If $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$

Question

Show that if $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$.

Definition of a minimal prime ideal of a ring $R$: Let $P,Q$ be prime ideals of $R$. We say that $P$ is a minimal prime ideal of $R$ if $Q \subseteq P \Rightarrow Q=P$.

I have proved that if $P$ is a minimal prime of $R$, then $P[x]$ is a minimal prime of $R[x]$.

So it remains to show that if $Z$ is a minimal prime of $R[x]$, then $Z$ is of the form $P[x]$ where $P$ is a minimal prime ideal of $R$.

This is in fact where the hypothesis of $R$ being noetherian enters.

My attempt:

Let $Z = P_1 \subsetneq \cdots \subsetneq P_n \subsetneq \cdots$ be a chain of prime ideals of $R[x]$.

Since $R$ is noetherian, by Hilbert Basis Theorem $R[x]$ is noetherian.

Then the above chain is stationary.

All that I have tried next leads nowhere.

I also tried with a primary decomposition of $Z$.

Can someone give a hint? Thanks!

Answer 1

Hint: The Noetherian hypothesis is unnecessary and irrelevant. Instead, just take $P=Z\cap R$ (note that if $Z$ were in fact of the form $P[x]$, then $P$ would have to be $Z\cap R$, so this is a good way to find the $P$ that will work). Use minimality of $Z$ to prove that $P[x]=Z$ and $P$ is a minimal prime of $R$.

Answer 2

Let $0=q_1\cap q_1\cap...\cap q_n$ be a reduced primary decomposition of $0$ in $R$ with $q_i$ are $p_i-$primary. Clearly $0=q_1[x]\cap q_1[x]\cap...\cap q_n[x]$ is a reduced primary decomposition of $0$ in $R[x]$ with $q_i[x]$ are $p_i[x]-$primary. Since the minimal prime ideals of a noetherian ring are the minimal elements of $Ass(0)$. This complete the proof.