In Nicholson's Introduction to Abstract Algebra the text gives an example of an Integral Domain in which the ACCP fails. He writes:
For example, consider $R=\{n+xf\mid n \in \mathbb{Z}, f \in \mathbb{Q[x]} \}$, the set of all polynomials in $\mathbb{Q}[x]$ whose constant term is in $\mathbb{Z}$. Then $R$ is an integral domain (subring of $\mathbb{Q}[x]$), but $\langle x\rangle \subset \langle \frac {1} {2} x\rangle \subset \langle \frac {1}{4}x\rangle \dotsb$
I'm not sure what the notation means here. Is $\langle x\rangle = \{ xf\mid x \in \mathbb{Q}[x] \}$ (so the constant term would be $0$)?
I am not sure how to show that there is an ascending chain, insights appreciated.
If $g\in R$, then $\langle g\rangle=\{gf\mid f\in R\}$ is the principal ideal generated by $g$ in $R$.
Clearly $\langle\frac{1}{2^n}x\rangle\subseteq \langle\frac{1}{2^{n+1}}x\rangle$, because $$ \frac{1}{2^n}x=2\frac{1}{2^{n+1}}x $$ and $2\in R$. On the other hand, there is no element $h\in R$ such that $$ \frac{1}{2^{n+1}}x=\frac{1}{2^n}xh $$ because we can cancel $x$ and evaluating at $0$ would yield $h(0)=1/2$. Therefore $$ \frac{1}{2^{n+1}}x\notin\Bigl\langle\frac{1}{2^n}x\Bigr\rangle $$ and the inclusion $\langle\frac{1}{2^n}x\rangle\subset \langle\frac{1}{2^{n+1}}x\rangle$ is proper.