If $P$ is a prime ideal of $R$, then $P[x]$ is a prime ideal of $R[x]$, for $R$ a commutative ring. 
Prove that if $P$ is a prime ideal of $R$ then $P[x]$ is a prime ideal of $R[x]$.

This is homework. I have been trying to assume that there is an $fg$ in $P[x]$ such that neither $f$ nor $g$ is in $P[x]$. Hence $f$ and $g$ have at least one coefficient not in $P$. I was trying to show that $fg$ would then have a coefficient not in $P$, obtaining a contradiction. But I don't see how to control the terms. If $f$ and $g$ had only one coefficient not in $P$, then I think I could use the properties of the ideal to complete the proof. The problem is not being able to know for certain which if any of the coefficients of $f$ and $g$ are in $P$.
Perhaps my approach is wrong to begin. Please help. Even a hint in the right direction will much appreciated.

I assume that $R$ is commutative, because the ring of polynomials with a non-commutative ring of coefficients is a very rare beast.
Can you combine the following pieces?

*

*An ideal $P$ of a commutative ring $R$ is prime, iff the quotient ring $R/P$ is an integral domain.

*The polynomial ring $R[x]$ is an integral domain if and only if $R$ is.

*There is an isomorphism $(R/P)[x]\cong R[x]/P[x]$.


Alternatively, following your own line of attack: Assume that $f$ and $g$ are two polynomials, neither in $P[x]$. Let $f(x)=\cdots +a x^m+\cdots$ and $g(x)=\cdots+ bx^n+\cdots$. Assume that the highlighted terms $ax^m$ and $bx^n$ are the highest degree term with the property that the coefficients don't belong to the ideal $P$. What can you say about the term of degree $m+n$ in the product $fg$?

Note for comments: notation has been changed: ideal $I$ is now $P$.
