Field importance in a Vector Space I know the definition of a vector space.
But heuristically, what is the meaning of defining a vector space over a field ? What is the intuition behind the field in a vector space?
 A: The short answer is that if $2\vec x$ is in our vector space, then we also want $\tfrac{1}{2}\vec x$ to be in the vector space. Vector spaces scale nicely under multiplication, and things behave weirdly if this scaling isn't true. Let me give one example of weird behavior.
One of the nice aspects of a vector space is that the dimension of a subspace is the same as the number of linearly independent vectors providing a basis for that subspace. Proving this result is usually done with a transferrence-type theorem that says something like the following:

Suppose $V$ is a vector space of dimension $n$, with $v_1, \ldots, v_n$ as a basis. Suppose $w_1, \ldots, w_m$ are $m$ linearly independent vectors in $V$. Then (perhaps after reordering), $w_1, \ldots, w_m, v_{m+1}, \dots, v_n$ is also a basis for $V$.

This allows one to compare the sizes of two bases by progressively replacing one with the other.
The key element of the proof is evident even when there is only one $w_1$ vector of interest. Since $w_1 \in V$, there are coefficients $c_i$ such that
$$ w_1 = \sum_{i \leq n} c_i v_i.$$
As $w_1 \neq 0$, at least one of the $c_i$ are not zero. WLOG (otherwise, reorder), $c_1 \neq 0$. Then we can solve for $v_1$ and see that
$$ v_1 = c_1^{-1}\bigg(\sum_{2 \leq i \leq n} c_i v_i - w_1 \bigg).$$
Thus any vector written in terms of $v_1, \ldots, v_n$ can be written in terms of $w_1, v_2, \ldots, v_n$.
In this proof, note that we used $c_1^{-1}$. We used that generic inverses exist! This is essential.

We can give a direct example as well. Consider $\mathbb{Z}$ as a $\mathbb{Z}$-module. (If you are not familiar with modules, then you can think of this as trying to define a vector space over a ring instead of a field).
Then $\mathbb{Z}$ has a simple "basis": $1$. Every integer in $\mathbb{Z}$ can be written as an integer multiple of $1$. In analogy to vector spaces, can we say that the "dimension" of $\mathbb{Z}$ is $1$?
Note that $\mathbb{Z}$ has a different "basis", $\mathbb{Z} = \langle 2, 3\rangle$ --- or rather, every integer can be written as an integer multiple of $2$ added to an integer multiple of $3$. But each individual "basis vector" ($2$ or $3$) are independent, in the sense that $2$ is not a multiple of $3$ (and vice versa). So is the "dimension" of $\mathbb{Z}$ actually $2$?
No --- the concept of dimension is just messier when there isn't a field lying underneath.
