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$\begingroup$

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.

2 Answers 2

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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.) ($+1$)2012-03-31
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    Dear @Pierre-Yves : yes, that's exactly what I wanted. A thousand thanks. [To users : I couldn't write the very short answer you are now seeing because the software required that I write at least 30 characters]2012-03-31
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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.2012-03-31
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    Anyone could give a brief reason why 2R is free, while R is not free? I was thinking that R=Z/4Z is free since {[1]} is a basis? Thanks a lot.2012-03-31
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    You're confusing things: $2R$ is not free (for cardinality reasons, for example) while $R$ is free by definition.2012-03-31
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    anyone can briefly explain why 2R is not free? thanks a lot.2012-03-31
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    $2R$ has two elements, while a free module has a multiple of four elements.2012-03-31
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    Is that based on Lagrange's Theorem? I thought the "order of the subgroup divides the order of the group" so a free submodule has an order dividing four? (am very new to modules; beginner here)2012-03-31
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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.2012-03-31
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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.