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).

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