Proving Gal(Q(SQRT3)/Q) has Order 1

[calvino]

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.

 

[StatusX]

That's not true. You can send sqrt(3) to -sqrt(3).

 

[HallsofIvy]

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.

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

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

The Galois group has order 2, not 1.

(Generally, if a is algebraic of order n, then Gal(Q(a)/Q) has order n.)

 

[calvino]

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.

 

[HallsofIvy]

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

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