$I$ prime does not imply $I$ maximal I am attempting to prove the converse to $I$ maximal implies $I$ prime is not true.
$I$ prime $\iff A/I$ is an integral domain and $I$ maximal $\iff A/I$ is a field so I'm looking for a prime ideal that yields an integral domain but not a field. 
Obviously the integers fail (as the only ideals are $d\mathbb{Z}$ for some $d$) and these are prime iff $d$ is prime and the quotient ring modulo $p$ is a field.
I have proved $\langle p(x)\rangle$ is maximal iff $p(x)$ is irreducible in $K[x]$ where $K$ is a field, again tried looking for prime ideals that aren't generated by an irreducible polynomial but did not get far.
I imagine I should be considering the ring of diagonal matrices as this has zero divisors and I think zero divisors are important in the underlying examples.
 A: Your assertion about the integers is not quite correct.
It is true that $d\mathbb{Z}$ is a prime ideal if $d$ is a prime integer, and that these ideals are also maximal.
But you forgot one very important prime ideal of $\mathbb{Z}$: $0\mathbb{Z}$! In a commutative ring with unity, $(0)$ is a prime ideal if and only if the ring is an integral domain, and $\mathbb{Z}$ is definitely an integral domain (one might say it's the integral domain; it's why they are called "integral domains", after all...). So you can get lots of examples just by taking integral domains that are not fields.
For examples with zero divisors, consider a product of two rings: $R\times S$. Notice that if $I$ is an ideal of $R$ and $J$ is an ideal of $S$, then $I\times J$ is an ideal of $R\times S$, and $(R\times S)/(I\times J) \cong (R/I)\times (S/J)$. So you can get further examples this way. You can think of the case of diagonal matrices that way, since the ring of diagonal $n\times n$ matrix with coefficients in $R$ is isomorphic to the direct product of $R$ with itself, $n$ times.
A: HINT $\rm\ \ A\:$ is a field $\iff\: (0)\:$ is maximal; $\rm\ A\:$ is a domain $ \iff\: (0)\:$ is prime.  
Therefore $(0)$ is a non-maximal prime in any domain that is not a field.  
As I mentioned in your question yesterday  on the converse, factoring out by the prime $\rm\:P\:$ to reduce to the domain case $\rm\:P = 0\:$ is a technique that is very useful in algebra.  Again I highly recommend reading the masters (Kaplansky).
A: Another class of examples:  Take any integral domain $R$ that isn't a field.  Then the ideal of the polynomial ring $R[x]$ generated by $x$ is prime but not maximal, since $R[x]/(x)\cong R$ is an integral domain but not a field.
