Generalising norms over arbitrary fields 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?

The trivial real-valued absolute value $$|a|_{tr}=\cases{1 \ if \ a\ne 0\\ 0 \ otherwise}$$ works over any field $F$. We can then consider the trivial real-valued norm $$\| v\| = \sup_j |v_j|_{tr}, \qquad v\in F^n$$
When $char(F)=0$ you can think to $\|.\|$ as being $F$-valued, taking its values in the ordered subfield $\Bbb{Q}$.
