I have a curiosity question on a fundamental difference between vector spaces and general modules over rings.
Some of the fundamental facts of linear algebra:
(1) A finitely generated vector space has a basis.
(2) Minimal generating (spanning) sets of a vector space are linearly independent and therefore form a basis.
I recently took a course on modules. One basic example discussed: Let $R = K[x,y]$, where $K$ is a field, and let $I = \langle x,y \rangle $. We consider $I$ as a module over $R$.
$I$ is a finitely generated module, however it is not free (does not contain a basis). This is because the smallest generating set has size $2$, and no matter what generating set you choose, you can write a non-trivial $R-$linear combination of the elements of that set that equals $0$.
If $S = \{ f(x,y), g(x,y) \} $ so that $I = \langle S \rangle $, then $g(x,y)f(x,y) + (-f(x,y))g(x,y) = 0 $. A similar argument can be made for any finite generating set for $I$.
This example shows that those fundamental facts of vector spaces are not necessarily true for modules over general rings.
My question is what is it about the scalars coming from a field that makes these facts true but not so when the scalars come from a general ring? I never got a chance to ask my professor during the class. I tried reading proofs from linear algebra texts but I cannot see where the underlying scalar FIELD makes the difference.
Any clarification would be very helpful.