Is $M$ necessarily a submodule of $M\times N$? Suppose $M$ and $N$ are Abelian groups and $M \times N$ is the direct product of $M$ and $N$.  Now suppose $M \times N$ is an $R$-module.  Are $M$ and $N$ necessarily submodules of $M \times N$?  
$M$ and $N$ are subgroups of $M \times N$,
but is $M \times \{0\}$ closed the action of $M \times N$?
Is it the case the action of every $M \times N$ module has an action that acts on $M$ and $N$ component-wise?  This seems really simple to me, but I have been stuck trying to prove it.

No, this is usually not the case.  For instance, let $R=\mathbb{Z}[x]$ and $M=N=\mathbb{Z}$.  Then the abelian group $M\times N=\mathbb{Z}\times\mathbb{Z}$ can be made an $R$-module in many different ways: all you have to do is specify an endomorphism which is how $x$ acts.  For instance, one way to make $\mathbb{Z}\times\mathbb{Z}$ an $R$-module is to say $x(m,n)=(n,m)$.  With this $R$-module structure, $\mathbb{Z}\times\{0\}$ is not a submodule.
