Is a vector space over a finite field always finite?

Question

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:

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $F$.

Answer 4

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