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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?

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    $\begingroup$ 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? $\endgroup$ – DonAntonio Jun 13 '14 at 21:11
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    $\begingroup$ 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). $\endgroup$ – RghtHndSd Jun 13 '14 at 21:18
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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 $$m\otimes(n_1+n_2)\ \longmapsto\ (m\otimes n_1,m\otimes n_2).$$

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  • $\begingroup$ 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 $\endgroup$ – user149343 Jun 13 '14 at 21:45
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    $\begingroup$ Yes; this map is the tensor product of isomorphisms (the identity on $M$ and the isomorphism I gave), so it is again an isomorphism. $\endgroup$ – Servaes Jun 13 '14 at 21:49

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