Timeline for What structure do you get if you adjoint a root of $z \bar{z} = -1$ to the complex numbers?
Current License: CC BY-SA 3.0
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| Dec 15, 2016 at 19:27 | comment | added | მამუკა ჯიბლაძე | 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. | |
| Dec 15, 2016 at 16:44 | comment | added | WhatsUp | @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). | |
| Dec 15, 2016 at 16:25 | comment | added | მამუკა ჯიბლაძე | @SimonHenry I agree completely, that was exactly my suggestion in mathoverflow.net/q/248241/41291 and I am still thinking on it. | |
| Dec 15, 2016 at 16:11 | comment | added | Simon Henry | 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 $z$. | |
| Dec 15, 2016 at 15:31 | comment | added | Anton Fetisov | 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. | |
| Dec 15, 2016 at 14:47 | history | edited | WhatsUp | CC BY-SA 3.0 |
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| Dec 15, 2016 at 14:40 | history | edited | WhatsUp | CC BY-SA 3.0 |
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| Dec 15, 2016 at 14:31 | history | answered | WhatsUp | CC BY-SA 3.0 |