Timeline for Injective ring homomorphism from $\mathbb{Z}_p[[x,y]]$ to $\mathbb{Z}_p[[x]]$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8 at 7:37 | vote | accept | kindasorta | ||
| Apr 8 at 2:20 | answer | added | Will Sawin | timeline score: 34 | |
| Apr 8 at 0:26 | comment | added | Kevin Casto | @DenisT I guess I was referring to $\mathbb C$-algebra homomorphisms (taking 1 to 1), which are thus the identity on $\mathbb C \cdot 1$ and so can't embed the domain's copy of $\mathbb C$ in a strict subfield of the codomain's copy of $\mathbb C$. | |
| Apr 7 at 23:44 | comment | added | Denis T | @KevinCasto Of course, just embed formal series into the algebraic closure of the fraction field. It is abstractly isomorphic to complex numbers by categoricity of theory of alg. closed fields. | |
| Apr 7 at 23:36 | comment | added | kindasorta | Sure, I'd settle for that. | |
| Apr 7 at 23:34 | comment | added | Kevin Casto | Or for that matter (or an easy argument that there's not) from $\mathbb C[[x,y]]$ to $\mathbb C[[x]]$? | |
| Apr 7 at 23:07 | comment | added | user525759 | Is there such a homomorphism from $\mathbb{Z}[[x,y]]$ to $\mathbb{Z}[[x]]$? | |
| Apr 7 at 23:01 | comment | added | kindasorta | But then wouldn't $(y-x)\mapsto 0$ imply your map isn't injective? | |
| Apr 7 at 22:57 | comment | added | kindasorta | Yes, I would like it to be continuous, but would also be curious if there is an obvious incontinuous one. | |
| Apr 7 at 22:52 | comment | added | LSpice | Do you want the homomorphism to be continuous? (I'm not sure if that's automatic.) | |
| Apr 7 at 22:03 | history | asked | kindasorta | CC BY-SA 4.0 |