What fields can be used for an inner product space? The title is the question: What fields can be used for an inner product space?
This question has been discussed in Math Stack Exchange with no definitive resolution.  A similar question appeared here, and an answer was accepted, but someone pointed out a serious problem with the answer.
I am using the standard definition of inner product, which includes $\langle \mathbf{x}, \mathbf{x} \rangle > 0$ for all non zero vectors $\mathbf{x}$.
It seems to me that any field of prime characteristic does not make sense, because it does not have an order relation that respects addition.  It also seems to me that the field $\mathbb{F}$ can be any ordered field or any subfield of $\mathbb{C}$ that is stable under complex conjugation (for any non-algebraists like me, the word "stable" seems to be standard here. Anyone else would use the word "closed").  I do not know if any other fields are possible.  Of course, an ordered field may or may not be a subfield of $\mathbb{R}$.
It seems to be rare for people, even mathematicians, to use any field other than $\mathbb{R}$ or $\mathbb{C}$ for an inner product space.
Can anyone clear this up?
EDIT: Mark Grant alerted me to a Wikipedia page that addresses this question (http://en.wikipedia.org/wiki/Inner_product_space#Definition).  Let me quote the relevant part :
"There are various technical reasons why it is necessary to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition. Briefly, the basefield has to contain an ordered subfield[citation needed] (in order for non-negativity to make sense) and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. The basefield has to have additional structure, such as a distinguished automorphism. More generally any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$ will suffice for this purpose, e.g., the algebraic numbers…."
The Wikipedia article fails to explain why the basefield has to have additional structure.  They do not define a "distinguished automorphism" or provide a link to a definition.  I am not an algebraist.  I Googled the term and I could not find a definition of "distinguished automorphism".  I did find links to papers and books that probably do contain a definition.  The article states it is "necessary" to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition but then contradicts itself by at least suggesting that the basefield can be  any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$.
 A: Of course if you insist on condition $\langle \mathbf{x},\mathbf{x}\rangle > 0$, and not merely $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$, then you must have an order.  
Let $F$ be a formally real field.  Then
$$
\langle \mathbf{x}, \mathbf{y}\rangle = \sum_{j=1}^n x_j y_j
$$
can be a reasonable inner product on $F^n$.  According to an ordering for $F$ (indeed, any ordering, since there may be more than one) we have $\langle \mathbf{x}, \mathbf{x}\rangle > 0$ if $\mathbf x \ne \mathbf 0$.  
Another part that you quote is what would be required for metric completeness.    Do you want that? If $F$ is a proper subfield of $\mathbb R$, then even the one-dimensional space is not complete.  
Something weaker than completeness will be enough to carry out the Gram-Schmidt process.  It requires only that square-roots of $\langle \mathbf{x}, \mathbf{x}\rangle$ exist.
A: The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".
Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.
In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.
In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.
For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.
In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have an orthonormal basis.

In answer to your question, the appropriate setting is $*$-fields, and not fields. And all $*$-fields can admit inner product spaces.
