Proving Abelian Property of Galois Group in Q(z) over Q

  • Thread starter jeffreydk
  • Start date
  • Tags
    Group
In summary: Adjoining a single root does not necessarily mean it is the only element being extended. In this case, we are given that z, z2, z3, ..., zn=1 are all distinct roots of xn-1. This means that Q(z) must also contain all of the other roots, making it a Galois extension. Therefore, the Galois group of Q(z) over Q cannot be just the identity map, and is in fact abelian.
  • #1
jeffreydk
135
0
I am trying to show that if z, z2, z3, ..., zn=1 are n distinct roots of xn-1 in some extension field of Q (the rationals), then GalQQ(z) (the galois group of Q(z) over Q) is abelian. Would I be wrong to say that since the galois group we're talking about here only involves an extension field with one of the roots, namely z, then the only map we could have in the group would be the identity map and therefore it is abelian? Something feels wrong about this but I'm not sure how else there would be other automorphisms in the group.
 
Physics news on Phys.org
  • #2
To say that GalQ Q(z) is abelian requires an imlicit assumption: that GalQ Q(z) actually exists. For that to happen. Q(z) must actually be a Galois extension. Also, if Q(z) is a Galois extension of Q, then #GalQ Q(z) = [Q(z) : Q] -- so if GalQ Q(z) consists only of the identity, then z must actually be a rational number... which is only true if n < 3.

The problem secretly told you that Q(z) is a Galois extension of Q, and you know that automorphisms of Q(z) map roots of z^n - 1 to roots of z^n - 1. You made an unwarranted assumption that led you to conclude that the only automorphism of Q(z) is trivial -- what was it?


If you're still confused, maybe it's worth first working out the Galois group of the entire splitting field of x^n - 1, rather than just of Q(z).
 
  • #3
Yes that makes sense that its a Galois extension, because we're told that each of the roots are distinct (and there are n of them) so it is a splitting field and has characteristic 0, thus it's Galois. Although your argument makes perfect sense, saying that |GalQQ(z)|=[Q(z):Q]=1 if the group is trivial, I am having trouble noticing where my unwarranted assumption was, because to me, if the field is only extending z and not the rest of the roots, there are no combinations of isomorphisms that can be made other than some f:z-->z. I'll try working out the group for the entire splitting field.
 
  • #4
jeffreydk said:
Yes that makes sense that its a Galois extension, because we're told that each of the roots are distinct (and there are n of them) so it is a splitting field and has characteristic 0, thus it's Galois.
The splitting field is Galois -- but is Q(z)? Typically, adjoining a single root of a polynomial does not give a Galois extension, because you need to include all of its conjugates to make the extension Galois.

if the field is only extending z and not the rest of the roots,
Actually, that is the unwarranted assumption.
 

Related to Proving Abelian Property of Galois Group in Q(z) over Q

1. What is the Galois Group?

The Galois Group is a mathematical concept that was developed by French mathematician Évariste Galois in the 19th century. It is a group of automorphisms (symmetries) that preserve the structure of a mathematical object, such as a field or a polynomial equation.

2. How is the Galois Group related to field theory?

The Galois Group is closely related to field theory, which is the study of algebraic structures known as fields. The Galois Group can be used to determine the potential solutions of a polynomial equation, as well as the structure and properties of a given field.

3. What is the significance of the Galois Group in mathematics?

The Galois Group has significant applications in mathematics, particularly in the field of algebra. It has been used to prove the unsolvability of certain mathematical problems, as well as to study the properties of fields and polynomial equations.

4. How is the Galois Group calculated?

The Galois Group is calculated by finding the automorphisms of a given mathematical object, such as a field or a polynomial equation. These automorphisms can then be combined to form a group, which is known as the Galois Group.

5. What are some real-world applications of the Galois Group?

The Galois Group has many practical applications, particularly in cryptography and coding theory. It is also used in engineering and physics to study symmetry and group theory. Additionally, the Galois Group has been applied to the study of algebraic curves and surfaces in geometry.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
926
  • Linear and Abstract Algebra
Replies
1
Views
703
  • Linear and Abstract Algebra
Replies
17
Views
4K
  • Linear and Abstract Algebra
Replies
1
Views
939
  • Linear and Abstract Algebra
Replies
6
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
2K
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
Back
Top