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Galois Theory: Morphisms

by wattsup03
Tags: galois, morphisms, theory
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Nov3-11, 07:20 AM
P: 6
1. The problem statement, all variables and given/known data

Let K = Q(21/4)

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 K-Homomorphism 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 K-homomorphisms 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.
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Nov3-11, 07:41 AM
Sci Advisor
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].
Nov3-11, 02:46 PM
Sci Advisor
P: 906
it is somewhat misleading to say the values of f depend on the values of f(21/4), f(21/2) and f(23/4).

f is a field automorphism, so (for example) f(23/4) = f((21/4)3)= (f(21/4)3,

so f ONLY depends on the value of f(21/4).

while it is true that f permutes the roots of x4-2, 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 x4-2 as α1234, where:

αj = αe(j-1)πi/4

then (α2 α4) yield a member of Aut(K/Q), but (α2 α3) does not.

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