Timeline for When is $\mathbb{G}_m(R)$ enough to determine $R$?
Current License: CC BY-SA 3.0
9 events
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Nov 1, 2011 at 11:15 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
added 8 characters in body
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Nov 1, 2011 at 6:13 | comment | added | Adam Hughes | Alex: yes, sorry, I messed up my notation, I meant the field extension without the "Gal" bit at the beginning. | |
Nov 1, 2011 at 3:59 | comment | added | Harry Altman | Surely we need that k has no nontrivial nilpotents? | |
Nov 1, 2011 at 2:31 | comment | added | Alex | If S is required to be a subfield, then of course you get uniqueness, but I still don't see a natural condition that would guarantee existence. Besides, you wrote that your motivating example if the absolute Galois group of $\mathbb{Q}$, but you would need your subgroup to be commutative if it has to be the group of units in a field. Maybe I totally misunderstood, but I don't really see what you're trying to do. | |
Nov 1, 2011 at 2:07 | comment | added | Adam Hughes | (and of course S should be required to be a subfield) | |
Nov 1, 2011 at 1:45 | comment | added | Adam Hughes | M Turgeon: if R is required to be a field that circumvents the problem with $k[x]$, so there is hope in that direction. | |
Nov 1, 2011 at 1:38 | comment | added | M Turgeon | @Adam: This is just my gut feeling, but I don't think many interesting hypotheses on your ring could circumvent Steven's counterexample to Uniqueness... | |
Nov 1, 2011 at 0:34 | comment | added | Adam Hughes | That's great for general rings, and I thought I said this, but I apparently did not: can I include assumptions on $R$ which will make it hold true. I'll edit the original question. Thanks! | |
Nov 1, 2011 at 0:26 | history | answered | Steven Landsburg | CC BY-SA 3.0 |