# Are there any examples of vector spaces over non-numerical fields? If not, why not?

By non-numerical vector spaces I mean vector spaces that do not have as their scalars some sort of easily discernible numerical fields (e.g. complex numbers, functions are usually maps from one numerical space to another, etc.).

Are there any examples of non-numerical vector spaces? If not, why not?

I know that this question asked for something similar, but its accepted answer gave "numerical vector spaces", and more importantly the motivation behind the question was pedagogical.

I on the other hand, am asking in order to determine if there would be any drawbacks to building an algebra that assumes its scalars are all reals, and that complex numbers, quarternions, etc. are built into its higher dimension forms (e.g. geometric algebra/Clifford algebra), as such an algebra might perhaps not be able to capture a useful vector space over a non-numerical field.

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Any vector space is a vector space over some field. Are you really asking if there are fields that don't seem to have any connection with any sort of numbers? – Harald Hanche-Olsen Jan 31 '14 at 17:51
@twirlobite $,$ is a comma, for separation. – AlexR Jan 31 '14 at 18:00
If I'm working over a field, I am inclined to think of its elements as numbers, no matter what the field is. – Hurkyl Jan 31 '14 at 18:09
@twirlobite Every field contains a prime subfield that is isomorphic to either $\mathbb{Q}$ or $\mathbb{Z}/p\mathbb{Z}$. So if you consider elements of these prime subfields to be "numbers" then all fields contain numbers. – SpamIAm Jan 31 '14 at 18:31
let's say: a numerical vectorspace is a vectorspace equipped with a total order. Then still I learned something here. This is my last comment on the subject. – drhab Jan 31 '14 at 18:33

The scalars of a vector space just have to be a field.

The examples $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, etc. which I guess is what you mean by "numeric" are not the only examples of fields.

For instance, there is also plenty of fields of functions: rational functions, meromorphic functions, in general, any quotient field of any however defined ring.

That is, you can just as well form a vector space with the scalars being rational functions instead of "just numbers".

Example:

Consider the space of all pairs $(f,g)$ of rational functions $f,g$ equipped with componentwise addition, i.e. $(f_1,g_1)+(f_2,g_2)=(f_1+f_2,g_1+g_2)$ and the scalar multiplication $f\cdot(g,h)=(fg,fh)$.

This is a two dimensional vector space over the space of rational functions.

But:

No matter how "abstractly" you define your field and how you denote its elements, you will always find "numbers" inside.

Consider the following. Let $K$ be a field of characteristic $0$. Let $e$ be the unit in $K$. Then refer to a sum of $n$ times $e$ as $n\cdot e$ ($n$ is a natural number $\ge 1$).

Since $K$ is a field, given any $n\cdot e$ there exists a multiplicative inverse, which we shall call $\frac{1}{n}\cdot e$ (here we use that $K$ is of characteristic zero).

Likewise, there exists an additive inverse of $n\cdot e$, which we shall call $-n\cdot e$.

Now, consider $n\cdot e\in K$ for some $n\in\mathbb{Z}$ and $\frac{1}{m}\cdot e$ for some $m\in\mathbb{N}$. We denote their product in $K$ by $\frac{n}{m}\cdot e$ (we tacitly named the additive neutral element (that is the "zero") $0\cdot e$).

What have we done here? We constructed an injective field homomorphism

$$\mathbb{Q}\hookrightarrow K$$

sending $\frac{n}{m}$ to $\frac{n}{m}\cdot e$ (you can check readily that it is well-defined and really an injective field homomorphism).

That means: You will always find a copy of $\mathbb{Q}$ in your field. You cannot escape "numbers".

Note: If $\mathrm{char} K=p$ instead, then you find a subfield isomorphic to $\mathbb{Z}/(p)$ by the same procedure.

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Your example uses functions over a numeric field. I edited my question a bit to explain why I didn't want such an example. – user89 Jan 31 '14 at 18:02
@twirlobite: Well, somewhere you're always gonna find "numbers"; e.g. for any field $K$ of characteristic $0$ there is an injective field homomorphism $\mathbb{Q}\hookrightarrow K$. Algebra is about operations between symbols, which satisfy certain relations. The algebra doesn't care whether you call the symbols numbers or not. – Your Ad Here Jan 31 '14 at 18:05
"Well, somewhere you're always gonna find "numbers"; e.g. for any field K of characteristic 0 there is an injective field homomorphism Q↪K" Ah! That is what I was hoping for! I was wondering if we could have fields that are not even enumerated by natural numbers, but if I understand what you are saying correctly, that is just not possible? – user89 Jan 31 '14 at 18:12
@twirlobite: No. I will explain upstairs. – Your Ad Here Jan 31 '14 at 18:13

When we want to construct a vector space "containing no numbers", we can proceed by a very simple method:
1. Chose an Alphabet $\mathfrak A$, i.e. an ordered set of tokens of some sort. examples: $$\{A,B,C,\ldots,Z,AA,AB,\ldots,ZZ\}$$ $$\mathfrak A =\{\text{car, boat, ship, elephant}\}$$ 2. Chose a prime number $p$, f.ex. $p=2$
3. if $\phi$ is an enumeration of our alphabet, we can define a Field by $$\mathcal A :=\{\phi(i) \in \mathfrak A | 0 \leq i<p\}$$ $$+: \mathcal A\times\mathcal A\to \mathcal A, \qquad a+b \mapsto \phi(\phi^{-1}(a)+\phi^{-1}(b))$$ $$\cdot: \mathcal A\times\mathcal A\to\mathcal A, \qquad a\cdot b \mapsto \phi(\phi^{-1}(a) \cdot \phi^{-1}(b))$$ 4. chose a dimension $n\in\mathbb N$ and consider $\mathcal A^n$ as your vector space (f.ex. $n=2$)

To elaborate on an example, we chose $\def\t#1{\text{#1}}$ \begin{align*} \mathfrak A & := \{\text{car, boat, ship, elephant}\} \\ p & = 3\\ \Rightarrow \mathcal A & = \{\text{car,boat,ship}\} \\ \text{multiplication:} & \begin{pmatrix}\cdot&\t{car}&\t{boat}&\t{ship}\\ \t{car}&\t{car}&\t{car}&\t{car}\\ \t{boat}&\t{car}&\t{boat}&\t{ship}\\ \t{ship}&\t{car}&\t{ship}&\t{boat}\end{pmatrix}\\ \t{addition:} & \begin{pmatrix} +&\t{car}&\t{boat}&\t{ship}\\ \t{car}&\t{car}&\t{boat}&\t{ship}\\ \t{boat}&\t{boat}&\t{ship}&\t{car}\\ \t{ship}&\t{ship}&\t{car}&\t{boat} \end{pmatrix} \\ n &=2 \\ 0\text{-vector} & = \begin{pmatrix}\t{car}\\\t{car}\\\t{car}\end{pmatrix}\\ \ldots \end{align*}

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Due to 3, don't we conveniently make the field equivalent to the natural numbers (except, to various natural numbers we assign various tokens, along with their usual interpretation)? – user89 Jan 31 '14 at 18:09
@twirlobite Almost, we take a finite field; but yes we do - still if you don't "tell" your students, only those who grasp the concept will see that I guess. – AlexR Jan 31 '14 at 18:13
Hmm, I was hoping for an example of a field that avoids that kind of connection with a numeric field via something as subtle as enumeration. I suppose what I am trying to get at is that I wonder if there are fields that cannot be enumerated? – user89 Jan 31 '14 at 18:15
@twirlobite If they are indexable, they are indexable by numbers, so you will always be able to find a representation which uses digits as tokens (take the reals as an index set, for example) – AlexR Jan 31 '14 at 18:17
Right! That's what I am somehow trying to get at. I made the following edit a while ago for instance: "I...am asking in order to determine if there would be any drawbacks to building an algebra that assumes its scalars are all reals, and that complex numbers, quarternions, etc. are built into its higher dimension forms (e.g. geometric algebra/Clifford algebra), as such an algebra might perhaps not be able to capture a useful vector space over a non-numerical field. – user89 Jan 31 '14 at 18:18

Example of a vector space over a non-numerical field: consider the set $k\left[ x \right]$ of polynomials with coefficients in the field $k$. Then $k\left[ x \right]$ is a vector spaces over $k$, with the usual operations of polynomials. This is true for every field. Then, if $k$ is a 'non-numerical field' then $k\left[ x \right]$ is the vector space that you are looking for. Now can you give an example of a 'non-numerical field'?

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