Can the property $z^*z\in [0,\infty)$ be generalized to other fields?

Question

Let $K$ be a field equipped with the following structure:

1. Let $*$ be an involution $*:K\to K,z\mapsto z^*,$ where $*$ preserves field structure and $(z^*)^* = z$ for all $z\in K$.
2. Assume that $z^*z\in[0,\infty)$ for all $z\in K$.

Two well-known structures satisfy these properties: $\mathbb R$ (with the identity map) and $\mathbb C$ (with complex conjugation). Any subfields of these examples should also work. Are there any other examples?

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Background: This property is crucial for defining the inner product over $\mathbb C$-vector spaces, and I am interested in generalizing inner products to vector spaces over other fields. But if no other "interesting" fields satisfy this property, then there isn't much point to this effort.

Answer 1

Any field that fulfils your conditions is isomorphic to a subfield of $\mathbb C$.

Proof:

First, note that obviously $z^*z\in K$, and your condition 2 implies that lso $z^*z\in\mathbb R$. Therefore $R:=K\cap\mathbb R$ is nonempty. Indeed, it is not hard to show that it is a field, too. Which in particular means that is is a subfield of $\mathbb R$. Since in particular $R$ is ordered, this also implies that $K$ is a field of characteristic $0$. This is important because it allows us to divide by $2$; I'll make use of that in the following. Also, since $R\subseteq\mathbb R$, it inherits the order of $\mathbb R$; this means we can compare elements in $R$.

Now let's take the fact that $z\mapsto z^*$ is an involution. If $z=z^*$ I'll call $z$ symmetric, if $z=-z^*$, I'll call it antisymmetric. This allows us to define a symmetric and an antisymmetric part as \begin{eqnarray} \operatorname{symm}(z)&=\frac12(z+z^*)\\ \operatorname{asym}(z)&=\frac12(z-z^*) \end{eqnarray} so that $z=\operatorname{symm}(z)+\operatorname{asym}(z)$. Note that in the complex numbers, the symmetric part is the real part, and the antisymmetric part is $\mathrm i$ times the imaginary part.

Now if $z=s+a$ where $z$ is symmetric and $a$ is antisymmetric, we have $$z^*z = (s+a)^*(s+a) = s^*s + s^*a + a^*s + a^*a = s^2 - a^2$$

Now let's define the set of symmetric elements, $S=\{z\in K: z=z^*\}$. It is not hard to check that it is a subfield of $K$, let's call it the symmetric subfield.

But that means that $S$ is a field in which the square of any element is a positive real number. Since the real numbers contain all the square roots of all the positive real numbers, this implies that $S$ is isomorphic to a subfield of $\mathbb R$.

Without loss of generality, in the following we can assume $S\subseteq R$. That is, we can decompose any $z\in K$ into a real number and an antisymmetric element of $K$. But for antisymmteric elements of $K$ we have $$(a^2)^* = (a^*)^2 = (-a)^2 = a^2$$ therefore the square of an antisymmetric element is symmetric, and therefore a real number. Moreover we have seen above that $-a^2$ is positive, therefore $a^2$ is negative. Thus the antisymmetric elements of $K$ square to negative real numbers. But that is exactly what the purely imaginary numbers do. Therefore we can map the antisymmetric elements to the imaginary numbers.

But that means that, up to isomorphism, every element of $K$ is the sum of a real number and an imaginary number, that is, a complex number.

And this means that $K$ is isomorphic to a subfield of $\mathbb C$.

Note: I've explicitly used the requirement stated in the comments to the question that the interval from condition $2$ is indeed the real interval. If it were not required to be the real interval, then other solutions would exist.

Answer 2

Here is an example. Fix a prime $p$. Let $K = \mathbb F_{p^2}$ and $\sigma$ the order $2$ automorphism of $K$ over $\mathbb F_p$, so the involution is $z^* = \sigma(z)$. Identify $\mathbb F_p$ with $\{0,1,...,p-1\}$ in $\mathbb R$ so that we can talk about its intersection with $\mathbb R$.

Then this clearly satisfies $(1)$ and $(2)$ - for $(2)$ just notice that $zz^*$ is the norm from $\mathbb F_{p^2}$ to $\mathbb F_p$ and so it lands in $\mathbb F_p$ which we have selected to be in $[0,\infty)\subseteq \mathbb R$.

If you put restrictions on how $K$ and $\mathbb R$ can intersect, then it seems very likely that you will just end up with subfields of $\mathbb C$ as the other answer explains. Condition (2) strongly biases your fields towards this case. I would think about replacing it by requiring the fixed field of the involution to be an ordered field and that $zz^*\geq 0$ according to that ordering. Even that I don't think is optimal e.g. it won't work in the $p$-adic setting, but that is a pretty interesting place to study vector spaces and analysis and so on.

Edit: to clarify a misunderstanding on the comments to another answer, I do not claim that the identification of $\mathbb F_p$ with $\{0,1,...,p-1\}$ preserves or has anything to do with the ring properties, as it certainly cannot (e.g. for characteristic reasons). This is just purely at the level of sets so that, as explained above, one can even talk about an intersection with $\mathbb R$ for the purposes of treating (2) in the question.