I'm having a little trouble understanding vector spaces over abstract fields and I was wondering are the integers a vector space over the field Z mod 2. thanks for any help
when referred to as a field Z2 is typically called F2 (that should look like this: $\displaystyle F_2$) or sometimes $\displaystyle GF_2$ (for "galois field").
this field is finite, so all finite-dimensional vector spaces over F2 are also finite (for example, F2 x F2 has four elements).
Z is infinite, so Z cannot possibly be a finite-dimensional vector space over F2.
well, you might say, how about an infinite-dimensional vector space over F2? again, here that is impossible.
suppose v is ANY vector in an infinite-dimensional vector space over F2.
then v+v = 1v + 1v = (1+1)v = 0. so our vector space cannot be the integers, because there is only ONE integer n for which n+n = 2n = 0,
and that is, the integer 0.
for a finite-dimensional vector space V over a field F, the most profitable way to think of V is: dim(V) copies of F.
for infinite-dimensional vector spaces the situation is more complicated, because these can display more variety,
in general, finding a basis for an infinite-dimensional vector space is a very complicated question, although some
infinite-dimensional spaces are "nice", like F[x], the space of all polynomials in F, which has the basis {1,x,x^2,x^3,.....etc.}
which is countable.
Something to always check, as Deveno well points out (although implicitly), is that if $\displaystyle V$ and $\displaystyle W$ are isomorphic as vector spaces, they are in particular, isomorphic as abelian groups. And, as Deveno pointed out, $\displaystyle \mathbb{Z}_2^{\lambda}$ for any cardinal $\displaystyle \lambda$ has the property that $\displaystyle 2\mathbb{Z}^\lambda=\{0\}$ which can't be true for $\displaystyle \mathbb{Z}$.
EDIT: I feel as though I should clarify, just in case this wasn't clear, the fact that I stated about $\displaystyle \mathbb{Z}^\lambda$ was not because any $\displaystyle \mathbb{Z}_2$-space is isomorphic to such a space but because any $\displaystyle \mathbb{Z}_2$-embeds into such a space. Any $\displaystyle \mathbb{Z}_2$-vector space is isomorphic to a vector space of the form $\displaystyle \mathbb{Z}^{\oplus\lambda}$.
personally, i'm a big fan of doing things in this order:
1). studying R^n and matrices for a little bit.
2). learning a little group theory.
3). learning linear algebra proper.
4). learning some more group theory, a little ring theory, and then tackling R-modules.
often the order of 2 and 3 is transposed, or 2 is omitted entirely. this often obscures some of the natural motivations for why we care about linear transformations, null spaces, subspaces and images spaces (span sets). abelian groups are very "nice" algebraic structures, and much of the regularity of linear algebra comes from this. in fact, i feel too much emphasis is placed on the underlying field in most people's first look at linear algebra, the only reason we want a field in the first place, is so we have enough algebraic rules for something like row-reduction to work. so even R has "too much structure", working over Q would be "good enough".
I would agree with that. I definitely think people have a tendency to play up all the features of a structure, forgetting that each component of the structure is important. Having done a lot of 'set theory for the working mathematician' I have definitely noticed that a lot of people forget to check cardinalities! You don't know how often someone will ask my why structure A and structure B aren't equivalent (in some concrete category ), and I'll say "They aren't even equipotent!". My anecdotal story is the student who couldn't figure out why $\displaystyle \mathbb{R}$ is cyclic. Or, there's also the classic why can't $\displaystyle \mathbb{R}$ be a finite dimensional $\displaystyle \mathbb{Q}$-space.