Proposition. Let $A$ be a local ring with maximal ideal $P$, and $f=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\in A[x]$ such that $a_i\in P$ for all $i=0,\dots, n-1$, with $a_0\notin P^2$. Consider the $A$-module $B := A[x]/(f)$. Show that B is a local ring.
My idea is to show that $f$ is irreducible. If I prove this, then I can use the Kummer's Lemma: I observe that $\bar{f}(x)=x^n$, then for the Kummer's lemma we have that the set of all maximal ideals of $A[x]/(f)$ is the singleton $\{J/(f)\}$ where $J\:=(P[x],x)$, i.e. $B$ is a local ring with maximal ideal $J/(f)$.
CAN ANYONE HELP ME IN PROVING THAT $f$ IS IRREDUCIBLE?
Edit: This answer only works for $A$ an integral domain.
This is a variant of the Eisenstein criterion and the proof idea carries over.
Suppose that $f=gh$, where $g$ and $h$ are not a unit. Now consider the reductions mod $P$: $x^n=\bar{f}=\bar{g} \bar{h}$. If $\bar{g}$ is constant, then one can show that $g$ is a unit by using the fact that $A$ is an integral domain. By analogy, $\bar{h}$ is not constant either. Using the fact that $(A/P)[x]$ is a UFD, one sees the the only divisors of $x^n$ in $(A/P)[x]$ are of the form $\lambda x^k$ for some $\lambda \in (A/P)^\times$ and $k \leq n$. Thus we obtain $\bar{g}=\lambda x^k$. As we have shown that $\bar{g}$ is not constant, we have $k \geq 1$. But this implies that the constant term of $g$ is in $P$. By analogy, so is the constant term of $h$. Multiplying out, we see that the constant term of $f=gh$ is contained in $P^2$, a contradiction.