# How do you count the elements in a finite vector space?

I was looking at the proof for Wedderburn's Little Theorem that skew fields are fields in the finite case. The proof involves viewing a ring $R$ as a vector space over its center $Z(R)$, and it claims rather nonchalantly that if the order of $Z(R)=q$ and the vector space $(R,Z(R))$ has dimension n, then the order of $R$ is $q^n$.

I've tried again and again to count them, I've asked my boss if he could count them, but could anyone explain why this is so? It seems like this shouldn't be a very deep claim at all, but I just can't seem to find them, because when I try I think that if $(R,Z(R))$ has dimension n and $Z(R)$ has order q, then we can write out every element in the space as $$(a_{k_1})(b_1) + (a_{k_2})(b_2)+ \ldots + (a_{k_n})(b_n)$$

Where each $a_{k_i}$ is in $Z(R)$ and the $b_i$ are in the basis. So then I say that the number of things in the space should be the number of unique ways to choose n elements out of q, or q elements out of n if the former is larger than the latter, but neither of those are the answer $q^n$.

I feel kind of silly not getting something this obvious but well my algebra course never covered field theory and linear algebra never covered finite vector spaces, so I guess it's not surprising really.

-
You're not counting correctly. Try the case $Z(R) = \mathbb{F}_2$ and small $n$ and explicitly write down all of the vectors. – Qiaochu Yuan Sep 20 '11 at 22:29

Every finite-dimensional vector space $V$ of dimension $n$ over a field $F$ is isomorphic to $F^n$ by choosing a basis. If $F$ has $q$ elements, then $V$ has $q^n$ elements.