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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
up vote 10 down vote accepted

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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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.

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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}

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Any vector space of finite dimension over a finite field has finitely-many elements:…. – Austin Mohr Nov 1 '12 at 2:34

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