Definition of a vector space:

Let $V$ be a set and $(\mathbb{K}, +, \cdot)$ a field.

$V$ is called a vector space over the field $\mathbb{K}$ if:

V1: $(V, +)$ is a commutative group

V2: $\forall \lambda, \mu \in \mathbb{K} \land \forall x, y \in V:$

  1. $1 \cdot x = x$
  2. $\lambda \cdot (\mu \cdot x) = (\lambda \cdot \mu) \cdot x$
  3. $(\lambda + \mu) \cdot x = \lambda \cdot x + \mu \cdot x$
  4. $\lambda \cdot (x + y) = \lambda \cdot x + \lambda \cdot y$

My question:

If you have a vector space over a finite field $\mathbb{K}$, is the set $V$ always finite?

My examples

An example for a finite vector space is

$V = (\mathbb{Z}/2\mathbb{Z})^n, n \in \mathbb{N}$ over the field $\mathbb{Z}/2 \mathbb{Z}$.

I've tried to find a infinite vector space (I mean the number of vectors should be infinite) over a finite field. I chose $\mathbb{Z}/2 \mathbb{Z}$ as my field and $V = \mathbb{R}^2$. But in this case V2.3 doesn't work:

$\lambda = \mu = 1, x = \begin{pmatrix}1\\2\end{pmatrix}$:

$(\lambda + \mu) \cdot x = (1+1)\cdot x = \begin{pmatrix}0\\0\end{pmatrix} \cdot x = 0 \neq \begin{pmatrix}2\\4\end{pmatrix} = 1 \cdot x + 1 \cdot x$

  • 8
    $\begingroup$ What do you think about $\mathbb{K}[X]$ if $\mathbb{K}$ is finite ? $\endgroup$
    – Ahriman
    Sep 7, 2012 at 10:35
  • 2
    $\begingroup$ I think this might the answer to my question. If $\mathbb{K}$ is finite, then you still have an infinite amount of polynomials, don't you? The basis of $\mathbb{K}[X]$ has still an infinite amout of vectors, doesn't it? $\endgroup$ Sep 7, 2012 at 10:39
  • $\begingroup$ Yes, $\{1, X, X^2, ..., X^n, ... \}$ is an infinite subset of $\mathbb{K}[X]$. $\endgroup$
    – Ahriman
    Sep 7, 2012 at 10:40

4 Answers 4


Yes to your comment below Ahriman's, Moose.

These are not the only examples, though: if $\,\Bbb F=\Bbb F_p\,$ is the prime finite field of order a prime $\,p\,$, then $\,\Bbb F\times \Bbb F\times...\,$ is an infinite vector space over $\,\Bbb F\,$.

In short: a vector space over a finite field is finite iff it is finite dimensional.

  • 3
    $\begingroup$ In fact, $\Bbb F\times\Bbb F\times...$ is isomorphic to $\Bbb F[X]$, is it not? $\endgroup$ Sep 7, 2012 at 12:54
  • $\begingroup$ Indeed it is, @Cameron...as $\,\Bbb F-\,$ vextor spaces, of course. Yet it is possible to take the direct limit indexing the field by any infinite set, not only a countable one. $\endgroup$
    – DonAntonio
    Sep 7, 2012 at 13:29
  • $\begingroup$ Fair point. I've only seen that use of the ellipsis in the context of a countable collection, personally. $\endgroup$ Sep 7, 2012 at 13:44
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    $\begingroup$ Actually $\Bbb F\times \Bbb F\times...$ does not exist in my vocabulary. It looks to me like $\prod_{i\in\mathbf N}\mathbb F$ is intended, which is not isomorphic to $\mathbb F[X]$ as a vector space, but to $\mathbb F[[X]]$. $\endgroup$ Sep 7, 2012 at 13:46
  • 3
    $\begingroup$ @DonAntonio: no, the direct product of countably many copies of $\mathbb{F}$ is uncountable-dimensional. You want the direct sum. $\endgroup$ Sep 7, 2012 at 17:19

The direct sum $F=\bigoplus_{i\in I} \Bbb F$ is a vector space over $\Bbb F$ of dimension equal to the cardinality of I. Thus you can get a vector field of any dimension, for every field.


$\frac{\mathbb{Z}}{2\mathbb{Z}}[X]$ is a vector space but is not finite since it contains $1, X, X^2, ...$


If a commutative ring $R$ with unity contains a field $F$, then $R$ is a vector space over $F$. Thus it suffices to find an infinite commutative ring with unity that has a finite field as a subfield. One example is $F[x]$ for any finite field $F$.

  • $\begingroup$ Not every commutative ring with unity contains a field (e.g. $\mathbb{Z}$). Your suggestion works if you consider infinite $F$-algebras though. $\endgroup$
    – Sebastian
    Sep 7, 2012 at 14:48
  • $\begingroup$ Well, I meant commutative rings with unity that do contain a field. I'll edit to make it more clear $\endgroup$ Sep 7, 2012 at 14:51

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