Module is Direct Sum of Cyclic Modules

Question

Is every $R$-module the direct sum of cyclic submodules? It seems to me that the answer is yes, as for any $R$-module $M$ we have that $$ \bigoplus_{r\in M} Rr = M. $$

Containment in both directions is (seemingly) obvious. Is this correct?

Answer 1

Let's look at $R = \Bbb Z/3\Bbb Z$ and $M = \Bbb Z/3\Bbb Z$, the simplest example (it's even finite!).

Well, $\displaystyle \bigoplus_{r \in M} Rr = (\Bbb Z/3\Bbb Z)0 \oplus (\Bbb Z/3\Bbb Z)1 \oplus (\Bbb Z/3\Bbb Z)2 = 0 \oplus (\Bbb Z/3\Bbb Z) \oplus (\Bbb Z/3\Bbb Z) = (\Bbb Z/3\Bbb Z)^2 = R^2$, but $M = R$!

In fact, $\displaystyle \left| \bigoplus_{r \in M} Rr \right| = 3 \times 3 \ne 3 = |M|$.

Answer 2

Of course, the answer is NO (see Kenny's answer). However this question is very interesting and there is a class of rings for which the answer is YES (for a certain choice of generators $r$, not all $r\in M$ of course).

See I. S. Cohen and I. Kaplansky, Rings for which every module is a direct sum of cyclic modules, in Cambridge, Mass. and Chicago, Ill. Mathematische Zeitschrift, Band 54, Heft 2, S. 97--101 (1951).