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Recall that a module $M_R$ ($R$ is a unital ring) is called an SIP-module if the intersection of any two summands of $M$ is a summand. The ring $R$ is called (left) right SIP-ring if the module (${}_RR$) $R_R$ is an SIP-module.

Is there a commutative ring (with unity) which is NOT an SIP-ring?

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    $\begingroup$ One can prove quite elementarily that for central idempotents $e, f$, we have $eR\cap fR=efR$. Since summands are precisely of that form, that settles things. $\endgroup$
    – rschwieb
    Commented Aug 14 at 14:41
  • $\begingroup$ Thanks. This is a good answer. But excuse me, I want to ask is every $R$-submodule of a commutative ring $R$ is an SIP-module? (i.e., is every ideal of $R$ is also an SIP-module?) $\endgroup$
    – Hussein Eid
    Commented Aug 14 at 14:52
  • $\begingroup$ Do I understand the definition of SIP correctly?: If $N_1$ and $N_2$ are complemented submodules of M then their intersection is a complemented submodule again $\endgroup$
    – Ali Taghavi
    Commented Aug 15 at 13:56

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The answer is no. If $R$ is commutative with unity and $S$ and $S'$ are summands of $R$ as an $R$-module then there are associated idempotent decompositions of the identity $1=e_1+e_2$ and $1=e'_1+e'_2$. So $S=e_1R$ and $S'=e'_1R$. Then multiplying, we get $1=e_1e'_1+e_1e'_2+e_2e'_1+e_2e'_2$. The idempotent $e_1e'_1$ projects onto $S\cap S'= SS'$, with complement $(e_1e'_2+e_2e'_1+e_2e'_2)R$.

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