Let $Q = \sum_{j=0}^n \alpha_j X^j$ be a polynomial with integer coefficients and $\alpha_n = 1$. Show that $Q$ is a prime element of $\mathbb{Z}[x]$ if and only if $Q$ is a prime element of $\mathbb{Q}[x]$.
My ideas: By definition, $Q$ is prime element means $Q \neq 0, Q \notin R^*$ and $Q = ab \Rightarrow Q|a$ or $Q|b$.
Now, I don't know how to continue.
(1) Because the coefficients of $\mathbb{Z}[X]$ are from an integral domain it can be proven that $\mathbb{Z}[X]$ is also an integral domain.
(2) A prime element of an integral domain is always irreducible. By definition, this means $Q = ab \Rightarrow a \in \mathbb{Z}[X]^*$ or $b \in \mathbb{Z}[X]^*$.
Thank you in advance!
The ring $\mathbb{Z}[X]$ is not only a domain, but a unique factorization domain (hence irreducible elements coincide with prime elements) for which Gauss’ lemma holds.
In the particular case of $a_n=1$, the polynomial cannot have noninvertible integer factors, so by Gauss’ lemma it is irreducible in $\mathbb{Z}[X]$ if and only if it is irreducible in $\mathbb{Q}[X]$.