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

  • Thread starter calvino
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In summary, the Galois group of Q(\sqrt{3})/Q has an order of 2, not 1. This is because any automorphism in the set must either send \sqrt{3} to itself or to -\sqrt{3}. However, with the equation x^3=3, there is only one real root, ^3\sqrt{3}, and only that root is in Q(^3\sqrt{3}). Therefore, the Galois group of Q(^3\sqrt{3})/Q has an order of 1.
  • #1
calvino
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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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  • #2
That's not true. You can send sqrt(3) to -sqrt(3).
 
  • #3
calvino said:
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. [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.)
 
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  • #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.
 
  • #5
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].
 

1. How do you prove that Gal(Q(SQRT3)/Q) has Order 1?

In order to prove that Gal(Q(SQRT3)/Q) has Order 1, we need to show that there is only one automorphism in the Galois group. This can be done by demonstrating that the only possible automorphism is the identity map, which fixes all elements in the field extension.

2. What is the significance of proving Gal(Q(SQRT3)/Q) has Order 1?

Proving Gal(Q(SQRT3)/Q) has Order 1 is significant because it shows that the field extension Q(SQRT3)/Q is a trivial Galois extension. This means that the field extension is not Galois, and therefore does not possess the properties of a Galois extension, such as being a normal and separable extension.

3. What are the steps to prove Gal(Q(SQRT3)/Q) has Order 1?

The steps to prove Gal(Q(SQRT3)/Q) has Order 1 are:

  1. Show that the field extension Q(SQRT3)/Q is a finite extension.
  2. Determine the minimal polynomial of SQRT3 over Q.
  3. Show that the Galois group is isomorphic to the group of roots of the minimal polynomial.
  4. Determine the possible automorphisms in the Galois group.
  5. Finally, demonstrate that the only possible automorphism is the identity map.

4. Can Gal(Q(SQRT3)/Q) have a different order?

No, Gal(Q(SQRT3)/Q) can only have Order 1. This is because the field extension Q(SQRT3)/Q is a simple extension, meaning that it is generated by a single element (SQRT3). In simple extensions, the Galois group can only have Order 1.

5. What are some real-world applications of proving Gal(Q(SQRT3)/Q) has Order 1?

One real-world application of proving Gal(Q(SQRT3)/Q) has Order 1 is in the field of cryptography. The concept of Galois groups is used in cryptography to create secure codes and ciphers. By proving that a field extension is not Galois, we are able to identify which fields can be used to create secure codes and which are vulnerable to attacks.

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