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Galois Theory: Morphisms 
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#1
Nov311, 07:20 AM

P: 6

1. The problem statement, all variables and given/known data
Let K = Q(2^{1/4}) Determine the automorphism group Aut(K/Q) 2. Relevant equations An automorphism is an isomorphism from a Field to itself Aut(K/Q) is the group of Automorphisms from k/Q to K/Q Definition: A KHomomorphism from L/K to L'/K is a homomorphism L> L' that is the identity on K 3. The attempt at a solution I am completely at a loss really. I have calculated there are four homomorphisms from K to C and think from there if I know how many are Khomomorphisms then that'll be the number of automorphisms, because a homomorphism from a field to itself is an automorphism (Please correct me if I'm wrong on this). Then that'll give me the set of Automorphisms. My problem is that I don't know how to go from the number of homomorphisms to the actual homomorphisms. I think it has a relation to the roots of 2^{(1 /4)} in C (which I have calculated to be 2^{(1 /4)},  2^{(1 /4)}, i*2^{(1 /4)}, i2^{(1 /4) }) Please help, this lack of understanding is preventing me from moving forward with other questions and my notes from lectures completely gloss over how to do this. 


#2
Nov311, 07:41 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

[itex](2^{1/4}^2= 2^{1/2}[/itex] and [itex](2^{1/4}^3=2^{3/4}[/itex] are irrational but [itex](2^{1/4})^4= 2[/itex] is rational so any number in [/itex]Q(2^{1/4})[/itex] is of the form [itex]a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4}[/itex] for rational numbers a, b, c, d. For any automorphism, f, [itex]f(a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4})= a + bf(2^{1/4})+ cf(2^{1/2})+ df(2^{3/4})[/itex] so the possible values of f depend entirely upon the possible values of [itex]f(2^{1/4})[/itex], [itex]f(2^{1/2})[/itex] and [itex]f(2^{3/4})[/itex].
Further, [itex](f(2^{1/4})^4= f(2)[/itex] is rational so [itex]f(2^{1/4}[/itex] must be a fourth root of 2. Also, [itex]f(2^{1/2})^2= f(2)[/itex] so [itex]f(2^{1/2}) must be [itex]2^{1/2} (or [itex]2^{1/2}[/itex] which is just a rational number times [itex]2^{1/2}. Each f permutes the fourth roots of 1 while fixing [itex]2^{1/2}[/itex]. 


#3
Nov311, 02:46 PM

Sci Advisor
P: 906

it is somewhat misleading to say the values of f depend on the values of f(2^{1/4}), f(2^{1/2}) and f(2^{3/4}).
f is a field automorphism, so (for example) f(2^{3/4}) = f((2^{1/4})3)= (f(2^{1/4})^{3}, so f ONLY depends on the value of f(2^{1/4}). while it is true that f permutes the roots of x^{4}2, it is not true that any such permutation yields an "f". f takes conjugate pairs to conjugate pairs, as well (the square of a fourth root of 2 must be a square root of 2). so if one regards the 4 roots of x^{4}2 as α_{1},α_{2},α_{3},α_{4}, where: α_{j} = αe^{(j1)πi/4} then (α_{2} α_{4}) yield a member of Aut(K/Q), but (α_{2} α_{3}) does not. 


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