TITLE: Why is every vector space free, but not every module?
QUESTION [5 upvotes]: For every vector space $V$ over a field $K$ we have that $V \cong K^n$ for some natural number $n$, and so every vector space has a basis (of $n$ elements in this case). Therefore every vector space is free. 
However, for modules, this is not the case. Most modules are not free. Let $R$ be a ring, and $M$ and $R$-module, then we don't necessarily have $M \cong R^n$ for some $n$, not even $M \cong R^{(I)}$ for some set $I$. 
Why is this the case, conceptually? The only different is that we are working over a ring instead of a field. To me, the vector space case seems weird, since it is normally not the case that every algebra in some class is free. For example we also don't have that every group is a free group, etc.

REPLY [2 votes]: As GreginGre mentioned, fields are very particular rings. There are many situation where having a field yields more properties and that is pretty much all the time due to the invertability of every non-zero element.
Another example is that $R[T]$ is a pid iff $R$ is a field.