If $M,N$ are $R$-modules, then every submodule of $M \times N$ is the form of $U\times V,$ where $U,V$ are submodules of $M,N \ ?$

Question

Let $M,N$ be $R$-modules and suppose $U$ and $V$ are submodules of $M$ and $N$ respectively. I have shown that $U \times V$ is submodule of $M \times N.$ May I know how do I prove/disprove that the converse is true?

Thank you.

Answer 1

Consider the counterexample of the diagonal in $M \times M$. More concretely, let $M=N=R=\mathbb{R}$, then the vector space $\mathbb{R}\left(1,1\right)$ is a sub vector space of $\mathbb{R}^{2}$ which does not come from a "product" construction. Anyway you can do this construction for any module $M$ over a ring $R$ and this provides a counterexample, just use $\lbrace \left( x,x \right)|x \in M\rbrace$.