# 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 $k\in \mathbb{N}$ ($k\neq 0$).

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Infact, the prime-powers are the only $n's$ for which you can find a vector space with those many elements! –  user21436 Jan 23 '12 at 23:47

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?

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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 \end{align}