# Examples of vector spaces defined on fields other than field of numbers: $\Bbb Q,\Bbb R,\Bbb C$ etc.?

I'm using the textbook "Linear Algebra Done Right" by Sheldon Axler. The author mentions that vector spaces can be generally defined over any field. But the only examples that can I spot in the textbook are of vector fields defined over $\Bbb R$ or $\Bbb C$.

Could someone please give me some common examples of vector spaces which are not defined on field of numbers: $\Bbb Q,\Bbb R,\Bbb C$ etc.? I tried searching on the net, but couldn't find anything relevant.

• 1) $\Bbb N$ and $\Bbb Z$ are not fields 2) You can look at $\Bbb F_p$-vector spaces, $\Bbb Q_p$-vector spaces 3) If $R$ is any ring, you can look at left $R$-modules. Mar 4, 2018 at 15:17

The set of polynomials over a field, $$\Bbb F[x]$$, has many properties of a field, but doesn't have multiplicative inverses. If we define a set $$\Bbb W$$ which has elements of $$\Bbb F[x]\,mod\,h(x)$$ where $$h(x)$$ is an irreducible non constant polynomial in $$\Bbb F[x]$$ (can't be split into non constant factors), then we can show this is a field. For example, $$(x+1)(x^2+1)+3\ mod\ (x^2+1) = x+4$$

All the addition axioms are obvious and simple to check with the addition identity being the $$0$$ polynomial and the inverse of $$f(x)$$ given by $$-f(x)$$.

We use the division algorithm for polynomials which states that $$\forall f(x),\ g(x) \in \Bbb F[x], \exists\ a(x),\ b(x) \in \Bbb F[x]$$ such that $$f(x)=a(x)g(x)+b(x)$$ and $$deg(b(x)) < deg(g(x))$$ where $$g(x) \neq 0$$. Using this with $$g(x) = h(x)$$ we see that multiplication is closed as in $$\Bbb W$$, $$f(x)$$ would be $$b(x)$$. Multiplication is clearly commutative, associative and has identity $$1$$.

Using the division algorithm it is possible to show that $$\forall f(x),\ g(x) \in \Bbb F[x] \backslash \{0\},\ \exists\ a(x),\ b(x) \in \Bbb F[x]$$ such that $$a(x)f(x) + b(x)g(x) = gcd(f(x), g(x))$$. We can use this to show the existence of a multiplicative inverse.

Take $$f(x) \in \Bbb W$$ and assume $$f(x)=c$$ is a constant function. Then it's inverse is given simply by $$\frac 1c$$. For the non trivial case assume $$f(x) \neq constant$$. As $$f(x) \in \Bbb W$$ we know that h(x) doesn't divide $$f(x)$$, and as $$h(x)$$ is irreducible it can't divide $$f(x)$$. This means that $$gcd(f(x), h(x)) = 1$$, and we use the result above to get $$a(x),\ b(x)$$ such that $$a(x)f(x)+b(x)h(x)=1\\a(x)f(x)=1-b(x)h(x)\\a(x)f(x)=1$$ as $$b(x)h(x)=0$$ in $$\Bbb W$$. This means that $$a(x)$$ is a multiplicative inverse of $$f(x)$$.

The distributive laws also hold and are simple to check, and so $$\Bbb W$$ is a field. We could find some trivial vector spaces such as $$\Bbb W^n$$ or if you wanted something more interesting then maybe the set $$\Bbb W[y]$$ which is the set of polynomials with coefficients in $$\Bbb W$$, or the set of matrices with elements in $$\Bbb W$$. When checking the vector space axioms you don't use any specific properties of the field, you only need the axioms of a vector space, and so any vector space you can think of will work under any field.

To define a vector space over a field other than the rationals, the reals or the complex numbers you have to start with such a field.

For example, the set $\{0,1\}$ with arithmetic modulo $2$ is a field. Then the set of $n$-tuples ("vectors") with coefficients in that field is a vector space over that field.

For example, the set of eight vectors $(a,b,c)$ where each of the entries is either $0$ or $1$ is a three dimensional vector space over that two element field. (Just remember that you do arithmetic modulo $2$, so $(1,1,0) + (1,0,1) = (0,1,1)$.)

There are many other fields, some finite, some not. With any of them you can form vector spaces of any finite dimension in just this way. See https://en.wikipedia.org/wiki/Field_(mathematics) . (The integers and the natural numbers aren't fields.)

• Could you give an example of n-tuple with coefficients in $\{0,1\}$ ? I'm not sure what you meant by n-tuple there. Do you mean something like $(1,0,1,1,..)$ ?
– user400242
Mar 4, 2018 at 15:29
• Awesome. Thank you very much for that example. So I can see that the vector space of $(a,b,c)$ is closed. But one more question: How would scalar multiplication behave for it? Would $0.(a,b,c)=(0,0,0)$ and $1.(a,b,c)=(a,b,c)$ ?
– user400242
Mar 4, 2018 at 15:37
• Yes, exactly. With this two element field scalar multiplication is pretty boring. Less so with other fields. Play with the three element field $\{0,1,2\}$, arithmetic modulo $3$. The two dimensional vector space has $9$ elements, the three dimensional vector space has $27$. Mar 4, 2018 at 15:40