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

[Elwin.Martin]

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 \omega = cos (2\pi /7) + i sin (2\pi / 7)

Why is it that:
"Since x^7 -1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1),
|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]

 

[DonAntonio]

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 \omega = cos (2\pi /7) + i sin (2\pi / 7)

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


*** We have that, in fact, \,\,\omega=e^{2\pi i/7}\,\,,\,\,so\,\,w^7-1=0\Longrightarrow \omega\,\, is a root of \,\,x^7-1=(x-1)(x^6+...+x+1)\Longrightarrow |Gal(\mathbb{Q}(\omega)/\mathbb{Q}|\leq [\mathbb{Q}(\omega):\mathbb{Q}]\leq 6\,\,since \,\,\omega\,\, 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 \,\,\{1,\omega,\omega^2,...,\omega^5\}\,\, , 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]

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