Its been years since I've studied this stuff, but I'd assume its no coincedence that the powers of alpha that show up in the galois group are precicely those that are coprime with 8?
For your first question, no. One of the roots is i and the other is -i. If you evaluate x^2+1 on -1, you get (-1)^2+1 =1+1=2. For your second question, σ^2(√2) = σ(σ(√2)) = σ(-√2) = σ(-1)σ(√2) [since σ is an automorphism σ(ab) = σ(a)σ(b)] = (-1)(-√2) [since σ is a Galois automorphism of Q(√2,i)/Q, σ(c) = c for all c in Q] = √2.
thank you, nice example for me to understand cyclotomic extension
What software do you use for making this kind of videos (handwritten math)?
This was the iPad app Notability and the iPad's native screen recorder. Today, I use OneNote or Xournal recorded with Camtasia.
thank you
Its been years since I've studied this stuff, but I'd assume its no coincedence that the powers of alpha that show up in the galois group are precicely those that are coprime with 8?
Yes, Gal(Q(a)/Q) is group isomorphic to Z^x_8 multiplicative group mod 8
Sorry , aren't the roots of x^2+1 ,i and -1. Also i dint get why σ squared of sqrt2 is σ(-1)σ(sqrt2) shouldn't it be 2
For your first question, no. One of the roots is i and the other is -i. If you evaluate x^2+1 on -1, you get (-1)^2+1 =1+1=2.
For your second question, σ^2(√2) = σ(σ(√2)) = σ(-√2) = σ(-1)σ(√2) [since σ is an automorphism σ(ab) = σ(a)σ(b)] = (-1)(-√2) [since σ is a Galois automorphism of Q(√2,i)/Q, σ(c) = c for all c in Q] = √2.
@@AdamGlesser oh my god ..i was tired from studying. I see it clearly now..thanks for your time