What happens when vector spaces and its field are different? Most of the times you have an $\mathbb{K}$-vector space over $\mathbb{K}$
What happens when you have an $\mathbb{K_1}$-vector space over $\mathbb{K_2}$ (two different vectorspaces)
I had some thoughts here is my summary. Correct me where I'm wrong.


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*$\mathbb{R}$-vector space over $\mathbb{Q}$


This one statisfies the vector space axioms. The familiy $\left \{  1,2\right \}$ is not linear dependent but the family $\left \{1, \sqrt{2}  \right \}$ is linear independent.
What about the dimension? I can't figure out a basis fot this one.
Is it just $\left \{z,z^1,z^2,z^3,...  \right \} ; z \in \mathbb{Q}$? You get $\mathbb{Q_p}$ which is the closest to $\mathbb{R}$ ?
What happens with $\mathbb{Q}$-vector space over $\mathbb{R^2}$ Propapbly the same? 


*$\mathbb{Q}$-vector space over $\mathbb{R}$


This isn't a vector-space because it doesn't statisfy scalar multiplication.
Is it impossible to make a vectorspace over a field with more elements than the vectorspace? (This would help with 5.)


*$\mathbb{C}$-vector space over $\mathbb{Q}$


I know that $\mathbb{C}$-vector space over $\mathbb{R}$ is a valid thing and  $\left \{1,i \right \}$ is a basis. Can you say the same thing with $\mathbb{Q}$
It propably has the same dimension as a $\mathbb{R}$-vector space over $\mathbb{Q}$


*$F_{p_1}$-vector space over $F_{p_2}$


Is $F_3$ a vector space over $F_5$? Or something similar?


*Is there any way to connect a finite field with an infinite vector-space or the other way around?


$\mathbb{K}$-vector space over $F_{p}$ when $\mathbb{K}$ is infinite doesnt make sense. What is with $F_p$-vector space over $\mathbb{K}$ when $\mathbb{K}$ is infinite? Is there any example or any way to make this happen?
 A: If $\Bbb K_1$ is a subfield of $\Bbb K_2$ (such as $\Bbb Q$ is a subfield of $\Bbb R$ and  $\Bbb R$ is a subfield of $\Bbb C$), and $V$ is a $\Bbb K_2$-vector space, then we can naturally view it also as a $\Bbb K_1$-vector space. In particular, $\Bbb K_2$ is a $\Bbb K_1$-vector space. For dimensions, we have
$$\dim_{\Bbb K_1}V=\dim_{\Bbb K_2}V\cdot \dim_{\Bbb K_1}\Bbb K_2.$$
Indeed, if $\{\,v_i\mid i\in I\,\}$ is a basis of $V$ as $\Bbb K_1$-space and $\{\,a_j\mid j\in J\,\}$ is a basis of $\Bbb K_2$ as $\Bbb K_1$-space, then one readily verifies that $\{\,a_jv_i\mid (i,j)\in I\times J\,\}$ is a basis of $V$ as $\Bbb K_1$-space.
So for example if $V$ is a three-dimensional $\Bbb C$-space, it can be viewed as a six-dimensional $\Bbb R$-space (because $\Bbb C$ is a two-dimensional $\Bbb R$-space). Also note that $\dim_{\Bbb Q} \Bbb R$ is infinite, more precisely, $\dim_{\Bbb Q} \Bbb R=|\Bbb R|$ so it is "more infinite" than for example the vector space $\Bbb Q[X]$ of polynomials with rational coefficients. And there is a principal obstacle to explicitly listing a basis: The very existence of such a basis relies on the Axiom of Choice and thus makes it non-constructive.

You also ask about "vector space over $\Bbb R^2$". This does not make sense, as $\Bbb R^2$ is not a field (though $\Bbb C$ is a field that has an additive structure isomorphic to $\Bbb R^2$).

You ask whether it is possible to make a $\Bbb Q$-vector space into an $\Bbb R$-vector space, for example. No, it isn't in general. In particular, if $V$ is a $\Bbb K$-space and $\dim_{\Bbb K} V>0$ then $|V|\ge|\Bbb K|$ (as $a\mapsto av$ is an injection $\Bbb K\to V$ for any $v\ne0$ we pick). Hence the only vector space with $|V|<|\Bbb K|$ is the zero-dimensional space $\{0\}$.
But you may learn later that we can produce a different vector space that in some way is the simplest extension of a $\Bbb K_1$-space $V$ that also is a $\Bbb K_2$ space. This is called tensor  product, written $V\otimes_{\Bbb K_1}\Bbb K_2$. To give you a feeling, if $V=\Bbb Q[X]$ then $V\otimes_{ \Bbb Q}\Bbb R$ is surprisingly nothing else but $\Bbb R[X]$; but keep in mind that this is really a different object.

You also ask about finite fields. Here, $\Bbb F_{q_1}$ is a vector space over $\Bbb F_{q_2}$ if and only if $q_2\mid q_1$. (And the reasons are quite obvious, just check the axioms).

Finally, even a finite field allows an infinite vector space, for example the vector space $\Bbb F[X]$ over a field $\Bbb F$ always has countably infinite dimension.
More generally, if $\Bbb F$ is any field and $S$ is a set of a cardinality that is both infinite and larger than $\Bbb F$, then the set of maps $S\to \Bbb F$ naturally forms an $\Bbb F$-vector space of dimension $|S|$.
