A non-domain ring in which every non-zero annihilating ideal is a prime ideal, is of the form $F_1 \bigoplus F_2$, $F_1$, $F_2$ are fields or has only one non-zero proper ideal.
Note: Here, an ideal $I$ is known as an annihilating ideal, if $\exists$ a non-zero ideal $J$ s.t. $IJ = 0$. Also $Z(R)$ is the set of all zero divisors of R.
Let $R$ be a non-domain ring and suppose every annihilating ideal of $R$ is a prime ideal.
Let $z \in R $ be a non-zero zero divisor , then $Q =zR$ is a prime ideal. ($z$ is a non-zero zero divisor $\implies$ $Ann(z)$ $\neq$ $0$ )
Case(i) : $Q = Q^2$
Then , $z^2R = zR$ $\implies $$zR(1-z)R = 0$ $\implies$ $N = (1-z)R$ is a prime ideal.
Also neither $Q$ nor $N$ can have proper subideals. (???).........(1)
Hence in this case $R = F_1 \bigoplus F_2 $ for fields $F_1$ , $F_2$. (???)........(2)
Case(ii) : $Q \neq Q^2$
Then, $Q^2 = 0$ (this holds by (1) , since $Q^2$ is a proper subideal of $Q$)
Thus, we can assume that $z^2=0$ for all non-zero $z \in Z(R)$.
since $zR$ is prime, $zR = Z(R)$ (???)..........(3)
Also if $r \in R-Z(R)$ , then $rzR = zR$ (again it follows from ( 1) , as $rzR$ is a non-zero subideal of $zR$ therefore it must be equal to zR) and hence $r$ is a unit.(???) .............(4)
i.e $rR = R$
Thus $Z(R)$ is the only non-zero proper ideal of $R$
(1) $Q$ & $N$ will have no proper subideals.
I can start with a subideal $I$ of $Q$ , then since $Q$ is an annihilating ideal, $I$ is also. So $I$ is a prime ideal. This also imply that $I \neq 0$, since $R$ is not a domain.
I don't know how to go further to show $I=Q$.
(2) $R = F_1 \bigoplus F_2 $ for fields $F_1$ , $F_2$.
I am able to see since $R = Q \bigoplus N$ , they are concluding this. But I am not able to write properly the reason for this or explicitly find $F_1$ , $F_2$.
(3) $zR = Z(R)$
Clearly $zR \subset Z(R)$ , but how to show the backward inclusion?
(4) $r$ is a unit.
Since cancellation law does not hold in a non-domain ring, I am not able to see how to prove this.
Thanks in advance!