Dimension of vector spaces with finitely many elements Background -
I am currently taking an introductory linear algebra course and we were introduced to vector spaces. Now, at first it seemed pretty much 'obvious' that the only vector space with finitely many elements is the trivial $\{ 0 \}$ as all other finite sets seemed unable to satisfy closure.
However I recently realized that using 'remainders', i.e. modular arithmatic would allow me to sidestep this problem. So, I experimented a bit and came up with the following general vector space with $(n + 1)$ elements (over the field $\mathbb{R}$)-
$\{ V , +, \cdot \}$
$V= \{ a \cdot i : a \in \mathbb{R}; n \in \mathbb{N} , 0 \leq i \leq n \}$
$\forall \ v_1, v_2 \in V \ \ v_1 + v_2 = ((\frac{v_1 + v_2}{a}) \ (\text{mod} n)) \cdot a$
$\forall \ \alpha \in \mathbb{R}, v \in V \ \ \alpha \cdot v = (\vert [ \alpha \cdot \frac{v}{\pi} ] \vert (\text{mod} n)) \cdot a$, where $[ ... ]$ denotes the greatest integer function.
It obeys closure, commutes and is associative. I had the addditive and multiplicative identities as $0$ and $1$ and I was able to define the additive inverse of $v$ as $(n-v)$.

Question -
My question relates to the fact that the the dimension of a vector space is defined as the number of elements in its basis. Before I constructed the above space, I understood that vector spaces ${\mathbb{R}}^n(\mathbb{R})$ have dimension $n$, $\mathbb{Z} (\mathbb{Z})$ has dimension 1. While $\{ 0 \}$ and {by extension) all finite element vector spaces have dimension $0$.
However, I am now unsure whether that is the case. After all is $\{ 1 \}$ not a basis for the vector space I defined? It has $1$ element.
So, my question is -

What would be the dimension of a vector space with finitely many elements like the one I defined?

It would be great if if the answer could provide some background/intuitive justification for the choice made in the case in addition to stating it.
 A: Good for you for being curious. That said, your definition does not make a vector space. I suggest you fix $n=1$ and actually write down a few "vectors" in your vector space and try adding them and multiplying by scalars. In particular, in a vector space over the real numbers, $av \ne bv$ whenever $a \ne b$ and $v \ne 0$. Is that true for your operations?
You can make finite vector spaces using modular arithmetic - if you use "remainders" for the scalars. For example, suppose the scalars are from the field $\mathbb{Z}_3$ with elements $\{0, 1, 2\}$ and arithmetic modulo $3$. Then the two dimensional vector space of pairs of scalars has $9$ elements. Sample vector arithmetic:
$$
(1, 0) + (2,1) = (0,1)
$$
and
$$
2(1,2) = (2,1) .
$$
Technical note. You can do this for any $n$, but you get a vector space only when $n$ is prime. For composite values the vector space axioms still all work but the object is called a "module".
A: A vector space $V$ over the field $ \mathbb R$ with $V \ne \{0\}$ has infinitely many elements! Reason: let $v \in V$ with $v\ne 0$. Then
$$\{tv: t \in \mathbb R\} \subseteq V$$
and $\{tv: t \in \mathbb R\}$ has infinitely many elements.
