When associated prime ideals are comaximal

Question

Let $R $ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. I want to know is there any equivalent conditions under which any two distinct associated prime ideals are comaximal, that is, their sum is $R $?

Answer 1

This is not going to be a completely general answer for the following reason:

This is obviously not true of rings with embedded primes, but there are nonreduced rings with no embedded primes, e.g. $k[x]/(x^2)$, which has a single prime ideal, $(x)$, and hence satisfies your properties.

However, to forbid embedded primes, let's assume $R$ is reduced. Let's also assume $R$ is Noetherian, since I'm not sure how much of what I'm going to say applies to non-Noetherian rings without checking some references. Then the associated primes of $R$ are precisely the minimal prime ideals.

If $R$ is reduced and Noetherian, the following are equivalent:

1. The associated primes are pairwise comaximal.
2. $R\cong \prod_{P\in\operatorname{Ass} R} R/P$
3. Every associated prime is the annihilator of an idempotent
4. Every associated prime is principal, generated by an idempotent

Geometrically these all correspond to $\newcommand\Spec{\operatorname{Spec}}\Spec R$ being the disjoint union of its irreducible components.

Proof:

$1\implies 2$: If every pair of associated primes is comaximal, then Chinese remainder theorem guarantees that $$R\cong \prod_{P\in\operatorname{Ass} R} R/P,$$ since the intersection of the associated primes is $0$, since $R$ is reduced.

$2\implies 3$: If $R$ is isomorphic to that product, then $P$ is the annihilator of the idempotent corresponding to the preimage in $R$ of the identity in $R/P$.

$3\implies 4$: If every associated prime is the annihilator of an idempotent, then if $P=\newcommand\ann{\operatorname{ann}}\ann e$, $P=R(1-e)$, since every element of $R(1-e)$ is in $\ann e$, and if $fe=0$, then $f(1-e)=f-fe=f$, so $f\in R(1-e)$. Thus $P$ is principal, generated by the idempotent $1-e$.

$4\implies 1$: Finally, if $P$ and $Q$ are distinct primes with $P=Re$ and $q=Rf$ for idempotents $e$ and $f$, then consider $R/P$. In this ring the image of $f$ is still idempotent, but $P$ is prime, so the image of $f$ is either $0$ or $1$. If the image of $f$ is $1$, then $P$ and $Q$ are comaximal, since this implies that $1-f$ lies in $P$, so $1-f+f = 1 \in P+Q$. On the other hand, if the image of $f$ is $0$, $f\in P$, so $Q\subseteq P$, and considering the image of $e$ in $R/Q$ we see that either the image of $e$ is $1$, in which case $P$ and $Q$ are comaximal, or the image of $e$ is $0$ also, so $P\subseteq Q$ as well, contradicting that $P$ and $Q$ were distinct.