I want to prove the following result:
Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.
I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:
- $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal;
- If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification);
- The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.
I'll appreciate any comment.