When a prime ideal is a maximal ideal In a commutative ring with unit every maximal ideal is prime. Under what conditions does the converse occur? 
 A: When the ring has Krull dimension equal to zero.  If we're talking about integral domains then every prime ideal of $R$ is maximal if and only if $R$ is a field (since $0$ is a prime ideal in any integral domain).
A: When the ring contains no elements that are neither units nor zero divisors.  Irritatingly, I know of no short name for this third category of elements of rings.
Polynomials are a typical example:  Let $R$ be an integral domain and consider $R[x]$.  Polynomials (of positive degree) are not units (they don't have multiplicative inverses), nor are they zero divisors.  In such a ring, there are prime ideals that are not maximal ideals.  For instance $(x)$, which has $R[x]/(x) \cong R$.  Since $R$ had no zero divisors, neither does this quotient, so $(x)$ is prime.  However, unless $R$ was already a field, this quotient is not a field so $(x)$ is not maximal.
A: A ring $R$ with identity is artinian if and only if it is noetherian and every prime ideal is a maximal ideal by Hopkins theorem.
