Example of a ring $R$ such that some maximal ideal of $R[x]$ contracts to a non-maximal ideal of $R$. Can you give me an example of a polynomial ring $R[x]$ and a maximal ideal $M$ of $R[x]$ such that 
$M\cap R$ is not a maximal ideal of $R$?
 A: You might be having a hard time finding a counterexample because the property that maximal ideals of $R[x]$ contract to maximal ideals of $R$ defines exactly the class of Jacobson (aka Hilbert) rings, which includes PIDs which maximal ideals intersect to $0$, in particular fields.  Equivalently you might encounter Jacobson rings by the defining property that every prime ideal is an intersection of maximal ideals, or that $R$ satisfies the Zarisiki lemma.
For the particular angle on Jacobson rings relevant to your question, a classic reference is section 1-3 in Kaplansky's Commutative Rings (starting at Theorem 18).
A simple way to produce a non-Jacobson ring stems from the realization that local rings are Jacobson precisely when they have Krull dimension $0$ (which follows immediately from the 'primes are intersections of maximals' characterization).  
So, for a simple example, any (Discrete Valuation Ring) will do.  To be concrete, we can localize $\mathbb{Z}$ at any nonzero prime $p$, and we'll get the ring $\mathbb{Z}_p$ which has prime ideals $0$ and $p\mathbb{Z}_p$.  Every element in $\mathbb{Z}_p$ can be written as $\frac{a}{b}p^n$ where $p$ doesn't divide $a,b$.  Thus the quotient field of $\mathbb{Z}_p$ is obtained by adjoining just one element, the inverse of $p$, i.e. $\mathbb{Q} = \mathbb{Z}_p[p^{-1}]$.  This gives us a surjective homomorphism $\mathbb{Z}_p[x] \twoheadrightarrow \mathbb{Q}$ by sending $x \rightarrow p^{-1}$.  By the first isomorphism theorem, the kernel of this homomorphism must be a maximal ideal $\mathfrak{m}$ of $\mathbb{Z}_p[x]$, and since the homomorphism restricts to an injection on $\mathbb{Z}_p$, we have that $\mathfrak{m} \cap \mathbb{Z}_p = 0$.  (Note that this is just a concrete version of Kaplansky's Theorem 24.)
