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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.

Proof:-

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 $R$

My Doubts:

(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!

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  • $\begingroup$ By your definition of "annihilating ideal," the zero ideal is always an annihilating ideal. If all annihilating ideals in your ring are prime, the zero ideal is prime, and your ring is a domain. Presumably you didn't mean for this to happen. What did you intend? Every nonzero annihilating ideal prime? $\endgroup$ – rschwieb Jul 28 '16 at 13:15
  • $\begingroup$ Yes you are right. Sorry for this mistake. I meant this thing for nonzero annihilating ideals $\endgroup$ – Shivani Goel Jul 28 '16 at 13:47

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