## Binary tensor products essentially commute with arbitrary direct sums

Let $R$ be a ring, let $I$ be a nonempty set, let $M$ be a right $R$-module, and let $\{N_i\}_I$ be a family of left $R$-modules. Prove that as abelian groups, $M \otimes_R (\bigoplus_I N_i) \cong \bigoplus_I (M \otimes_R N_i)$.

We will make use of the universal properties of tensor products and direct sums to prove this without too many fine details.

Recall that for each $i$, we have a canonical injection $\iota_i : N_i \rightarrow \bigoplus_I N_i$. Thus for all $i$ we have a group homomorphism $1 \otimes \iota_i : M \otimes_R N_i \rightarrow M \otimes_R (\bigoplus_I N_i)$. By the universal property of direct sums, we have a (unique) group homomorphism $\Phi = \bigoplus_I (1 \otimes \iota_i) : \bigoplus_I (M \otimes_R N_i) \rightarrow M \otimes_R (\bigoplus_I N_i)$ such that $\Phi((m \otimes n_i)) = \sum m \otimes \iota_i(n_i)$.

Now define $\psi : M \times \bigoplus_I N_i \rightarrow \bigoplus_I (M \otimes_R N_i)$ by $(m,(n_i)) \mapsto \sum_{n_i \neq 0} m \otimes n_i$. This mapping is well defined because for a fixed $(n_i)$, finitely many $n_i$ are nonzero. Evidenly $\psi$ is bilinear, and so induces a group homomorphism $\Psi : M \otimes_R (\bigoplus_I N_i) \rightarrow \bigoplus_I (M \otimes_R N_i)$.

We claim that $\Phi$ and $\Psi$ are mutual inverses. To see this, note that $(\Phi \circ \Psi)(m \otimes (n_i)) = \Phi(\sum m \otimes n_i) = m \otimes (n_i)$ and $(\Psi \circ \Phi)((m \otimes n_i)) = \Psi(\sum m \otimes \iota_i(n_i))$ $= \Psi(m \otimes \sum \iota_i(n_i))$ $= (m \otimes n_i)$.

Thus both $\Phi$ and $\Psi$ are group isomorphisms, and we have $M \otimes_R (\bigoplus_I N_i) \cong \bigoplus_I (M \otimes_R N_i)$.