I was wondering about this problem. If we consider $\mathbb{R}^2$ as a $\mathbb{R}$ vector space it has dimension 2. What if we consider it a $\mathbb{F}_3$ vector space? I thought about the following thing:
If we want to generate $\begin{pmatrix} 7 \\ 8\end{pmatrix}\in \mathbb{R}^2$ using the standard basis $\left\{ \begin{pmatrix} 1 \\ 0\end{pmatrix},\begin{pmatrix} 0 \\ 1\end{pmatrix}\right\}$, a sum like $\sum_{i=0}^2 a_i e_i$ with $a_i\in\mathbb{F}_3$ will not be sufficient since we the maximum we can obtain is clearly $\begin{pmatrix} 2 \\ 2\end{pmatrix}$. So in my opinion the dimension should be greater than 2.
I don't know why, but I got the feeling that maybe $\text{dim}_{\mathbb{F}_3}\mathbb{R}^2=\text{dim}_{\mathbb{R}}\mathbb{R}^2\cdot\text{dim}_{\mathbb{R}}\mathbb{F}_3$.
Thank you in advance
A.
You can't have $\Bbb R^2$, with regular addition, be a vector space over $\Bbb F_3$. It's simply not possible to fulfill the axioms.
Specifically, take the axiom that says
Now let $a = 1, b = 2$, and see what happens.
If you try to salvage this by saying "we define scalar multiplication by $2$ to make this work", then you run into trouble with $a = b = 1$ instead. And you can't salvage it by changing how the scalar multiplication with $1$ works, because scalar multiplication by $1$ is uniquely defined by another axiom.