Maximal ideals in $R[x]$
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.
In (2), your argument is valid because of the following more general fact:
Let $f : A \to B$ be a surjective ring homomorphism. If $P$ is a prime ideal of $A$ that contains $\ker f$, then $f(P)$ is a prime ideal in $B$.
Proof: We have $\ker f \subseteq P \subseteq A$. So by the third isomorphism theorem, we have that $P/\ker f$ is an ideal of $A/\ker(f)$. Furthermore, we have that
$$(A/\ker f)/(P/\ker f) \cong A/P.$$
The latter is an integral domain because $P$ is a prime ideal, this proves that $P/\ker f$ is a prime ideal in $A/\ker f$. Furthermore by the first isomorphism theorem you know that because $f$ is surjective,
$$A/\ker f \cong B.$$
It follows that because $P/\ker f$ is a prime ideal in $A/\ker f$ that $f(P)$ is a prime ideal in $B$.
$\hspace{6in} \square$
Once you know this, you just need to apply it with $A = R[x]$, $B = R[x]/M[X]$, $f = \pi$ where $\pi$ is the canonical projection of $R[x]$ onto the quotient $R[X]/M[x]$ and its pre-image $\pi^{-1}$ define bijective correspondences between prime ideals in $R[x]$ that contain $M[x]$ and prime ideals in $B$. The correspondence is as follows:
$$\{P' \in \textrm{Spec}(B) \implies \pi^{-1}(P) \text{ a prime ideal in $A$ that contains} \hspace{2mm} M[x]\}$$ $$\updownarrow$$ $$\{P \hspace{2mm} \text{a prime ideal in $A$ that contains $M[x]$} \implies \pi(P) \in \textrm{Spec}(B) \}$$