I am curious about the structure we require on arbitrary fields if we want to generalise the notion of norms on vector spaces over fields other than either $\mathbb{R}$ or $\mathbb{C}$. Specifcally, given a vector space $V$ over $\mathbb{F}$, what structure must be imposed on $\mathbb{F}$, such that we have a map $\|\cdot\|: V \to \mathbb{F}$ s.t. for any $a \in \mathbb{F}$ and any $\bar u, \bar v\in V$,
$$\text{(i) }\|\bar u+ \bar v\| \leq \|\bar u\|+\|\bar v\|$$
$$\text{(ii) }\|\bar v\| = e \to \bar v = \boldsymbol{0}, \text{where } e \text{ is the additive identity of } \mathbb{F}$$
$$\|a\bar u\| = |a|\|\bar u\|$$
From this question here, it seems that positive definiteness requires that we have an ordered field. The accepted answer implies this is enough structure to generalise the inner product.
Where I am getting somewhat confused is how the generalisation works for norms. Specifically, is it really enough to have just have an ordered field in order to satisfy the norm property of $\|a\bar x\| = |a|\|\bar x\|$?
I know that having an ordered field allows us to generalise (i) and (ii), but how do we define what the modulus of $a$ is? I think the linked question is relevant since if we generalise the inner product over an arbitrary field, and we get a norm out of those inner products, we have made some progress.
How does ordering the field allow us to define the modulus of scalars in a vector space over an ordered field? Do we need to specify additional structure?
Best Answer
In order to define a norm in an $F$-vector space for a field $F$, you need to have a modulus defined on $F$. Otherwise, $\lVert \alpha v\rVert=\lvert \alpha\rvert\lVert v\rVert$ does not make much sense.
If you want the familiar identities to hold, you should also want the codomain of $\lvert\cdot\rvert$ and $\lVert\cdot\rVert$ to be the same and be equipped with multiplication, addition and an order, since you want $\lvert\cdot\rvert$ to be multiplicative and subadditive, you want the modulus of $0$ to be $0$, etc.
In general, you can consider $\lvert\cdot\rvert\colon F\to K$ for an ordered field $K$, making $F$ into a "$K$-normed field", and then you can consider "$K$-normed vector spaces" defined in the obvious way. If you have a linear order on $F$, then you have a canonical choice of $\lvert\cdot\rvert$, but I believe that is the extent of it.
This can be further generalised by requiring $F,K$ to be only rings, possibly non-commutative, etc. The main issue with using $F$ and $K$ distinct from $\mathbf R,\mathbf C$ is that if $K$ is not the field of reals (or a subfield), then the resulting topology will be (I think always, but definitely sometimes) non-metrisable, and even if $K$ is the reals, then $F$ might be incomplete (e.g. if it is the rational with the standard modulus) or non-Archimedean (e.g. if it is the p-adics with the p-adic modulus), which immediately transfers to properties of finite-dimensional vector spaces you might consider pathological.
For example, I believe there should be no funny business happening if you consider real-normed vector spaces over the quaternions (beyond the non-commutativity itself, of course), or real-normed vector spaces over the p-adics (beyond the usual non-Archimedeanness). And as pointed in the other answer, you can also consider the trivial discrete norm on any field and all its vector spaces, which will give you the trivial topology on all of them.
There is also a related notion of a valued field and a valued vector space, in which you have maps into an ordered abelian group (plus infinity or 0, depending on how you order your value groups), which are, to my knowledge, much more common than normed fields with norms taking values outside of the reals.