# Submodules of Free modules counter-example: $Z/4Z$

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.

-
add comment

## 1 Answer

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

-
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
show 4 more comments