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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.

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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.

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