Let $p$ be a prime and $P(t) \in \mathbb{Z}_{p}[t]$ a monic polynomial whose image in $\mathbb{F}_{p}$ modulo $p$ (which we also denote by $P(t)$ if there is no ambiguity) is irreducible. In this posting, we give a proof of the following fact, which we learned from some lectures notes by Andrew Sutherland (Lemma 10.13 and Example 10.16).
Theorem. Keeping the notation as above, the quotient ring $R = \mathbb{Z}_{p}[t]/(P(t))$ is a DVR whose maximal ideal is generated by $p$.
Proof. We first show that $R$ is a local ring whose maximal ideal is $(p)$. First, note that $$R/(p) \simeq \mathbb{F}_{p}[t]/(P(t))$$ is a field because we assumed that $P(t)$ is irreducible in $\mathbb{F}_{p}[t].$ Hence $(p)$ is indeed a maximal ideal of $R.$ Let $\mathfrak{m}$ be any maximal ideal of $R.$ We want to show that $p \in \mathfrak{m}.$
For contradiction, suppose that $p \notin \mathfrak{m}.$ Then since $\mathfrak{m}$ is a maximal ideal, we must have $\mathfrak{m} + (p) = R.$ Since $R$ is Noetherian, we have $$\mathfrak{m} = (r_{1}, \dots, r_{l}) = Rr_{1} + \cdots + Rr_{l}$$ for some $r_{1}, \dots, r_{l} \in R.$ We have $$R = \mathbb{Z}_{p} \oplus \bar{t} \mathbb{Z}_{p} \oplus \cdots \oplus \bar{t}^{d-1} \mathbb{Z}_{p},$$ where $d = \deg(P),$ so $R$ is finite over $\mathbb{Z}_{p}.$ Since $\mathfrak{m}$ is finite over $R,$ we see that $\mathfrak{m}$ is finite over $\mathbb{Z}_{p}.$ That is, we have some $z_{1}, \dots, z_{n} \in \mathfrak{m}$ such that $$\mathfrak{m} = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n}.$$ Since $\mathfrak{m} + (p) = R,$ it follows that the images $\bar{z}_{1}, \dots, \bar{z}_{n}$ of $z_{1}, \dots, z_{n} \in R$ generate $R/(p)$ as an $\mathbb{F}_{p}$-vector space. Since $\mathbb{Z}_{p}$ is a local ring with the maximal ideal $(p),$ by Nakayama's lemma, this implies that $R$ is generated by $z_{1}, \dots, z_{n}$ as a $\mathbb{Z}_{p}$-module, but then this implies that $$R = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n} \subset Rz_{1} + \cdots + Rz_{n} = \mathfrak{m},$$ which is a contradiction. Thus, we conclude that $p \in \mathfrak{m},$ showing that $R$ is a local ring with the maximal ideal $(p).$ Since $R$ is a Noetherian local domain that is not a field, whose maximal ideal is principal, we conclude that $R$ is a DVR. $\Box$
