Examples of vector spaces defined on fields other than field of numbers: $\Bbb Q,\Bbb R,\Bbb C$ etc.? I'm using the textbook "Linear Algebra Done Right" by Sheldon Axler. The author mentions that vector spaces can be generally defined over any field. But the only examples that can I spot in the textbook are of vector fields defined over $\Bbb R$ or $\Bbb C$. 
Could someone please give me some common examples of vector spaces which are not defined on field of numbers: $\Bbb Q,\Bbb R,\Bbb C$ etc.? I tried searching on the net, but couldn't find anything relevant.
 A: The set of polynomials over a field, $\Bbb F[x]$, has many properties of a field, but doesn't have multiplicative inverses. If we define a set $\Bbb W$ which has elements of $\Bbb F[x]\,mod\,h(x)$ where $h(x)$ is an irreducible non constant polynomial in $\Bbb F[x]$ (can't be split into non constant factors), then we can show this is a field. For example, $(x+1)(x^2+1)+3\ mod\ (x^2+1) = x+4$
All the addition axioms are obvious and simple to check with the addition identity being the $0$ polynomial and the inverse of $f(x)$ given by $-f(x)$.
We use the division algorithm for polynomials which states that $\forall f(x),\ g(x) \in \Bbb F[x], \exists\ a(x),\ b(x) \in \Bbb F[x]$ such that $f(x)=a(x)g(x)+b(x)$ and $deg(b(x)) < deg(g(x))$ where $g(x) \neq 0$. Using this with $g(x) = h(x)$ we see that multiplication is closed as in $\Bbb W$, $f(x)$ would be $b(x)$. Multiplication is clearly commutative, associative and has identity $1$.
Using the division algorithm it is possible to show that $\forall f(x),\ g(x) \in \Bbb F[x] \backslash \{0\},\ \exists\ a(x),\ b(x) \in \Bbb F[x]$ such that $a(x)f(x) + b(x)g(x) = gcd(f(x), g(x))$. We can use this to show the existence of a multiplicative inverse.
Take $f(x) \in \Bbb W$ and assume $f(x)=c$ is a constant function. Then it's inverse is given simply by $\frac 1c$. For the non trivial case assume $f(x) \neq constant$. As $f(x) \in \Bbb W$ we know that h(x) doesn't divide $f(x)$, and as $h(x)$ is irreducible it can't divide $f(x)$. This means that $gcd(f(x), h(x)) = 1$, and we use the result above to get $a(x),\ b(x)$ such that $$a(x)f(x)+b(x)h(x)=1\\a(x)f(x)=1-b(x)h(x)\\a(x)f(x)=1$$ as $b(x)h(x)=0$ in $\Bbb W$. This means that $a(x)$ is a multiplicative inverse of $f(x)$.
The distributive laws also hold and are simple to check, and so $\Bbb W$ is a field. We could find some trivial vector spaces such as $\Bbb W^n$ or if you wanted something more interesting then maybe the set $\Bbb W[y]$ which is the set of polynomials with coefficients in $\Bbb W$, or the set of matrices with elements in $\Bbb W$. When checking the vector space axioms you don't use any specific properties of the field, you only need the axioms of a vector space, and so any vector space you can think of will work under any field.
A: To define a vector space over a field other than the rationals, the reals or the complex numbers you have to start with such a field. 
For example, the set  $\{0,1\}$ with arithmetic modulo $2$ is a field. Then the set of $n$-tuples ("vectors") with coefficients in that field is a vector space over that field.
For example, the set of eight vectors $(a,b,c)$ where each of the entries is either $0$ or $1$ is a three dimensional vector space over that two element field. (Just remember that you do arithmetic modulo $2$, so $(1,1,0) + (1,0,1) = (0,1,1)$.)
There are many other fields, some finite, some not. With any of them you can form vector spaces of any finite dimension in just this way. See https://en.wikipedia.org/wiki/Field_(mathematics) .
(The integers and the natural numbers aren't fields.)
