Let $R$ be a Noetherian ring and $x\in R$ an $R-\mathrm{regular}$ element. Show that $\mathrm{Ass}_R(R/(x^n))=\mathrm{Ass}_R(R/(x))$ for every $n\geqslant 1$.
Let $M$ be an $R-\mathrm{module}$. An element $a\in R$ is called a non-zero-divisor on $M$ or $M-\mathrm{regular}$ if whenever $x\in M$ and $ax=0$, then $x=0$.
$\def\Ass{\operatorname{Ass}}$There is a short exact seequence $$0\to\frac{(x^k)}{(x^{k+1})}\to\frac{R}{(x^{k+1})}\to\frac{R}{(x^{k})}\to0$$ and, since $x$ is regular, $(x^k)/(x^{k+1})$ is isomorphic to $R/(x)$, so that $$\Ass(R/(x^{k+1})\subseteq \Ass(R/(x^k))\cup \Ass(R/(x)).$$
It follows by induction that $\Ass(R/(x^k))\subseteq \Ass(R/(x))$.
On the other hand, $(x^{k-1})/(x^k)$ is a submodule of $R/(x^k)$ which is isomorphic to $R/(x)$, so that $$\Ass(R/(x))=\Ass((x^{k-1})/(x^k))\subseteq\Ass(R/(x^k)).$$
