Let $R = \mathbb{Z} + x\mathbb{Q}[x] \subset \mathbb{Q}[x]$. Find the irreducibles of $R$.
Show that the irreducible elements in $R$ are $\pm p$ for prime integers $p$ and the irreducible polynomials $p(x) \in \mathbb{Q}[x]$ whose constant coefficient is $\pm 1$. Prove these irreducibles are prime in $R$.
The first part is pretty clear. ($\pm p$ for prime integers $p$), if $\alpha \in R$ s.t. $\deg(\alpha)=0$ then $\alpha$ is a constant polynomial thus in $\mathbb{Z}$. If $\alpha$ is a prime then it is irreducible.
Having trouble showing that for $p(x) \in R$ s.t. $\deg(p(x)) > 0$ with constant term $\pm 1$ are irreducible.
It's obvious that an irreducible polynomial $p\in\mathbb Q[x]$ with $p(0)=\pm 1$ remains so in $R$.
The question is to prove that these are the only irreducible elements of $R$ of degree $\ge 1$.
If $k=|p(0)|>1$, then $p(x)=kq(x)$ with $q\in R$, so $p$ is reducible, a contradiction.
If $p(0)=0$, then $p(x)=xq(x)$ with $q\in\mathbb Q[x]$. Write $q(0)=\frac mn$, notice that $p(x)=(\frac 1nx)(nq(x))$ and $\frac 1nx,nq(x)\in R$, a contradiction.
Now suppose $p(0)=\pm1$, but $p$ is reducible in $\mathbb Q[x]$. Thus $p(x)=u(x)v(x)$ with $u,v\in\mathbb Q[x]$ with $\deg u,\, \deg v\ge1$. In particular, $u(0)v(0)=\pm1$, so we can assume $u(0)=v(0)=\pm1$ (why?). This shows that $p$ is reducible in $R$, a contradiction.