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Nov 2, 2022 at 12:53 history edited Robert Bryant CC BY-SA 4.0
Clarified the remark about the irreducibility of $f$.
Nov 2, 2022 at 9:50 history edited Robert Bryant CC BY-SA 4.0
Put in a correct proof of the correct answer
Nov 1, 2022 at 14:12 comment added Robert Bryant @Gro-Tsen: You are completely right. In fact, my claimed answer is completely wrong. I did not compute the Galois group and I should have, because it is easy. I will change my answer.
Nov 1, 2022 at 11:11 comment added Gro-Tsen While this is a plausible argument, it should be made clear that this does not constitute a full proof of the nonexistence of a formula in radicals: such a proof would most likely involve computing the Galois group of the extension defined by the nuclear norm over the space of (sufficiently general) matrices, probably by monodromy considerations.
Oct 30, 2022 at 12:00 vote accept zhamao dra
Oct 30, 2022 at 10:04 history answered Robert Bryant CC BY-SA 4.0