# Using $\mathbb{Z}/4\mathbb{Z}$ to construct a counterexample to submodules of free modules being free

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