Let $A$ be a ring, $E$ a right $A$-module, $F$ a left $A$-module, $M$
a submodule of $E$ and $N$ a submodule of $F$. Suppose that $M$ is a
direct factor of $E$ and $N$ is a direct factor of $F$. Then the
canonical homomorphism $M\otimes_A N\rightarrow E\otimes_A F$ is
injective and the image of $M\otimes_A N$ under this homomorphism is a
direct factor of the $\mathbf{Z}$-module $E\otimes_A F$.
Let $M'$,$N'$ be submodules of $E, F$, respectively, such that $E$ is a direct sum of $M,M'$ and $F$ is a direct sum of $N,N'$. Let $\phi:M\oplus M'\rightarrow E$ and $\psi:N\oplus N'\rightarrow F$ be the associated $A$-linear isomorphisms.
Let $i:M\rightarrow E$ and $j:N\rightarrow F$ be the canonical injections. On the other hand, let $p:M\oplus M'\rightarrow M$ and $q:N\oplus N'\rightarrow N$ be the canonical surjections. Then $(p\circ\phi^{-1})\otimes(q\circ\psi^{-1})$ is a retraction of $i\otimes j$; thus, $i\otimes j$ is injective.
Furthermore, the mapping
$$g:(M\oplus M')\otimes_{A}(N\oplus N')\rightarrow(M\otimes_{A}N)\oplus(M\otimes_{A}N')\oplus(M'\otimes_{A}N)\oplus(M'\otimes_{A}N')$$
such that $g((m,m')\otimes(n,n'))=(m\otimes n,m\otimes n',m'\otimes n, m'\otimes n')$, for $(m,m')\in M\oplus M'$ and $(n,n')\in N\oplus N'$, is a $\mathbf{Z}$-module isomorphism.
I now have to show that there exists a sub-$\mathbf{Z}$-module $X$ of $E\otimes_A F$ such that $E\otimes_A F\simeq\text{Im}(i\otimes j)\oplus X$ via the canonical mapping. I know that the mapping
$$\phi^{-1}\otimes\psi^{-1}:E\otimes_A F\rightarrow(M\oplus M')\otimes_{A}(N\oplus N')$$
is a $\mathbf{Z}$-module isomorphism. This means that
$$E\otimes_A F\simeq(M\otimes_{A}N)\oplus(M\otimes_{A}N')\oplus(M'\otimes_{A}N)\oplus(M'\otimes_{A}N').$$
However, I am not sure how to proceed at this point. Any suggestions?
Edit:
The sequence of $\mathbf{Z}$-linear mappings
$$0\xrightarrow{}M\otimes_AN\xrightarrow{i\otimes j} E\otimes_A F\xrightarrow{}(E\otimes_A F)/\text{Im}(i\otimes j)\xrightarrow{}0$$ is exact. Since $(p\circ\phi^{-1})\otimes(q\circ\psi^{-1})$ is a $\mathbf{Z}$-linear retraction of $i\otimes j$, it follows that $\text{Im}(i\otimes j)$ is a direct factor of $E\otimes_A F$. Is this enough?
Best Answer
To say that $M$ is a direct summand of $E$, is equivalent to the existence of an homomorphism $p:E\to M$ such that $p\circ i=1_M$ that's $M\hookrightarrow E\xrightarrow p M$ is the identity on $M$. Similarly, $q\circ j=1_N$ that's $N\hookrightarrow F\xrightarrow q N$ is the identity on $N$. Since $(i\otimes j)\circ(p\otimes q)=(i\circ p)\otimes(j\circ q)$, the composition $$M\otimes N\to E\otimes F\xrightarrow{p\otimes q}M\otimes N$$ is the identity on $M\otimes N$, hence $M\otimes N\to E\otimes F$ is injective and $M\otimes N$ is a direct summand of $E\otimes F$.