Recall that a module $M_R$ ($R$ is a ring with unity) is called SIP if the intersection of any two summands of $M$ is also a summand. I asked before if there exists a commutative ring which is not an SIP ring, and the answer is NO. Now, I ask if there exists a commutative ring $R$ and a proper right ideal $A\subset R$ such that the module $A_R$ is not an SIP.
Thanks for your help.
Now, I ask if there exists a commutative ring š¯‘… and a proper right ideal $A\subset R$ such that the module $A_R$ is not an $\textsf{SIP}$.
Yes. Start with any commutative ring $S$ that has a non-$\textsf{SIP}$ module $N$, then take $R$ to be the Nagata idealization $S\oplus N$ and take $A = 0\oplus N$.
For example, take $S=\mathbb Z$ and $N=\mathbb Z_4\oplus \mathbb Z_4$. $N$ is non-$\textsf{SIP}$ since the cyclic submodules $N_1=\langle (1,0)\rangle$ and $N_2=\langle (1,2)\rangle$ are direct summands of $N$, but their intersection has size $2$ and $N$ has no direct summand of size $2$. Now construct $R$ so that additively it is $S\oplus N$, while multiplicatively define $(s,n)(s',n')=(ss',sn'+s'n)$. $R=S\oplus N$ is commutative, $A=0\oplus N$ is an ideal of $R$, every $R$-submodule of $A$ is induced by an $S$-submodule of $N$, and $A$ has the same submodule structure as $N$.
