Galois Theory: Degree of Q(ω)/Q & Why 6 Basis Vectors?

In summary, the dimension of a field extension E over F can be considered as the degree of the extension, denoted as [E:F]. In most books, this notation is used to represent the dimension. Defining \omega = cos (2\pi /7) + i sin (2\pi / 7), we can see that it is the root of a rational polynomial of degree 6, making the degree of the extension |Gal(Q(ω)/Q)|=[Q(ω):Q]≤ 6. The vector space of Q(ω) over Q can be represented by 6 basis vectors, excluding the sixth power which can be expressed as a rational linear combination of the other five. This is why the
  • #1
Elwin.Martin
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So if we have an extension of E of F, then we can consider E as a vector space over F.
The dimension of this space is the degree of the field extension, I think most people use [E:F].
This is correct in most people's books, right?

Defining [itex] \omega = cos (2\pi /7) + i sin (2\pi / 7) [/itex]

Why is it that:
"Since [itex] x^7 -1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1) [/itex],
|Gal(Q(ω)/Q)|=[Q(ω):Q]≤ 6"
?
With another piece of information he gets that the degree is 6. Since the vector space of Q(ω) over Q has degree 6, we should be able to represent it with 6 basis vectors...but this just feels weird since ω is a seventh root of unity and it feels like we should have something like:
a+bω+cω2+...gω6 a,b,...,g in Q, right?

What am I missing? I have no reason to doubt Gallian, I even looked up the general method for calculating degrees of Q(n-th roots of unity)/Q and I know that 6 is correct. . .I just don't see why and I can't construct 6 basis vectors.

[This is from Gallian's Abstract Algebra book, 7th edition, in Ch. 32 on page 550]
 
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  • #2


Elwin.Martin said:
So if we have an extension of E of F, then we can consider E as a vector space over F.
The dimension of this space is the degree of the field extension, I think most people use [E:F].
This is correct in most people's books, right?

Defining [itex] \omega = cos (2\pi /7) + i sin (2\pi / 7) [/itex]

Why is it that:
"Since [itex] x^7 -1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1) [/itex],
|Gal(Q(ω)/Q)|=[Q(ω):Q]≤ 6"
?


*** We have that, in fact, [itex]\,\,\omega=e^{2\pi i/7}\,\,,\,\,so\,\,w^7-1=0\Longrightarrow \omega\,\,[/itex] is a root of [tex]\,\,x^7-1=(x-1)(x^6+...+x+1)\Longrightarrow |Gal(\mathbb{Q}(\omega)/\mathbb{Q}|\leq [\mathbb{Q}(\omega):\mathbb{Q}]\leq 6\,\,[/tex]since [itex]\,\,\omega\,\,[/itex] is a root of a rational pol. of degree 6 (why?) ***


With another piece of information he gets that the degree is 6. Since the vector space of Q(ω) over Q has degree 6, we should be able to represent it with 6 basis vectors...but this just feels weird since ω is a seventh root of unity and it feels like we should have something like:
a+bω+cω2+...gω6 a,b,...,g in Q, right?


*** Yes, right...and this is almost exactly what we have in that field. ***

What am I missing? I have no reason to doubt Gallian, I even looked up the general method for calculating degrees of Q(n-th roots of unity)/Q and I know that 6 is correct. . .I just don't see why and I can't construct 6 basis vectors.


*** But you have...almost! A basis of this space over the rational indeed is [itex]\,\,\{1,\omega,\omega^2,...,\omega^5\}\,\,[/itex] , without the sixth power which can

be expressed as a rational linear combination of the other five (why?)

DonAntonio ***



[This is from Gallian's Abstract Algebra book, 7th edition, in Ch. 32 on page 550]

...
 

1. What is the degree of the field extension Q(ω)/Q in Galois Theory?

The degree of a field extension is the dimension of the extension as a vector space over the base field. In the case of Q(ω)/Q, where ω is a complex root of a polynomial with rational coefficients, the degree is equal to the degree of the minimal polynomial of ω. This is because Q(ω) can be viewed as the field extension of Q obtained by adjoining ω, and the minimal polynomial determines the elements of Q(ω).

2. How is the degree of Q(ω)/Q related to the Galois group?

In Galois Theory, the degree of a field extension is directly related to the size of the Galois group. Specifically, the Galois group of Q(ω)/Q has order equal to the degree of the extension. This means that the number of automorphisms of Q(ω) that fix Q as a subfield is equal to the degree of the extension.

3. Why is the degree of Q(ω)/Q sometimes referred to as the degree of the field extension?

The degree of a field extension is often used to refer to the degree of the extension as a vector space. In the case of Q(ω)/Q, this is equivalent to the degree of the minimal polynomial of ω, as mentioned in the answer to the first question. This terminology is used to emphasize the connection between the dimension of the extension and the size of the Galois group.

4. What is the significance of having 6 basis vectors in Galois Theory?

In Galois Theory, the basis vectors correspond to the automorphisms of the field extension. The number of basis vectors is equal to the degree of the extension, which in the case of Q(ω)/Q is equal to the degree of the minimal polynomial of ω. Having 6 basis vectors means that there are 6 distinct automorphisms of Q(ω) that fix Q as a subfield, and these automorphisms play a crucial role in understanding the Galois group of the extension.

5. How does the degree of Q(ω)/Q relate to the roots of the minimal polynomial?

The degree of the minimal polynomial is equal to the degree of the field extension Q(ω)/Q. This means that if the minimal polynomial has degree n, then the field extension has degree n. Additionally, the roots of the minimal polynomial correspond to the basis vectors of the field extension and the automorphisms of Q(ω). This relationship between the degree of the extension, the minimal polynomial, and its roots is a fundamental concept in Galois Theory.

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