6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Would $\mathbb{C}[X]/(X^2 -1)$ be an option? $\endgroup$ – quid Aug 24 '16 at 14:54
  • 1
    $\begingroup$ 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
6
$\begingroup$

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 $(c,0)$.

$\endgroup$
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.