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Is it possible to use the ring $R=Z/4Z$ to construct a counter-example that submodules of free modules are not necessarily free?

Thanks a lot.

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$2R{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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Dear Georges: I hope I did what you wanted. (Otherwise you can make a new edit.) ($+1$) –  Pierre-Yves Gaillard Mar 31 '12 at 8:26
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To Georeges: you're welcome. To users: The trick is to add something like S{}{}{}$ with enough pairs of curly brackets. –  Pierre-Yves Gaillard Mar 31 '12 at 8:42
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You're confusing things: $2R$ is not free (for cardinality reasons, for example) while $R$ is free by definition. –  t.b. Mar 31 '12 at 8:57
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$2R$ has two elements, while a free module has a multiple of four elements. –  t.b. Mar 31 '12 at 9:11
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A free module is a module isomorphic to $R^n$, this latter has $4^n$ elements since $R$ has four of them. –  t.b. Mar 31 '12 at 10:07
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