# Tensor product and direct sums

I have an integral domain $R$, and $R$-modules $M$, $N_1$, $N_2$. I know that there is an $R$- module isomorphism $$M\otimes_R (N_1\oplus N_2)\cong (M\otimes_R N_1)\oplus(M\otimes_R N_2).$$ where $m\otimes (n_1,n_2)\to (m\otimes n_1,m\otimes n_2).$

What is the corresponding isomorphism in the case that $N_1$ and $N_2$ are submodules of $M$ and we regard $N_1\oplus N_2$ as an internal direct sum and how does it follow from the one above?

• I'm not sure I understand: the above works in any case : why would you expect it to be otherwise if $\;N_i\le M\;$ ? What difference you think there could be if the direct sum was internal o external in this fact? – DonAntonio Jun 13 '14 at 21:11
• I think perhaps the confusion may arise from not knowing the following fact: internal and external direct sums are isomorphic (when the former makes sense). – RghtHndSd Jun 13 '14 at 21:18

The external direct sum of $N_1$ and $N_2$ is canonically isomorphic to their internal direct sum by $$N_1\oplus N_2\ \longrightarrow\ N_1+N_2:\ (n_1,n_2)\ \longmapsto\ n_1+n_2,$$ whenever the internal direct sum exists, of course. So nothing changes except for some notation; for the isomorphism you could in stead write $$m\otimes(n_1+n_2)\ \longmapsto\ (m\otimes n_1,m\otimes n_2).$$
• So is the map $m\otimes (n_1,n_2) \mapsto m\otimes (n_1+n_2)$ an $R$-isomorphism from $M\otimes (N_1\oplus N_2)$ into $M\otimes (N_1+N_2)$? Regards – user149343 Jun 13 '14 at 21:45
• Yes; this map is the tensor product of isomorphisms (the identity on $M$ and the isomorphism I gave), so it is again an isomorphism. – Servaes Jun 13 '14 at 21:49