Prove that if $I$ is a prime ideal of $R$ then $I[x]$ is a prime ideal of $R[x]$.
This is homework. I have been trying to assume that there is an $fg$ in $I[x]$ such that neither $f$ nor $g$ is in $I[x]$. Hence $f$ and $g$ have at least one coefficient not in $I$. I was trying to show that $fg$ would then have a coefficient not in $I$, obtaining a contradiction. But I don't see how to control the terms. If $f$ and $g$ had only one coefficient not in $I$, 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 $I$.
Perhaps my approach is wrong to begin. Please help. Even a hint in the right direction will much appreciated.