TITLE: Is a vector space over a finite field always finite?
QUESTION [30 upvotes]: 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: $\forall \lambda, \mu \in \mathbb{K} \land \forall x, y \in V:$

$1 \cdot x = x$
$\lambda \cdot (\mu \cdot x) = (\lambda \cdot \mu) \cdot x$
$(\lambda + \mu) \cdot x = \lambda \cdot x + \mu \cdot x$
$\lambda \cdot (x + y) = \lambda \cdot x + \lambda \cdot y$

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

REPLY [3 votes]: $\frac{\mathbb{Z}}{2\mathbb{Z}}[X]$ is a vector space but is not finite since it contains $1, X, X^2, ...$