Can we view a vector space as a field extention?

Question

We can view field extentions as vector spaces over a field (an idea that in my experience has not really been explained but I get it more or less).

But can any vector space over a field also be seen as an extention of that field? I think no, since vector spaces are not fields. And frankly I don't see how any $r$ dimension vector space $V$ over a field $K$ of the form $\{\alpha_1v_1+...+\alpha_rv_r\ :\alpha_i\in K, v_i\in B\}$ and $B$ is some $K$-basis of $V$ can be a field...

Answer 1

A vector space over a field can be regarded as an extension of that field as long as you can equip the vector space with an appropriate concept of product that satisfies the field axioms.

For example, equipping $\mathbb{R}^2$ with the product $(a,b) \times (c,d) = (ac-bd, ad+bc)$ creates a field which is an extension of $\mathbb{R}$. Since $(0,1) \times (0,1) = (-1,0)$ we realise that we can identify $(0,1)$ with $i=\sqrt{-1}$ and we have created the field extension $\mathbb{R}(i)$ which is the field of complex numbers.

Or we can equip $\mathbb{Q}^2$ with the product $(a,b) \times (c,d) = (ac+2bd, ad+bc)$. Now $(0,1) \times (0,1) = (2,0)$ so we can identify $(0,1)$ with $\sqrt{2}$ and we have created the field extension $\mathbb{Q}(\sqrt{2})$.

Defining a concept of product that satisfies the field axioms is not always possible. For $n>2$ there is no product rule that makes $\mathbb{R}^n$ a field. It is possible to create interesting mathematical structures in this way, but they are not fields - for example, the quaternions can be defined by equipping $\mathbb{R}^4$ with a product, but this product is not commutative.

Answer 2

To your first question, definitely yes. Number fields are exactly this kind of object: if you adjoin $\alpha \in \mathbb{C}$ to $\mathbb{Q}$, written $\mathbb{Q}(\alpha)$, then $\mathbb{Q}(\alpha)$ is a vector space over $\mathbb{Q}$ with scalar multiplication defined as in $\mathbb{Q}(\alpha)$. The basis for some extensions is easy to find: if $d$ is a cube-free integer different from $0$ and $1$, then $\mathbb{Q}(\sqrt[3]{d})$ as a vector space over $\mathbb{Q}$ has basis $1, \sqrt[3]{d}, \sqrt[3]{d}^{2}$.

As for your second question, I'd say no for a different reason. Consider the vector space $\mathbb{R}^{n}$ over $\mathbb{R}$ for $n \geq 3$. Then $\mathbb{R}^{n} = \mathbb{R} \times \mathbb{R} \times \cdots \times \mathbb{R}$ is not a finite extension of $\mathbb{R}$, it's the Cartesian product of $n$ copies of $\mathbb{R}$.