Every non zero ideal prime then is every prime ideal maximal In a commutative ring with 1 if every non zero ideal is prime then is every prime ideal maximal? I make no other assumptions. Please help if further assumptions like noetherianness make the problem easier. However I would like an answer without any assumptions.
 A: It is true without further assumptions: If $P \subsetneq M$ are two prime ideals, we have that $M/P$ is non-zero prime ideal of the integral domain $R/P$. In particular the square of any element generates a non-prime ideal in $R/P$, which gives rise to non-prime ideal in $R$.
A: Another way to look at this:
Suppose $R$ is such a ring and let $N \subseteq R$ be the nilradical. Let $x \notin N$ be a non-unit. Then $(x^2)$ and $(x)$ are both prime ideals, and $x \cdot x \in (x^2)$ then implies there is some $a \in R$ such that $ax^2=x$. It is clear that this property still holds if $x$ is a unit, so $R/N$ is absolutely flat (or von Neumann regular). This is equivalent to every prime ideal of $R$ being maximal (see Atiyah-MacDonald, Chapter 3, Exercise 11). 
A: 
Lemma: If every proper ideal of a commutative ring $R$ is prime, $R$ is a field.

${}$

Proposition: If every nontrivial ideal of a commutative ring $R$ is prime, the ring is either field, a product of two fields, or a uniserial ring with exactly three ideals.

Proof of proposition: In case $\{0\}$ is prime, the Lemma immediately yields that it is a field. Suppose hereafter that nontrivial ideals exist.  
Let $I$ be any nontrivial ideal of $R$. Then $R/I$ satisfies the lemma, and $I$ is a maximal ideal. It should also be apparent that $I$ is a minimal ideal: if there is another nontrivial ideal $I'\subseteq I$, then $R/I'$ being a field implies $I'=I$. 
Thus every nontrivial ideal of $R$ is maximal and minimal. Obviously we can conclude now that $R$ is Artinian.
If there is only one maximal ideal $I$, then obviously we are in the three-ideal uniserial ring case.
If there are at least two maximal ideals, $I, I'$, then by minimality and maximality of $I$, $I+I'=R$ and $I\cap I'=\{0\}$. The Chinese remainder theorem yields that $R\cong I\oplus I'$, where each of $I$ and $I'$ are fields.
N.B. Thanks to Moos for suggesting a simplification. My original take on it was to consider the Jacobson radical. Either it was maximal (and then the ring was uniserial) or it was $\{0\}$, and the ring was semisimple, decomposing into only two fields. I like that too but Moos' suggestion is a little more elementary.

Corollary: In the context of the proposition, all prime ideals are maximal, and moreover there are only $0$, $1$, or $2$ nontrivial ideals.

A: 
Lemma If every proper ideal of $R$ is prime then $R$ is a field.

Since the zero ideal is prime, $R$ is a domain. If $x \ne 0, \in R$ then either $x$ is invertible or $(x)$ and $(x)^2$ are both prime. Thus $(x) \subseteq (x)^2$. It follows that $x = yx^2$ for some $y \in R$ and hence $1 = yx$ since $R$ is a domain.

Corollary If every nonzero, proper ideal of $R$ is prime then $R$ is Artinian (every prime ideal is maximal)

If $I$ is a nonzero, proper ideal of $R$ then every proper ideal of $R/I$ is prime by the correspondence theorem. Thus $R/I$ is a field by the Lemma. So $R$ is Artinian.
