Let $R$ be a commutative ring with identity and let $\{M_\alpha\}$ be a family of $R$-modules and $N$ another $R$-module. I've tried to show that
$$\left(\bigoplus_\alpha M_\alpha\right)\otimes N\simeq \bigoplus_\alpha (M_\alpha\otimes N).$$
I've tried to simply generalize the proof I've seem in the book for the case $\alpha=1,2$. For that, I've defined $\varphi : (\bigoplus_\alpha M_\alpha)\times N\to \bigoplus_\alpha (M_\alpha\otimes N)$ by
$$\varphi((m_\alpha), n)=(m_\alpha\otimes n),$$
this is well defined, since being $m_\alpha = 0$ unless for finite indices we have $m_\alpha\otimes n$ also well zero unless for finite $\alpha$. Also, $\varphi$ is clearly bilinear, so that the universal property grants that there is an $R$-module homomorphism $\phi : (\bigoplus_\alpha M_\alpha)\otimes N\to \bigoplus_\alpha (M_\alpha\otimes N)$ such that $\phi((m_\alpha)\otimes n)=(m_\alpha\otimes n)$.
Then, let $i_\alpha : M_\alpha \to \bigoplus_\alpha M_\alpha$ be the canonical injection, that is
$$(i_\alpha(m_\alpha))_\beta=\begin{cases}0 & \alpha\neq \beta, \\ m_\alpha & \alpha=\beta.\end{cases}$$
Then we define for each $\alpha$ the map $f_\alpha : M_\alpha\times N\to (\bigoplus_\alpha M_\alpha )\otimes N$ by $f_\alpha(m_\alpha,n)=i_\alpha(m_\alpha)\otimes n$, then $f_\alpha$ is bilinear and by the universal property induces $g_\alpha : M_\alpha\otimes N\to (\bigoplus_\alpha M_\alpha)\otimes N$ an $R$-module homomorphism such that $g_\alpha(m_\alpha\otimes n)=i_\alpha(m_\alpha)\otimes n$. Define then $\psi : \bigoplus_\alpha (M_\alpha\otimes N)\to(\bigoplus_\alpha M_\alpha)\otimes N$ by
$$\psi((m_\alpha\otimes n))=\sum_\alpha g_\alpha(m_\alpha\otimes n)$$
Now here's the point. I'm unsure $\psi$ is well-defined, since the sum can be infinite. My thought on that is: since just finitely many of the $m_\alpha\otimes n$ are nonzero and $g$ is an $R$-module homomorphism, this sum should become finite and we can continue the proof. Is this reasoning correct?
After that I would show that $\phi$ and $\psi$ are inverses. This sounds easy, because we have
$$\psi(\phi((m_\alpha)\otimes n)))=\psi((m_\alpha\otimes n))=\sum_\alpha g_\alpha(m_\alpha\otimes n)=\sum_\alpha i_\alpha(m_\alpha)\otimes n=(m_\alpha)\otimes n,$$
and similarly we have
$$\phi(\psi(m_\alpha\otimes n_\alpha))=\phi\left(\sum_\alpha i_\alpha(m_\alpha)\otimes n_\alpha\right)=\sum_\alpha \phi(i_\alpha(m_\alpha)\otimes n_\alpha)=(m_\alpha\otimes n_\alpha),$$
and so $\psi$ and $\phi$ are inverses $R$-module homomorphism and hence the isomorphism exists.
Is the proof allright? I'm mainly unsure with the reasoning about the sum.
Thanks very much in advance.
Best Answer
Your reasoning is right and $g$ is well-defined (and hence a homomorphism as the $g_\alpha$ are). Given $(m_\alpha \otimes n) \in \bigoplus_\alpha M_\alpha \otimes N$, there are only finitely many $\alpha$ with $m_\alpha \otimes n \ne 0$, so for only finitely many $\alpha$ we have $g_\alpha (m_\alpha \otimes n) \ne 0$ (as $g_\alpha(0) = 0$). So the sum $\sum_\alpha g_\alpha(m_\alpha \otimes n)$ has only finitely many non-zero terms.