I am well aware of the distributivity of tensor product over direct sum for $R$-modules over a commutative ring. I see how it works for direct sums and tensor products of the form
$$(\oplus_\alpha M_\alpha)\otimes N\cong \oplus_\alpha (M_\alpha\otimes N)$$
However, what happens if $N$ is also a direct sum depending on $\alpha$? For instance, let $M=\oplus_{\alpha\in\mathbb{Z}}M_\alpha$ and similarly $N=\oplus_{\alpha\in\mathbb{Z}}N_\alpha$. Consider the following direct sum of tensor products for a fixed integer $k$:
$$\oplus_{\alpha\in\mathbb{Z}} (M_\alpha\otimes N_{k-\alpha})$$
If I had independent indices for $M$ and $N$ I could simply use the above isomorphism and the analogue on the right to write the sum of tensor products as a tensor product of sums, obtaining something like
$$(\oplus_{\alpha\in\mathbb{Z}} M_\alpha)\otimes (\oplus_{\beta\in\mathbb{Z}} N_{\beta})$$ But in this case, $\beta=k-\alpha$, so it doesn't seem to make sense to write the tensor in that way. If I wrote
$$(\oplus_{\alpha\in\mathbb{Z}} M_\alpha)\otimes (\oplus_{\alpha\in\mathbb{Z}} N_{k-\alpha})$$ each $\alpha$ is free, so $k-\alpha$ would take any value independently of the value of the $\alpha$ in the left factor.
Is there a way I can write the sum
$$\oplus_{\alpha\in\mathbb{Z}} (M_\alpha\otimes N_{k-\alpha})$$
as tensor product of sums? Or some way of using distributivity in this case?