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 $k\in \mathbb{N}$ ($k\neq 0$).
 A: Let V be a finite vector space over a field F.
Since any two distinct scalars multiplied to a non zero vector will give distinct vectors,and since V has only finite number of elements, F must be finite. Hence the number of elements of F must be the n-th (say) power of some prime p.
Since V is finite, it must be m- dimensional for some m. So V is isomorphic to the direct sum of m copies F. So the number of elements of V must be the mn-th power of p.
Therefore, the number of elements of a non zero finite vector space also must have the number of elements a power of a prime number.
A: 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?
A: 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 &= (1,0,0,0,\ldots)\\
e_2 &= (0,1,0,0,\ldots)\\
e_3 &= (0,0,1,0,\ldots)\\
\vdots & = \vdots
\end{align}
