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If F is a skew field then are the arbitrary direct sums of F-modules isomorphic to their direct products (over the same index).

I mean, if R is a division ring, and $\{M_i\}_{i\in I}$ some familly of R-modules, then is there necesarilyl an isomorphism of R-modules $\underset{i\in I}{\prod} M_i \rightarrow \underset{i\in I}{\bigoplus} M_i$?

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    $\begingroup$ This is false by defintion if I is infinite.... $\endgroup$ – AIM_BLB Feb 23 '14 at 2:06
  • $\begingroup$ Dear @CSA : what definition are you referring to? $\endgroup$ – rschwieb Feb 23 '14 at 23:46
  • $\begingroup$ Well the direct sum of modules can be identified with the submodule of the direct product such that only a finite terms in the product are nonzero.... This will force the second module to be a quotient of the free R-module on I generators, where as the product may be much much bigger, say if each $M_i$ is isomorphic then their product would be of dimension $Card(M_i)^{I}$, which is atleast as big as the bigger of $M_i$ and $2^I$. $\endgroup$ – AIM_BLB Feb 24 '14 at 1:57
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    $\begingroup$ @CSA I agree completely with that :) (That doesn't really fall under the category of "by definition" though.) $\endgroup$ – rschwieb Feb 24 '14 at 18:43
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No, just think for a moment along the lines of $M=\oplus_{i\in \Bbb N} \Bbb Q$ and $N=\prod_{i\in \Bbb N} \Bbb Q$.

The first is countable (being a countable union of countable subsets), and the second is uncountable (you can just adapt Cantor's diagonal argument.)

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