Let $R$ be a ring. An $R$-module $M$ is a multiplication module if for each submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N = IM$.
An element $a$ of a ring $R$ is called multiplicatively idempotent if $a^2 = a$.
The ring $R$ is multiplicatively idempotent if every element of $R$ is multiplicatively idempotent.
Now, I get this example and proof of multiplication module below from article.
Let $R$ be a multiplicatively idempotent ring. Then every ideal of $R$ is a multiplication $R$-module.
Proof.
Let $J$ be an ideal of $S$ and $I \subseteq J$. If $x \in I$, then $x = x^2 \in IJ$. Therefore $I = IJ$ and hence $J$ is a multiplication $R$-module.
My questions:
(1) To prove $J$ is a multiplication module, we must show for all submodule of $J$ there exist ideal $I$ such that submodule is equal to ideal multiplied with module $J$. Why $I$ in the proof is not ideal, just subset of $J$? Why $I$ is not submodule? In the proof, $J$ is an ideal. Is true $J$ a module?
(2) In the proof in this example, to prove $I=IJ$, we must show $I\subseteq IJ$ and $IJ\subseteq I$. The author just write the proof of $I\subseteq IJ$. I confused to proof $IJ\subseteq I$. Let $x\in IJ$, how to get $x\in I$?
Thank you very much.
For your first question, the $R$-submodules of a ring $R$ are precisely the ideals of $R$, so if $I\subseteq J$ is a submodule of an ideal $J$, then it is also an ideal. Then we are proving that $J$ is a multiplication module by saying that if $I$ is a submodule of $J$ (that is, an ideal contained in $J$), then $I=IJ$ and the definition is satisfied.
For your second question, any element of $IJ$ is of the form $a=\sum_{i=1}^n x_iy_i$ for some $x_i\in I$, $y_i\in J$, $n\geq1$. But since $I$ is an ideal we have $x_iy_i\in I$ for all $i$, and so $a\in I$. Then $IJ\subseteq I$.