If $R$ is $\text{PID}$ and $x \in R$ is irreducible, then $R/(x^k)$ is a local ring.

Question

Suppose $R$ is $\text{PID}$ and let $x \in R$ be irreducible. Let $k \in \mathbb{Z}_{>0}. $Could anyone advise me on how to prove $R/(x^k)$ has a unique maximal ideal?

Hints will suffice, thank you.

Answer 1

Since $R$ is a PID and $x$ is irreducible, $(x)$ is maximal.

Given any commutative ring $R$ with maximal ideal $M$, then $M$ is the only maximal ideal that can contain $M^k$ no matter what positive integer $k$ is.

Any maximal ideal $N$ is prime, and if this ideal $N$ contains $M^k$, it has to contain $M$ by definition of primeness. But $M$ is already maximal, so $N=M$.

Answer 2

1. The correspondence theorem gives a bijection between the ideals of $R/(x^k)$ and the ideals of $R$ containing $(x^k).$

2. $R$ is a PID, so these ideals are of the form $(x^k) \subseteq (y).$ Rephrase this in terms of divisbility.