302.S8C: Automorphisms of Normal Extensions

Finite normal extensions have precisely as many automorphisms as their degree. And, only their base field remains fixed by the whole Galois group. Three examples illustrate this theorem.

[@PunmasterSTP]

Automorphisms? More like "Amazing lectures that can lead to wisdom!" 👍

[@jaeimp]

@11:05 "And finally, the term with i radical 2 in it is going to flip signs under the automorphism c." However, the automorphism c flips signs for the basis i and square root of 2, which would leave their product unchanged.

[@alexheaton2]

Why can't you send sqrt(2) to it's negative? At about 7:45, you are talking about all the possible automorphisms of Q adjoin the fourth root of 2 over base field Q. sqrt(2) is not in the base field, so it need not be fixed. Can't we switch out sqrt(2) with its negative? What am I missing that makes that not possible?

[@kenflorek7189]

Alex Heaton You probably have figured it out by now, but just in case: First, Automorphism "a" sends +sqrt(2) to +sqrt(2). Secondly, if there was an Automorphism, say "b", which sends sqrt(2) to -sqrt(2), then we'd have -sqrt(2) = b(sqrt(2)= b(2^1/4)b(2^1/4), which is not possible in Q adjoin (2^1/4.). .

[@ortollj4591]

there is an error on the first example at 4:0 poly min is not t^4 + 1
Hello Alex Heaton
I'm asking myself the same question than you, I suppose it has to do with the fact
that the third example is not a splitting Field (p=x^4 -2 ) does not split completely in linear terms in R)I remark that in the first example with Q{i,sqrt(2)}, all the roots are in the Field
p1=(x + sqrt(2))*(x - sqrt(2))*(x + i*sqrt(2))*(x - i*sqrt(2))=(x^2 - 2)*(x^2 + 2)= x^4 -4
==> roots:sqrt(2),-sqrt(2),i*sqrt(2),-i*sqrt(2)
vector space base= a + b*i + c*sqrt(2) + d*i*sqrt(2)
there exist for any roots a linear way to pass from one roots to another.
(multiply by -1) or (multiply by i) or (multiply by -1 and multiply by i ) or multiply by 1(do nothing)
;
but in the third example sqrt(2) is in the field but it is not a linear root of p3=x^4 - 2
p3=(x-2^(1/4) ) *(x +2^(1/4) ) *(x-i*2^(1/4) ) *(x + i*2^(1/4) )
p3=(x-2^(1/4) ) *(x +2^(1/4) ) *(x^2 + sqrt(2))
==> roots:2^(1/4),-2^(1/4), i*2^(1/4),-i*2^(1/4)
vector space base= a + b*2^(1/4) + c*2^(1/2) + d*2^(3/4)
Ken Florek remarked here, that there are no linear way to pass from one of the other roots to -sqrt(2)
but sqrt(2) is not a polynomial root ! I don't understand.
;
In the second example I'm also lost !
p2=1+x+x^2+x^3+x^4 = (x-ζ_5(1))*(x-ζ_5(2))*(x-ζ_5(3))*(x-ζ_5(4))
==> roots:(-1)^(1/5),(-1)^(2/5),(-1)^(3/5),(-1)^(4/5)
with ζ_5(n)=e^(2*π*i*n/5) n in{1,2,3,4}
==> roots: i, -i, 1/2 + i*sqrt(3)/2, 1/2 - i*sqrt(3)/2
vector space base= a + b*i + c*sqrt(3)/2 + d*i*sqrt(3)/2
maybe the different automorphisms are theses below ?
as all roots modules=1, and I Imagine passing to another roots from ζ_5(1)
arg ζ_5(1) = arg ζ_5(1)
arg(ζ_5(2))= arg(ζ_5(1)*ζ_5(1))=arg(ζ_5(1)) + arg(ζ_5(1))
arg(ζ_5(3))= arg(ζ_5(1)*ζ_5(2))=arg(ζ_5(1)) + arg(ζ_5(2))
arg(ζ_5(4))= arg(ζ_5(2)*ζ_5(2))=arg(ζ_5(2)) + arg(ζ_5(2))
I'm lost here maybe somebody can help me ?

[@Mrpallekuling]

sqrt(2) is in the base field, as he says 06:02

[@gaaraofddarkness]

for the ζ5 case, why dont we consider the aut from ζ^2 to ζ^3 and so on?

[@rocky171986]

By the definition of an automorphism, the action of automorphism on zeta completely determines what happens to zeta squared. Note phi(zeta^2)=phi(zeta*zeta)=phi(zeta)*phi(zeta). This applies to other powers of zeta as well.

[@rubberubertuber]

Maybe water that poor plant on the windowsill?

[@PunmasterSTP]

Oh yeah, I felt bad after you pointed that out and I saw it too 😥

[@studywinner5190]

Nice sir