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This half-serious question is inspired by the answer to my previous one, Want something like Cayley formula for unitary matrices

The equation $z^2=-1$ does not have solutions in $\mathbb R$; adding a solution produces $\mathbb C$.

The equation $z\bar z=-1$ does not have solutions in $\mathbb C$; adding a solution produces what?

Update - having learned more thanks to the comments and the answer, I've now posted a question on MO with hopefully more serious and interesting content in it: https://mathoverflow.net/q/248241/41291

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    $\begingroup$ What properties should conjugation preserve/retain? $\endgroup$ – quid♦ Aug 24 '16 at 14:47
  • $\begingroup$ Would that make $\mathbb C[z]$ into a ring? $\endgroup$ – Lehs Aug 24 '16 at 14:47
  • $\begingroup$ @quid @ Lehs Good questions, thanks. Say extend signature from rings to rings-with-antiinvolution. Or alternatively rings-with-an-automorphism (these will probably have different outcomes; choose by your own taste). $\endgroup$ – მამუკა ჯიბლაძე Aug 24 '16 at 14:48
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    $\begingroup$ Would $\mathbb{C}[X]/(X^2 -1)$ be an option? $\endgroup$ – quid♦ Aug 24 '16 at 14:54
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    $\begingroup$ Sorry, I did not write what I meant to write. It should be $C[x]/(x^2 + 1)$ so that a solution is $x$. $\endgroup$ – quid♦ Aug 24 '16 at 15:10
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A way to construct something like this is to consider the polynomial ring $\mathbb{C}[X]$, where conjugation is extended by imposing $\overline{X}=X$.

Then consider the quotient $R = \mathbb{C}[X]/(X^2+ 1)$. This is a ring with unity, yet not a domain of course.

Put differently, consider $\mathbb{C}^2$ with coordinatewise addition and conjugation, and multiplication given by $(a_1,b_1)(a_2,b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1)$. Identify the complex numbers with the elements $(c,0)$.

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    $\begingroup$ By the Chinese remainder theorem your ring $R$ is then isomorphic to a direct sum of two copies of $\Bbb{C}$. As $X^2+1=(X+i)(X-i)$ and $X\pm i$ are coprime, we get $$R\simeq \Bbb{C}[X]/(X+i)\oplus\Bbb{C}[X]/(X-i).$$ Edit: And unless I made a mistake your extension of conjugation then interchanges the two components. $\endgroup$ – Jyrki Lahtonen Aug 24 '16 at 15:38
  • $\begingroup$ Yes, but I feel this is less intuitive than the explicit description of the ring structure on $C^2$ that I gave in the final paragraph. // On the "edit" That's a nice property though of which I did not think. $\endgroup$ – quid♦ Aug 24 '16 at 15:43
  • $\begingroup$ Yours is in the spirit of Cayley constructions, sure! $\endgroup$ – Jyrki Lahtonen Aug 24 '16 at 15:47
  • $\begingroup$ @JyrkiLahtonen ...in other words it is isomorphic to ${\mathbb C}\oplus{\mathbb C}$ (with multiplication $(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2)$, that is), right? $\endgroup$ – მამუკა ჯიბლაძე Aug 24 '16 at 16:07
  • $\begingroup$ ...with $\mathbb C$ embedded via $a\mapsto(a,a)$, and with conjugation $\overline{(a,b)}=(\bar b,\bar a)$, so that $(i,-i)\overline{(i,-i)}=-1$? $\endgroup$ – მამუკა ჯიბლაძე Aug 24 '16 at 16:15

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