# Whats the relation between the set of vector in a vector space to the field that this vector space is over?

We have a set of "vectors" (elements) in the set $V$ such that $(V,+,\circ)$ is said to be a vector space over some field $\mathbb{F}$. Let $F$ be the set of elements that consist the field $\mathbb{F}$.

I'm trying to figure out what rules $V$ must follow on its own and what is the relation between the set of elements in $V$ to the set of elements in $F$. For example, if we look at the set $V$ on its own, it must be an abelian group. What I'm asking is what are the elements in $V$ have to do with the field $\mathbb{F}$, specifically:

1. $V$ is said to be "over the field $\mathbb{F}$". It doesn't mean that the elements of $V$ are taken from the field $\mathbb{F}$, right?
2. What is the relation between $V$ and $F$. Should it be that $F\subset V$? For example, let $V=\mathbb{R}$ and let $\mathbb{F}=\mathbb{C}$ (the complex numbers). Then $V$ on its own is an abelian group as required; but if you take an element in $V$ and multiply it be an element in $\mathbb{F}$ the result might not be in $V$.

For example take $1\in V$ and $i\in\mathbb{F}$, then $1\cdot i \notin V$. Thus we get that $\mathbb{R}$ over $\mathbb{C}$ is not a vector space. However, $\mathbb{C}$ over $\mathbb{R}$ is indeed a vector space.

My question is: For a set of vectors $V$ be a vector space over the field $\mathbb{F}$ must it hold that the underlying elements that consist the vectors be from a field that is equal or containing the field $\mathbb{F}$?

For example if you take the set of vectors $V = \mathbb{R}^2$ then for $V$ to be a vector space over some field $\mathbb{F}$ it must hold that the elements that the vectors in $V$ are made of (i.e. $\mathbb{R}$) must be from a field $\mathbb{F}\subseteq\mathbb{R}$?

• There is no difference between $F$ and $\mathbb{F}$. Also $\mathbb{F}$ is the set of elements of the field $\mathbb{F}$. – Dietrich Burde Jun 30 '16 at 15:18
• For $V$ to be a vector space over a field $\Bbb{F}$, as an abelian group it must either be torsion free or a $p$-group for some prime $p$. Correspondingly the characteristic of $\Bbb{F}$ must be either $0$ or $p$. – Inactive - avoiding CoC Jun 30 '16 at 15:20

There is no need for $\mathbb{F}$ to be a subset of $V$, or for them to have anything to do with one another, other than there being an operation $\cdot: \mathbb{F}\times V \rightarrow V$ which we call "scalar multiplication" which as to satisfy some properties.

When you first see vector spaces, they tend to be "concrete" and the objects in them are frequently what we think of as vectors in, say, calculus. But then you learn that there are much more complicated vector spaces e.g. the dual vector space of $V$, $C^{n}[a,b]$ then $n$ times continuously differentiable functions on $[a,b]$.

In these examples, the vectors are actually functions, and although you can find things that look like the underlying field living in them (for instance, in the second example, you can find a copy of $\mathbb{R}$ by looking at constant functions), in general, $\mathbb{F}$ will not be a subset of $V$, nor the elements of $V$ built out of $\mathbb{F}$ in any nice way.

• I'm sorry, I didn't understand your counter example, can you further elaborate please? – Bush Jun 30 '16 at 15:24
• I didn't give a counter-example, as you didn't make a claim. The $n$ times continuously differentiable functions on $[a,b]$ form a vector space (over $\mathbb{R}$) with addition and scalar multiplication defined pointwise: $(f+g)(x) := f(x) + g(x)$ and $(\lambda f)(x) := \lambda f(x)$. In this example the vectors are functions with domain $[a,b]$ and range $\mathbb{R}$, but real numbers are very different objects to functions. – James Jun 30 '16 at 15:29
• So in this example, the underlying field of $V$ is $\mathbb{R}$ (the range of the function). If you change this with $\mathbb{C}$, this will no longer be a vector space over $\mathbb{R}$. Right? – Bush Jun 30 '16 at 15:43
• Change what with $\mathbb{C}$, the codomain of the functions? – James Jun 30 '16 at 15:45
• In the example you gave, the set of the $n$ times continuously differentiable functions on $[a,b]$ are defined with domain and codomain $\mathbb{R}$. This immediately means that when you refer to the vector space that this set (of functions) forms, you mean that this vector space over $\mathbb{R}$? Or you can talk about this set of function and refer to the vector space that it forms as a vector space over some arbitrary field (e.g. $\mathbb{C}$)? – Bush Jun 30 '16 at 16:55

This is an excellent question and it's something a lot of students don't think about properly. The problem is that to truly understand the relationship between a vector space V and it's field of scalars F, you have to understand the notion of a group action.

Def: Let S be a nonempty set and let G be a group. Then we define a left group action $\psi$:G x S $\rightarrow$ S to be a mapping with the following properties:

(i) For every $x\in S$, e*x = x.

(ii)For every $g,h \in G$ and $x\in S$, (gh)x = g(h*x).

A right group action is defined the same way. It's not hard to show that every left group action is a bijection on S where $\psi^{-1}$ is defined by x $\rightarrow$ $g^{-1}x$.

It's easy now to understand the relationship between the field of scalars F and it's corresponding vector space V. There is a left group action of the Abelian group of F $\psi$ : F x V $\rightarrow$ V into V such that for every scalar $\lambda$ and $v\in V$, $\lambda$*v = $\lambda v$. This mapping requires only that the image lie in V. It doesn't require any containment relations between F and V.

As Servaes very correctly pointed out in his comment,though, it's not as simple as this because certain algebraic conditions have to be met for a field F to be suitable as a scalar field for a given vector space V. Although $\mathbb R$ and $\mathbb C$ have the same characteristic, $\mathbb R$ is not a vector space over $\mathbb C$ precisely because the group action of the scalar multiplication fails to be closed. The converse IS a vector space because $\mathbb R$ is a subset of $\mathbb C$ and therefore,real multiples of complex numbers give complex numbers and the operation in this case IS closed.