I'm stuck on a detail in the proof of the following:
Let $R$ be a commutative ring with $1\neq 0$. Prove that if every proper ideal of $R$ is prime, then $R$ is a field.
Since $0 \neq 1$ we have that $\{0 \}$ is a proper ideal of $R$. Further, $R$ is an integral domain since if $a\cdot b \in \{0\}$, then either $a=0$ or $b=0$ as $\{0\}$ is prime. Now I would like to prove that $\{0\}$ is the only proper ideal of $R$ which would imply that $\{0\}$ is maximal and $F/\{0\} \cong F$ is a field. My idea is that if $A,B \subseteq R$ are two proper ideals then it must be that $A = B$, and therefore any proper ideal of $R$ is $\{0\}$. Notice that $AB \subseteq B$ so that $AB$ is also a proper ideal of $R$. Since $AB \subseteq AB$, and $AB$ is prime, we have that either $A \subseteq AB \subseteq A$ or $B \subseteq AB \subseteq B$, which implies that either that $B = AB \subseteq A$ or $A = AB \subseteq B$. Say $B = AB \subseteq A$. I'm now stuck trying to show that $A \subseteq B$.
Is there a better way to do this? I have also tried picking $r \in R\setminus\{0\}$, and showing that $r^{-1} \in R$ without success.
