2 ms·

You're doing fine. Do cheer up. No field of rational functions here, that's waaaay off given what Stillwell intends to do. Also, subfields of reals are a bit

• monfrere 9y ago
  Oh, right, subfield of the complex numbers. Now sigma is supposed to be an automorphism on that subfield. Which means sigma does take scalar values as arguments, contrary to what stablemap said...
  • ky3 9y ago
    > Oh, right, subfield of the complex numbers Right-o! > Now sigma is supposed to be an automorphism on that subfield. Which means sigma does take scalar values as arguments, contrary to what stablemap said... There's that ol' abuse of notation going on here. The first sigma is just a permutation on the roots: {x_i} -> {x_i}. In particular, this skinny sigma is not defined on scalars. It embiggens into a second sigma that's your automorphism: Q({x_i}) -> Q({x_i}). This fat sigma now maps scalars. Now identify the first and second sigmas, and the abuse is complete.