Let $M$ be a fixed faithful $R$-module over integral domain $R$. Is there any equivalent condition (on $R$ or on $M $) under which the annihilator of any nonzero submodule of $M$ to be a prime ideal of $R$?
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2$\begingroup$ Sorry if I misunderstand something obvious, but if you fix a single $M$ once and for all, It seems doubtful to ask for conditions on $R$ only, since the answer will probably depend on your $M$. Isn't the real question the following one? : Which are necessary and sufficients conditions on $R$ so that, for all faithful modules $M$, annihilators of submodules of $M$ are prime ideals ? $\endgroup$– GreginGreOct 27, 2021 at 8:28
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$\begingroup$ Dear GreginGre, M is a fixed module. $\endgroup$– ArastoeiOct 27, 2021 at 17:41
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$\begingroup$ Then one cannot say anything interesting, see my answer. $\endgroup$– GreginGreOct 27, 2021 at 17:43
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If you fix an $M$ once and for all, you cannot say anything interesting, because anything can happen. For example, for $M=R$, it is true for any integral domain $R$: for any nonzero submodule $N$, the annihilator is zero.
If you assume that $R$ is not a field, and if $a$ is nonzero and not a unit of $R$, then $M=R\times R/(a^2)$ is faithful (because an element of the annihilator need to annihilate $(1,\bar{0})$, but has a submodule isomorphic to $R/(a^2)$ whose annihilator is not prime (the annihilator being $(a^2)$. The fact is it is not prime uses the fact that $R$ is an integral domain and that $a$ is not a unit )
If the question is to find necessary and sufficient conditions on an integral domain $R$ such that, for all faithful modules $M$, annihilators of submodules of $M$ are prime, the answer is: $R$ is a field.
Indeed, if $R$ is a field, this is clear, and conversely, if $R$ is an integral domain but is not a field, then consider a nonzero element $a$ which is not a unit and take $M=R\times R/(a^2)$. As noted before, $M$ does not satisfy the required conditions.
