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Galois theory

  1. Jan 22, 2007 #1
    How does one show that the order of Gal(Q(SQRT3)/Q) is 1.

    I mean, I understand that all the elements of Q (irrationals) will stay fixed, and that really only leaves one mapping of SQRT3 to itself as well, but I don't know how one would construct a proof. any help would be great.
     
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  3. Jan 22, 2007 #2

    StatusX

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    That's not true. You can send sqrt(3) to -sqrt(3).
     
  4. Jan 22, 2007 #3

    HallsofIvy

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    You don't, it's not. [itex]Gal(Q(\sqrt{3})/Q)[/itex] is the group of automorphisms from [itex]Q(\sqrt{3})[/itex] to itself that "fixes" members of Q- that is f(r)= r if r is rational. Any number in [itex]Q(\sqrt{3})[/itex] can be written in the form [itex]a+ b\sqrt{3}[/itex]. If f is an automorphism in that set, then [itex]f(a+ b\sqrt{3})= f(a)+ f(b)f(\sqrt{3})[/itex] (since f is an isomorphism)[itex] = a+ bf(\sqrt{3})[/itex]. What can [itex]f(\sqr{3})[/itex] be? Well, [itex]\sqrt{3}[/itex] satisfies [itex](\sqrt{3})^2= 3[/itex] so, applying f to both sides, [itex]f((\sqrt{3})^2)= (f(\sqrt{3}))^2= f(3)= 3[/itex]. In other words, [itex](f(\sqrt{3}))^2= 3[/itex] so [itex]f(\sqrt{3})[/itex] must be either [itex]\sqrt{3}[/itex] or [itex]-\sqrt{3}[/itex].

    If [itex]f(\sqrt{3})= \sqrt{3}[/itex] then [itex]f(a+ b\sqrt{3})= a+ b\sqrt{3}[/itex]: f is the identity function.,
    If [itex]f(\sqrt{3})= -\sqrt{3}[/itex] then [itex]f(a+ b\sqrt{3})= a-b\sqrt{3}[/itex].

    The Galois group has order 2, not 1.

    (Generally, if a is algebraic of order n, then Gal(Q(a)/Q) has order n.)
     
    Last edited: Jan 22, 2007
  5. Jan 22, 2007 #4
    that's right. it has order 2. the actual question talked of the cube root of 3, and not the sqrt. I mistakingly thought It would be similar if I used the latter. So by the same method I can show that the cuberoot of 3 must be sent to the cuberoot of 3. Thank you. I will try it out now.
     
  6. Jan 23, 2007 #5

    HallsofIvy

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    Notice the difference. [itex]x^2= 3[/itex] has two real roots: [itex]\sqrt{3}[/itex] and [itex]-\sqrt{3}[/itex] both of which are in [itex]Q(\sqrt{3})[/itex].

    The equation [itex]x^3= 3[/itex] has 3 roots but only one of them, [itex]^3\sqrt{3}[/itex] is real and only that one is in [itex]Q(^3\sqrt{3})[/itex].
     
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