number of elements on a finite vector space I know that there is no vector space over any field $\mathbb{F}$ having precisely $6$ elements. I would like to know if there is a theorem which characterizes those $n's \in \mathbb{N}$ such that there is a vector space having exactly $n$ elements over a field $\mathbb{F}$. For example: $n=p^{k}$ where $p$ is a prime number and $k\in \mathbb{N}$ ($k\neq 0$).
I think you can put a proof together from the following ingredients. 

If $F$ is a finite field with $q$ elements, then $q$ is a power of a prime. 

If $V$ is a finite vector space over a field of $q$ elements, then it has a finite basis, which we'll call $\lbrace\,b_1,b_2,\dots,b_r\,\rbrace$, and every element has a unique expression as a linear combination of basis elements, and the number of such linear combinations is $q^r$. 

Are the "ingredients" things you already know?
You cannot have vector spaces with finite elements. Even if the field is finite having just finite elements
consider a vector space with basis $\{e_1, e_2, e_3,\ldots\}$
where
\begin{align}
e_1 &amp;= (1,0,0,0,\ldots)\\
e_2 &amp;= (0,1,0,0,\ldots)\\
e_3 &amp;= (0,0,1,0,\ldots)\\
\vdots &amp; = \vdots
\end{align}