I know from a course I'm taking on algebra that the following is true:
Let $R$ be a ring, let $M$ be an $R$-module, and let $M_1, M_2 \le M$. If $M_1 + M_2 = M$ and $M_1 \cap M_2 = \{0\}$, then $M \cong M_1 \oplus M_2$.
My question is, does the converse also hold? That is, does $M \cong M_1 \oplus M_2$ imply that $M_1 + M_2 = M$ and $M_1 \cap M_2 = \{0\}$? If so, can somebody please provide a brief proof or an intuitive explanation of why it is true? I think I'm missing something here about internal direct sums, since in the textbook I'm reading whenever we have $M \cong M_1 \oplus M_2$, the author seems to conclude that $M_1 \cap M_2 = \{0\}$, but I've never seen a theorem or an exercise stating directly that we can make that assumption. Thanks in advance for any assistance!
I suppose you mean that there is an isomorphism $\phi:M_1\oplus M_2\to M$ given by $\phi(m_1,m_2)=m_1+m_2$?
Well, $\phi$ is surjective, so we must have $M_1+M_2=\phi(M_1\oplus M_2)=M$. On the other hand, if $m\in M_1\cap M_2$, then $\phi(m,-m)=0$. Since $\phi$ is injective we must have $(m,-m)=(0,0)$, i.e. $m=0$. Hence, $M_1\cap M_2=(0)$.