Why do we call a vector space in terms of vector space over some field? I am getting a bit confused with the terminology here. I understand that a field means some set of scalars like real numbers but why do we need a field for a vector space? Are not the numerical values used to define a vector are inherent properties of the vector space? Why do we term it like "vector space over real field"? If not numbers, what are the other fields possible for a vector space because you obviously numbers to define the value of magnitude and direction of the vectors?
 A: First you cannot define magnitude and direction for all vector spaces: these notions correspond to so-called *normed and/or inner product vector spaces*.
Next one has very naturally to consider vector spaces over $\mathbf C$, even in problems concerning real spaces. I'll mention Fourier series, or second order linear differential equations.
In arithmetic, one has to consider vector spaces over $\mathbf Q$ or algebraic number fields or $p$-adic numbers.
I once heard a well-known mathematician say cryptography, a domain closely related to arithmetic, is but the theory of finite-dimensional vector spaces over finite fields.
A: 
Are not the numerical values used to define a vector are inherent properties of the vector space?

Yes, they are. If you use other numerical values (like complex numbers, or $\mathbb F_2$), you get another vector space (for example $\mathbb F_2^3$ instead of $\mathbb R^3$). However you can also get a different vector spaces using the same field (for example, $\mathbb R^2$ and $\mathbb R^3$). Now it turns out that vector spaces over the same field are in some sense compatible with each other, in a way that vector spaces over different fields generally are not. For example, it's no problem to define a linear function $f:\mathbb R^2\to\mathbb R^3$. However there are no linear functions $f:\mathbb F_2^2\to\mathbb R^2$. Therefore it makes sense to always explicitly say on which field the vector space is based.
A: A vector space is defined as a quadruple $(V,K,+,\cdot)$ where $V$ is a set whose elements are called ''vectors'' , $K$ is a field, $+\colon V\times V \rightarrow V$ and $\cdot\colon K\times V \rightarrow V$ are operations that satisfy a suitable set of axioms (see here).
For the same set $V$ we can define different vector spaces changing the field $F$ and the difference can be dramatic.
The simpler case is for $V=\mathbb{R}$. If we take $K=\mathbb{R}$ with the usual addition and multiplication, we have a vector space of dimension $1$, but if we take $K=\mathbb{Q}$ we have a vector space that has a uncountable basis so its dimension is infinite (Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?).
Another, less dramatic, example, is if $V=\mathbb{C}^n$: if we choose $K= \mathbb{C}$ than we have a vector space (with the usual operations) over $\mathbb{C}$ of dimension $n$, if we choose $K= \mathbb{R}$ we can obtain a space of dimension $2n$.
Again, if $K=\mathbb{Q}$ the change is more relevant and we have a space of infinite (not countable) dimension.
A: Not all vector fields have "magnitutes". In order to have a magnitude of a vector, you need to have a norm defined on the vector space, and typically, only vector spaces over $\mathbb R$ or $\mathbb C$ have norms defined on them.
We have to say "vector space over real field" because we need to be careful about what we are doing at all times. For example, $[1,0,0]$ is an element of both $\mathbb R^3$ which is a vector space over $\mathbb R$, and an element of $\mathbb C^3$ which is a vector space over $\mathbb C$.
Furthermore, the fact that $\mathbb C$ is itself a vector space over $\mathbb R$ means that any vector space $V$ over $\mathbb C$ can also be seen as a vector space over $\mathbb R$, but of a different dimension.
You can also have vector spaces over other fields, like $\mathbb Z_p$ where $p$ is prime, and not define any norm there.
