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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

Community
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მამუკა ჯიბლაძე
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2 Answers2

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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 $(c,0)$.

quid
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    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. – Jyrki Lahtonen Aug 24 '16 at 15:38
  • 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. – quid Aug 24 '16 at 15:43
  • Yours is in the spirit of Cayley constructions, sure! – Jyrki Lahtonen Aug 24 '16 at 15:47
  • @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? – მამუკა ჯიბლაძე Aug 24 '16 at 16:07
  • ...with $\mathbb C$ embedded via $a\mapsto(a,a)$, and with conjugation $\overline{(a,b)}=(\bar b,\bar a)$, so that $(i,-i)\overline{(i,-i)}=-1$? – მამუკა ჯიბლაძე Aug 24 '16 at 16:15
  • An explicit isomorphism from @quid's $\mathbb C^2$ to this one sends $(a,b)$ to $(a-ib,a+ib)$. – მამუკა ჯიბლაძე Aug 24 '16 at 18:47
  • Is not it isomorphic to bicomplex numbers? If so, this is strange, because I have checked for the solutions there. – Anixx Apr 28 '21 at 18:15
  • @Anixx I don't know what bicomplex numbers are. – quid Apr 28 '21 at 18:17
  • If you add to complex numbers a second complex unity such as $j^2=-1$ you get bicomplex numbers. They are isomorphic to the tessarines and I checked for solutions there, seems there is none. – Anixx Apr 28 '21 at 18:19
  • Maybe, the difference is in your definition of conjugation? – Anixx Apr 28 '21 at 18:21
  • Possibly @Anixx – quid Apr 28 '21 at 18:22
  • What does it mean, coordinate-wise conjugation? – Anixx Apr 28 '21 at 18:23
  • @Anixx $\overline{(c, d)} = (\overline{c}, \overline{d})$ – quid Apr 28 '21 at 18:43
  • Well, I tried to check your proposal ($X^2=-1$, $\overline{X}=X$), but also see no solution. – Anixx Apr 28 '21 at 18:43
  • @Anixx what do you mean with no solution? – quid Apr 28 '21 at 18:44
  • I tried your ring, and cannot find a number that satisfies $z\overline{z}=-1$. See my answer below (this is for a somewhat different case). But for your case I also get empty set. – Anixx Apr 28 '21 at 18:46
  • @Anixx but X clearly has the property. – quid Apr 28 '21 at 18:48
  • Wait... Yes. Let's see where I was wrong. – Anixx Apr 28 '21 at 18:50
  • Oh, I see the mistake, I used number -1 instead of equivalent diagonal matrix. There are solutions both in your system and tessarines. – Anixx Apr 28 '21 at 18:55
  • @Anixx thanks for the information; I'm glad that this is settled. – quid Apr 28 '21 at 19:03
  • Oh well, in tessarines there are no solutions (the equations give imaginary coefficients). In your system all works, but the only objection is that such conjugation looks non-canonical, because it contradicts the matrix transposition of bicomplex numbers. – Anixx Apr 28 '21 at 19:27
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It seems, the answer depends on the definition of conjugation. The answer by @quid seems to be isomorphic to bicomplex numbers or tessarines, except for the different definition of conjugation.

If we consider the ring of tessarines, with conjugation being defined as changing the sign of all occurrences of $i$ to $-i$, there is no solution. This definition of conjugation is equivalent as transposition of tessarines represented as $4\times4$ real matrices or conjugate-transpose of tessarines represented as $2\times2$ complex matrices.

In other words, we add $j$ such that $j^2=1$, $\overline{j}=j$, $(ij)^2=-1$, $\overline{ij}=-ij$

Code in Mathematica, proving there is no solution:

Solve[Dot[( {
     {w0, -w1, w2, -w3},
     {w1, w0, w3, w2},
     {w2, -w3, w0, -w1},
     {w3, w2, w1, w0}
    } ), Transpose[( {
      {w0, -w1, w2, -w3},
      {w1, w0, w3, w2},
      {w2, -w3, w0, -w1},
      {w3, w2, w1, w0}
     } )]] == ( {
    {-1, 0, 0, 0},
    {0, -1, 0, 0},
    {0, 0, -1, 0},
    {0, 0, 0, -1}
   } ), {w0, w1, w2, w3}, 
 Assumptions -> 
  w0 \[Element] Reals && w1 \[Element] Reals && w2 \[Element] Reals &&
    w3 \[Element] Reals]

Out={ }
Anixx
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