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

Thanks a lot.

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up vote 6 down vote accepted


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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
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
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
$2R$ has two elements, while a free module has a multiple of four elements. – t.b. Mar 31 '12 at 9:11
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

For any commutative ring $R$ with unity, $R$ is a free module over itself. If $0\ne a\in R$ is a zero-divisor, then the principal ideal generated by $a$ as a submodule of $R$ is not free. So your example works.

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