# Homework Help: Galois theory

1. Jan 22, 2007

### 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.

2. Jan 22, 2007

### StatusX

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

3. Jan 22, 2007

### HallsofIvy

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.)

Last edited by a moderator: Jan 22, 2007
4. Jan 22, 2007

### 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.

5. Jan 23, 2007

### 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 $Q(^3\sqrt{3})$.