Timeline for Two (other) rings...are they isomorphic?
Current License: CC BY-SA 3.0
11 events
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| Nov 19, 2014 at 0:19 | comment | added | Nicholas Proudfoot | It turns out that there's an even easier isomorphism that works in any characteristic: just send $x_1$ to $\frac{x_1}{1+x_1}$, and send every other $x_i$ to itself! (This is the map from $R$ to $S$.) | |
| Nov 16, 2014 at 20:34 | comment | added | Vladimir Dotsenko | @NicholasProudfoot: Ah I see. Yes that extra property of your problem of course makes it natural to not only consider the $S_n$-invariant things but really just restrict yourself to the diagonal embedding of diffeomorphisms of the line (which indeed worked) :) Thanks for this explanation, it is good to know. And yes, it would be interesting to know what happens when the characteristic divides $n$, I would not be surprised if the answer actually is different! | |
| Nov 16, 2014 at 16:44 | comment | added | Nicholas Proudfoot | One thing that still bothers me is that my solution doesn't work when $\mathbb{C}$ is replaced by a field of characteristic 2. (Or, with $n$ variables, a field of characteristic dividing $n$.) I would still be interested to know if the rings are isomorphic in such a situation. | |
| Nov 16, 2014 at 16:42 | comment | added | Nicholas Proudfoot | In the more general situation that I care about, there will be a number of relations like this, with coefficients, each involving some subset of the variables. I realized that if I need to construct an isomorphism that deals with each of the equations simultaneously, it would be really helpful if $\varphi(x_i)$ were a power series in only $x_i$. Furthermore, because of the symmetry, I should try the same power series in each variable. Once I restricted the problem this much, it was easy to work out the first few terms when $n=3$ and guess the general solution. | |
| Nov 16, 2014 at 16:38 | comment | added | Nicholas Proudfoot | @VladimirDotsenko: My feeling is of course that your approach will work, but I don't have any further insight into why that is true. | |
| Nov 16, 2014 at 14:23 | comment | added | Vladimir Dotsenko | @NicholasProudfoot: this is a very nice argument, albeit somewhat specific for the problem you wanted to solve. Do you now have a feeling as to whether the condition that arose in both the approach of David Speyer and the approach I outlined (absence of $x_i^n$) can be maintained inductively if it was the case in the beginning? | |
| Nov 16, 2014 at 14:19 | comment | added | Vladimir Dotsenko | @WłodzimierzHolsztyński: I agree with Andy: I believe that the OP is quite right to have handled it like this once he figured out a complete solution (in case hearing this from one of the people who wrote incomplete answers above is more convincing). | |
| Nov 16, 2014 at 7:32 | comment | added | Andy Putman | @WłodzimierzHolsztyński: While of course one can overdo this, I think that what the OP does here is totally fine, and I see no reason to criticize him for it (implicitly or otherwise). | |
| Nov 16, 2014 at 1:10 | history | edited | Nicholas Proudfoot | CC BY-SA 3.0 |
added 142 characters in body
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| Nov 15, 2014 at 22:37 | vote | accept | Nicholas Proudfoot | ||
| Nov 15, 2014 at 22:37 | history | answered | Nicholas Proudfoot | CC BY-SA 3.0 |