# Every module is the quotient of a free module.

Is every module a quotient of a free module by the following?

Suppose I have a module $M.$ Take the free module $F = \oplus_{m \in M}R.$ Construct a surjective module homomorphism by sending the element with 1 at the $m^{th}$ coordinate and 0 everywhere else to $m.$ Then, just $M = F/ker,$ hence the quotient of a free module?

• Yes: basically this is the argument. Commented Oct 22, 2017 at 22:23
• Ok but what if $M$ is uncountable. Doesn't coordinatizing imply ability to index? Commented Oct 22, 2017 at 22:25
• @ntntnt Why would $M$ being uncountable matter? I don't know what you mean by "doesn't coordinatizing imply ability to index," but the index set $I$ of $\bigoplus_{i\in I} R$ can be any set at all, no matter what the cardinality is. Commented Oct 23, 2017 at 13:08
• Oh because I think of $\oplus_{i = 1}^n R$ as $n-$degree coordinates for example. So if we can write all $m$ in the direct sum as a position in the coordinate then that gives each $m$ an index. Commented Oct 23, 2017 at 17:05
• @ntntnt Exactly: you can index with uncountable sets as well. Commented Oct 31, 2017 at 15:05