A vector space must satisfy closure under addition and multiplication. Sorry if this is obvious but does that mean that, assuming the normal rules of arithmetic and excluding the trivial examples like $V=\{0\}$ it is impossible for a vector space to have a finite number of elements? (since for every two elements $u,v \in V$ there must be another number $u+v \in V$)
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2Just work with a finite dimensional vector space over a finite field – daw Jun 29 '16 at 19:51
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2This all depends on the field it is a vector space over. If the field is finite then yes, otherwise no. – Tobias Kildetoft Jun 29 '16 at 19:51
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An example of finite field is $\mathbb{Z}/p\mathbb{Z}$ for prime $p$. Also, there are finite fields of order $p^n$ for any prime $p$ and integer $n\geq 1$. – Sungjin Kim Jun 29 '16 at 19:56
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Ah okay, but if the vector field is over $\mathbb{R}$ or $\mathbb{C}$ it must have an infinite number of elements correct? – orange_juice Jun 29 '16 at 19:59
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Yes, since it has to contain all scalar multiples of an element, and these are already infinitely many. – Phil Jun 29 '16 at 20:03
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related – Ben Grossmann Jun 29 '16 at 20:05
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Also related – Ben Grossmann Jun 29 '16 at 20:07
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Such a vector space would have to be finite-dimensional, of course.
But every finite-dimensional vector space $V$ over a field $\Bbb k$ is isomorphic to $\Bbb k^{\oplus n}$ for some $n$. This means that $V$ has finitely many vectors if and only if $V$ is finite-dimensional and $\Bbb k$ is a finite field.
For each prime $p$ and $n\geq0$ it is known that there is a unique field of order $p^n$. This classifies all possible vector spaces with finitely many vectors.
Brian Fitzpatrick
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