Timeline for Galois elements determined by action on $n$-th roots of rationals?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 22:40 | review | Close votes | |||
| Apr 26, 2019 at 9:17 | |||||
| Apr 25, 2019 at 20:11 | comment | added | LSpice | @alpoge, right, but my question is: which $\tau$? If we allow $\tau$ to range over the whole Galois group, then, as you point out, there is no such $\sigma$. | |
| Apr 25, 2019 at 20:10 | comment | added | alpoge | (Indeed I interpreted it as: \sigma is so distinguished if and only if (\tau agrees with \sigma on that set implies \tau = \sigma) holds.) | |
| Apr 25, 2019 at 20:07 | comment | added | LSpice | When there's already an ambient group acting on a set $X$, I understand what it means to say that a particular element of that group is distinguished from others by its action on some subset $X'$; but I don't know what it means to look at an element in isolation and ask if it is, or isn't, so determined (in order to consider the set of elements that are). Could you clarify? Do you mean to consider the set of all elements of the absolute Galois group whose restriction to the set you indicate is shared with no other element? If so, then I agree with @alpoge. | |
| Apr 25, 2019 at 19:50 | comment | added | alpoge | Can’t I multiply on the right by an element of \Gal(\Qbar/\Q^{ab}({p^{1/n}: p prime, n > 1}))? Why is the identity determined up to conjugacy by its action on these elements? I think I’m being dumb, so forgive me. | |
| Apr 25, 2019 at 19:25 | review | First posts | |||
| Apr 25, 2019 at 19:51 | |||||
| Apr 25, 2019 at 19:22 | history | asked | Shu | CC BY-SA 4.0 |