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Jun 23 at 6:55 history edited GH from MO CC BY-SA 4.0
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Jun 27, 2021 at 13:53 comment added Random Yes, that's true, but I think that the general statement is quite nice in and of itself.
Jun 27, 2021 at 9:30 comment added Wojowu This would be much simpler if you didn't try to prove the more general statement - assume $n=\sum\sqrt{n_i}$ and let $k$ be the degree of the extension we are taking trace over. Traces give $\sum tr(\sqrt{n_i})=kn=k\sum\sqrt{n_i}$, and $tr(\sqrt{n_i})$ is either $0$ or $k\sqrt{n_i}$ depending on if $n_i$ is a square. To get equality, all $n_i$ must thus be squares.
Jun 27, 2021 at 0:24 vote accept Mohammad Ali Nematollahi
Jun 26, 2021 at 22:50 history answered Random CC BY-SA 4.0