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Jul 4 at 4:19 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question has been bumped anyway)
Nov 25, 2009 at 17:08 comment added Tyler Lawson For Galois automorphisms of C, you actually need the axiom of choice for any to exist because (as an exercise) any others are Q-linear but not R-linear and hence are nonmeasurable functions. I am not sure if you actually need the axiom of choice to show that there are nontrivial Galois automorphisms of the closure of Q, but it may be similar - you need to make a choice of lift of the automorphism for every finite Galois extension F of Q in some compatible way.
Nov 25, 2009 at 14:06 comment added Mariano Suárez-Álvarez I have no idea what exactly she meant.
Nov 25, 2009 at 13:23 comment added Lior Bary-Soroker what do you mean by `write-down'?
Nov 25, 2009 at 13:12 history answered Mariano Suárez-Álvarez CC BY-SA 2.5