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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 $U(1)$.

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    $\begingroup$ It looks like you are identifying $\mathbb{C}$ with the group of diagonal matrices $$\left( \begin{array}{cc} z & 0 \\ 0 & z \\ \end{array} \right)$$ and then you are adjointing the Pauli matrix $$J:=\left( \begin{array}{cc} 0 & -i \\ i & 0\\ \end{array} \right).$$ $\endgroup$
    – Francesco Polizzi
    Commented Dec 15, 2016 at 14:08
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    $\begingroup$ In fact, we have $$J \bar{J}=\left( \begin{array}{cc} 0 & -i \\ i & 0 \\ \end{array} \right) \left( \begin{array}{cc} 0 & i \\ -i & 0 \\ \end{array} \right) = \left( \begin{array}{cc} -1 & 0 \\ \; \; 0 & -1 \\ \end{array} \right),$$ so $J$ is actually a root of $J \bar{J}=-1$. $\endgroup$
    – Francesco Polizzi
    Commented Dec 15, 2016 at 14:16
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    $\begingroup$ Thanks for pointing the terminology of a Pauli matrix. Reading the Wikipedia description, it seems that $SU(2)$ and the quaternions come into play. $\endgroup$
    – user94803
    Commented Dec 15, 2016 at 14:33
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    $\begingroup$ "... the complex numbers $\mathbf C$ were conceived by trying to adjoin a root of the equation $x^2=−1$ to the field of real numbers, or so we are told." Who told you that? The complex numbers were conceived because of the cubic formula for real solutions of cubic polynomials. When a cubic has all three roots in $\mathbf R$, the cubic formula has square roots of negative numbers. Nobody was inventing new number systems to solve problems nobody cared about (like being able to solve $x^2 = -1$), but rather to do something of interest: find a formula for real solutions of a real cubic. $\endgroup$
    – KConrad
    Commented Dec 15, 2016 at 15:41
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    $\begingroup$ Not exactly a duplicate of my question "Making extensions $L/K$ aware of the Galois group coming from $K/k$" but that one in turn was a followup of my math.SE question Adding a root of $z\bar z=-1$ to $\mathbb C$ with practically identical title :D Not that I have anything against, on the contrary, I am glad to see I was not alone in thinking about it $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 15, 2016 at 16:18

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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.

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    $\begingroup$ We can just take the quaternions with involution $i \mapsto -i, \ j \mapsto j, \ k \mapsto -k$. Then $j$ is such a solution and $\mathbb C$ is a subfield. $\endgroup$
    – Anton Fetisov
    Commented Dec 15, 2016 at 15:31
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    $\begingroup$ For $R$ commutative, there is a simple universal solution to this problem: It is the ring au laurent polynomial $\mathbb{C}[Z,Z^{-1}]$ with the involution $P \mapsto \overline{P}(-1/Z)$. So as I understand the question, this is what we get when one freely add such $z$. $\endgroup$
    – Simon Henry
    Commented Dec 15, 2016 at 16:11
  • $\begingroup$ @SimonHenry I agree completely, that was exactly my suggestion in mathoverflow.net/q/248241/41291 and I am still thinking on it. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 15, 2016 at 16:25
  • $\begingroup$ @SimonHenry Yes, I was aware of this example, which is why I added "$f$ algebraic" in the end (which doesn't quite make sense, though). $\endgroup$
    – WhatsUp
    Commented Dec 15, 2016 at 16:44
  • $\begingroup$ There is an interesting point here - although of course $Z$ in $\mathbb C[Z,Z^{-1}]$ is transcendental over the field $\mathbb C$, it is (by definition) algebraic over the ring-with-an-involution $(\mathbb C,\bar{\phantom z})$, once one extends complex conjugation to it via $\bar Z:=-1/Z$; my version in fact was to take $\mathbb C(Z)$, then algebraicity would hold in the signature of fields-with-an-involution. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 15, 2016 at 19:27

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