Be a direct sum implies zero intersection

Question

Let $M_1$ and $M_2$ be a submodule of $M$ and $M \cong M_1 \oplus M_2$ then $M_1 \cap M_2 = \{0\}$ can you please help.

Answer 1

The statement itself is false: it is entirely possible to have an isomorphism $g: M_1\oplus M_2\to M$ without having $M_1\cap M_2=0$. For instance, consider $M_1=M_2=M=\Bbb Z^{\Bbb N}$. If you are requiring that the map $s:M_1\oplus M_2\to M$, $s(x,y)=x+y$ is an isomorphism (in one way or another, this is what most authors refer to when they say that $M$ is the internal direct sum of $M_1$ and $M_2$), then $M_1\cap M_2=0$, because $(x,-x)\in \ker s$ for all $x\in M_1\cap M_2$.