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 $R$ 
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!
 A: 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 $R$.
A: Thanks to Eric Wofsey's helpful comment, I tried writing a solution.
Let $R$ be a commutative ring.
1) If $P$ is a minimal prime of $R$, then $P[x]$ is a minimal prime of $R[x]$.

Proof) Let $Q\subset P[x]$ with $Q$ prime. Then $R∩Q$ is a prime ideal in $R$ and $R∩Q\subset R∩(P[x])=P$. Then since $P$ is minimal, $R∩Q=P$. Then $P[x]=(R∩Q)[x]\subset Q\subset P[x]$. So $P[x]=Q$.

2) If $Q$ is a minimal prime of $R[x]$, then there is a minimal prime $P$ in $R$ such that $Q=P[x]$.

Proof) As above, $(R∩Q)[x]\subset Q$ and since $Q$ is minimal $(R∩Q)[x]=Q$. Let $P=R∩Q$ and now we show $P$ is minimal. Let $P'\subset P$. Then $P'[x]\subset P[x]=Q$. Since $Q$ is minimal, $P'[x]=P[x]$. Then $P'=R∩(P'[x])=R∩(P[x])=P$.

So, minimal primes of $R[x]$ are exactly the ideals of the form $P[x]$ where $P$ is a minimal prime of $R$.
A: 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.
