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Jul 8, 2013 at 3:24 history edited Chandan Singh Dalawat CC BY-SA 3.0
simplified the argument by point out that the local Kronecker-Weber theorem is not used
Jul 7, 2013 at 18:39 comment added Jeff Yelton There you go, that argument should have occurred to me. I guess that although all the square roots are in $K$, higher roots generally are not. Thanks!
Jul 7, 2013 at 18:37 vote accept Jeff Yelton
Jul 7, 2013 at 16:05 history answered Chandan Singh Dalawat CC BY-SA 3.0