Isomorphic finite rings?

Question

Let $p$ be a prime number, $P$ and $Q$ monic polynomial of degree $k\geq 2$, irreducible over $\mathbb Z_p$. Do we have $$\mathbb Z_{p^2}[x]/(P) \simeq \mathbb Z_{p^2}[x]/(Q)\ ?$$

Answer 1

For $f,g\in \Bbb{Z}[x]$ of same degree, with their leading coefficient not divisible by $p$, and $f,g\in \Bbb{Z}/(p)[x]$ irreducible.

• $R=\Bbb{Z}/(p^2)[x]/(f)$ has a unique maximal ideal $(p)$ and $R/(p)$ is the finite field with $p^{\deg(f)}$ elements

• by our knowledge of finite fields there is some $a\in R$ such that $g(a)=0\in R/(p)$,

• Hensel lemma : there is some $b\in R$ such that $g(a+pb)=0\in R$

  Proof: $g'\in \Bbb{Z}[x]$ then $g'(a) \in R/(p)^\times$ (as $\gcd(g,g')=1\in \Bbb{Z}/(p)[x]$) which means that $ g'(a) \in R^\times, g(a+pb) = g(a)+pb g'(a)\in R$, and since $g(a)= pc\in R$ then $ b = -g'(a)^{-1}c$ gives $g(a+pb)=0\in R$.

• The map $y\to a+pb, \Bbb{Z}/(p^2)[y]/(g)\to R$ is an isomorphism