Let $R$ be an integral domain containing a prime element $p$ such that $\cap_{n \ge 1} (p^n)=(0)$ ; if
$f : R \setminus \{0\} \to \mathbb Z$ defined as $f(x):=\max \{i : x \in (p^i)\}$ is a function such that
$f(x)\ge 0 , \forall 0\ne x\in R$ and for every $a,b \in R$ with $ 0\ne b , a \notin Rb$ , $\exists q,r \in R$ such that $a=qb+r$
where $f(r)<f(b)$ , then is it true that $R$ is a Local ring ?
Since $R$ is already given to be an Euclidean domain , so it is a PID , so a non-zero ideal is maximal iff it is prime iff it is generated by an irreducible element iff it is generated by a prime element . Now $p$ is already a given prime , so if we can show that any prime $b\in R$ is associated to $p$ , we are done . Consider $f(b)$ ; if $f(b)>0$ , then $p|b$ , since both $p,b$ are irreducibles , hence this implies $(p)=(b)$ ; now if $f(b)=0$ , then dividing $p$ by $b$ , by Euclidean algorithm , $\exists q,r \in R$ such that $p=bq+r$ , where $r=0$ or $f(r)<f(b)=0$ , but $f(x)\ge 0,\forall x \in R\setminus \{0\}$ , hence $f(r) <0$ is impossible , so $r=0$ , so $b|p$ , then again we get $(b)=(p)$ , thus there is a unique non-zero maximal ideal .