TITLE: What structure do you get if you adjoint a root of $z \bar{z} = -1$ to the complex numbers?
QUESTION [6 upvotes]: A pop-up is informing me that my question is likely to be closed.
Still, recall that the complex numbers $\mathbb{C}$ was conceived by trying to adjoint a root of the equation $x^2 = - 1$ to the field of real numbers $\mathbb{R}$, or so we are told. There are two such roots, now known as $i$ and $-i$, and the conjugation involution $z \mapsto \bar{z}$ is the field automorphism of $\mathbb{C}$ which fixes $\mathbb{R}$.
Suppose we entertain the fantasy to adjointing a root of the analogous equation $z \bar{z} = |z|^2 = -1$ to the complex numbers $\mathbb{C}$. What "structure" will we get? Now, given such a root, call it $f$ for the fantasy unit, then $z= e^{i\theta}\cdot f $ would also be a root of $z \bar{z} = 1$ for all $e^{i\theta}$ in the circle group $U(1)$.

REPLY [4 votes]: The first thing to realize is that the question must be properly interpreted, which probably can be done in several ways.
It seems reasonable to interpret it as follows:

Find a ring $R$ with an injective homomorphism $\mathbb{C} \rightarrow R$, such that:

there is an involution $\sigma: R \rightarrow R$, i.e. an endomorphism of order $2$, such that the restriction of $\sigma$ to $\mathbb{C}$ is the complex conjugation;
there is an element $f \in R$, such that $f \sigma(f) = -1$;
$R$ is generated by $\mathbb{C}$ and $f$.


The last condition can possibly be replace by: $R$ is generated by $\mathbb{C}$, $f$ and $\sigma(f)$.
But you probably also want to put other conditions on $R$.
For example, if you require $R$ to be commutative, then $R$ can be the ring $\mathbb{C} \oplus \mathbb{C}$, with $\mathbb{C}$ embedded diagonally, $\sigma(z, w)=(\overline{w}, \overline{z})$, and the element $f=(1, -1)$.
If you don't require $R$ to be commutative, then it can also be the biquaternion $R = \mathbb{C} \oplus \mathbb{C} i \oplus \mathbb{C} j \oplus \mathbb{C} k$.
For the moment I don't have an example where $R$ is still a division algebra and $f$ is algebraic.