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Homework Help: Galois Theory: Morphisms

  1. Nov 3, 2011 #1
    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.
  2. jcsd
  3. Nov 3, 2011 #2


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    [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].
  4. Nov 3, 2011 #3


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    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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