Show that $x$ is not a prime in $R=\mathbb{Z}+x\mathbb{Q}[x]$, the set of polynomials with rational coefficients whose constant term is an integer.

Question

The Problem: Show that $x$ is not a prime in $R=\mathbb{Z}+x\mathbb{Q}[x]$, the set of polynomials in $x$ with rational coefficients whose constant term is an integer.

Source: Abstract Algebra, $\mathit{3^{rd}}$ edition by Dummit and Foote.

Clearly we want to show that $x\mid a(x)b(x)$ but $x\nmid a(x)$ AND $x\nmid b(x)$ for some $a(x), b(x)\in R$. I struggle to come up with such a counterexample. Indeed, $a(x)$ and $b(x)$ cannot both have constant terms; but then one of the two is of the form $r_nx^n+\dots+r_1x$, which obviously can be divided by $x$. Any help would be greatly appreciated.

Answer 1

Consider $2$ and $\frac{x}{2}$. Clearly their product is $x$ and $2 \not\in (x)$. But $\frac{x}{2} \not\in (x)$ as well; by degree considerations, if it were then we would have some degree $0$ polynomial in $R$ that when multiplied by $x$ yielded $\frac{x}{2}$. But the constant polynomials in $x$ are integers, and there is clearly no integer $m$ such that $mx = \frac{x}{2}$.