Why define vector spaces over fields instead of a PID? In my few years of studying abstract algebra I've always seen vector spaces over fields, rather than other weaker structures. What are the differences of having a vector space (or whatever the analogous structure is called[module?]) defined over a principal ideal domain which is not also a field? What properties of vector spaces break down when defining it over a PID? Please give an example (apart from letting the vector space be the field/PID itself) where the vector space over a field has properties that the module over a PID does not.
 A: It is not just one way. Modules over $\mathbb Z$ are abelian groups, and have a rich structure theory, involving the existence of finite modules, and prime factorisation.
Passing to $\mathbb Q$ as field of fractions we find that every scalar is a unit and every non-trivial vector space is infinite.
So modules can have properties which vector spaces don't share.
A: Actually, someone has already asked a question about Pathologies in module theory, whose solutions will show you a lot about how module theory differs (in interesting ways) from vector space theory. In short, when you begin to lose nice properties, you begin to appreciate them more :)
After learning about vector spaces and their usefulness in linear algebra, they serve as a baseline for intuition about module theory. When you pass from a field to a less-nice ring, things begin to break down. This isn't really a bad thing: we just learn more about how the nice properties come about.
$\Bbb F[x]$ is an important principal ideal domain, and it turns out a lot of linear algebra can be recovered from module theory over these polynomial rings.
$\Bbb Z$ is also an important PID, and the study of its modules is the study of abelian groups.
But there are still lots of other questions to ask. For example: What rings have modules who are isomorphic to copies of the ring? (answer: division rings.)
For what rings are the modules direct sums of simple modules? (answer semisimple rings.
For what rings are the modules all flat? (answer: von Neumann regular rings)
So you can see that moving from fields to PIDs is just one step in a big web of moving outward from fields, exploring what their modules look like. You don't have to stop at PID's!
A: There is a very nice structure theory for finitely generated modules over a PID, which is frequently discussed here at MSE. 
However, the general theory is much more complicated already in the case of (not finitely generated) abelian groups, i.e., $\Bbb{Z}$-modules.
The first thing that's missing is that not all modules over a PID will have a base. Take for instance the $\Bbb{Z}$-module $\Bbb{Z}/2\Bbb{Z}$. When dealing with vector spaces, the only invariant is the dimension. With modules over a PID, this is not the case anymore..
