divides the gcd of the coefficients of $fg$. We wish to show that this is in fact an equality.

Divide $f$ by $m$ and $g$ by $n$, so that we need only consider the case $m=n=1$. It suffices to show that the gcd $d$ of the coefficients of $fg$ must be 1. If $d>1$, then let $p$ be some prime dividing $d$. Consider the equation $fg=0(\mathrm{mod}p)$. Since ${\mathbb{Z}}_p[x]$ is an integral domain, this means $f=0$ or $g=0$, implying that $p$ divides all the coefficients of $f$ or $g$, which is a contradiction. Thus $d=1$.

Now suppose $f=gh$ over $\mathbb{Q}[x]$. Find $m,n\in \mathbb{Q}$ such that $mg,nh\in \mathbb{Z}[x]$, and the gcd's of the coefficients of $mg,nh$ are 1. Then we have $(mg)(nh)=df$ for some $d$.