Finding a normal field extension of degree 7

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MathNugget

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I am supposed to find an extension [imath]k \subset K[/imath] that is normal and of degree 7. Neither k nor K are fixed (I can choose them). The hint I was given is to choose one of them to be [imath]F_7[/imath] (it should be the double line F, but I can't find the latex code for it). (the idea is that the field [imath]F_7[/imath] has characteristic 7, but I don't really know how to use that).

I'll define the notions used first, then I'll give my idea:

A field extension [imath]k \subset K[/imath] is normal if every irreducible polynomial in [imath]k[X][/imath] that has a root in K, has all its roots in K.
Let's suppose the degree of the field extension is finite. then [imath]K=k(\alpha, \beta, ...)[/imath], and the degree of the extension, noted [K : k], is equal to the minimum degree of the polynomial with coefficients in k and the roots \alpha, \beta, ... .

My idea: let [imath]k=\mathbb{Q}[/imath] and [imath]K=\mathbb{Q}(\theta_1,...\theta_7)[/imath], with [imath]\theta_k=\sqrt[7]{2}(cos(\frac{360k}{7})+isin(\frac{360k}{7}))[/imath], [imath]k=0,...6[/imath] (the roots of [imath]x^7=2[/imath]).

All that's left to prove now, is that a polynomial in [imath]\mathbb{Q}[X][/imath] having one of these [imath]\theta_k[/imath] as its root, must have the others too...

I suppose there is a polynomial [imath]f(x)=a_0+a_1X+...+a_6X^2[/imath], with [imath]a_0, ... a_6 \in \mathbb{Q}[/imath], with [imath]f(\theta_k)=0[/imath]. Equalize rational and irrational parts, [imath]a_0=0[/imath], divide by [imath]\theta_k \neq 0[/imath], induction etc etc we get the polynomial is 0.
Now I think I have to prove [imath](\theta_k)^j[/imath] is not rational, for [imath]j \in \{1,...6\}[/imath], which can be reduced to proving [imath]\sqrt[7]{2}^j[/imath] isn't rational, which is proven similar to the way [imath]\sqrt{2}[/imath] is proven to be irrational...

Does my sketch of the proof go in the right direction? I noticed this method is very messy (and doesn't seem complete), do you have any idea about how I could use the hint, to get a (supposedly) simpler example?
 
blamocur said:
Like [imath]\mathbb F[/imath] ? For that one can use \mathbb
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Damn... didn't realize \mathbb would make the F exactly as I wanted. Thanks :) . And I saw the message too late, and cannot edit the main post...unfortunate.
 
I am quite rusty on field extensions, but here is my question regarding your approach:
Consider extension [imath]\mathbb L[/imath] of [imath]\mathbb Q[/imath] defined by the polynomial [imath]x^7-1 = 0[/imath]. Am I right to believe that the degree of [imath]\mathbb L[/imath] over [imath]\mathbb Q[/imath] is 7? And that [imath]\mathbb Q \subset \mathbb L \subset \mathbb K[/imath] and [imath]\mathbb L \neq \mathbb K[/imath] ? Then this would mean that the degree of [imath]\mathbb K[/imath] must be greater than 7, wouldn't it?
 
blamocur said:
I am quite rusty on field extensions, but here is my question regarding your approach:
Consider extension [imath]\mathbb L[/imath] of [imath]\mathbb Q[/imath] defined by the polynomial [imath]x^7-1 = 0[/imath]. Am I right to believe that the degree of [imath]\mathbb L[/imath] over [imath]\mathbb Q[/imath] is 7? And that [imath]\mathbb Q \subset \mathbb L \subset \mathbb K[/imath] and [imath]\mathbb L \neq \mathbb K[/imath] ? Then this would mean that the degree of [imath]\mathbb K[/imath] must be greater than 7, wouldn't it?
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I am not convinced that besides 1, the roots of [imath]x^7-1=0[/imath] are in the field [imath]\mathbb{K}[/imath] I tried to create there, but that seems so painful to prove...
say [imath]cos(\frac{360}{7})+isin(\frac{360}{7})=\sqrt[7]{2}[a_1(cos(\frac{360}{7})+isin(\frac{360}{7})) +a_2(cos(\frac{360\cdot 2}{7})+isin(\frac{360\cdot 2}{7})) + ... +a_7(cos(\frac{360\cdot 7}{7})+isin(\frac{360\cdot 7}{7}))]+t[/imath], with [imath]a_1,...a_7, t[/imath] some rational numbers. I think we can create a system with 3 equations, the pure rationals, the irrationals that have imaginary part 0, and the pure imaginary... maybe I am missing something, but I dont think this has solutions in [imath]\mathbb{Q}[/imath]
 
[imath]x_1 = 2^{1/7} \in \mathbb K \wedge x_2 = 2^{1/7} e^{i\pi/7} \in\mathbb K \Longrightarrow x_1/x_2 = e^{i\pi/7} \in \mathbb K[/imath]
 
That is...strange. I thought, given I started with a polynomial with deg=7, the max degree of the field extension is 7. I guess something's wrong in my calculations...
 
MathNugget said:
That is...strange. I thought, given I started with a polynomial with deg=7, the max degree of the field extension is 7. I guess something's wrong in my calculations...
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I am confused myself :( I am guessing that my post #4 might be wrong and [imath]\mathbb L = \mathbb K[/imath], but at this point I don't know how to get there.
 
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blamocur said:
I am confused myself :( I am guessing that my post #4 might be wrong and [imath]\mathbb L = \mathbb K[/imath], but at this point I don't know how to get there.
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Sounds like we aren't passing the Galois theory exam on Tuesday :p.

Something's definitely wrong somewhere in the reasoning. I think that the property I was using to say the extension has degree 7 involved a property of the field, which maybe it doesn't have...

I'll open a new thread, there's 3 more problems (out of a total of....4) which I dont know how to solve.
 
Degree seven means that the Galois group is [imath] \mathbb{Z}_7, [/imath] i.e. cyclic. The usual candidates have either minus one or plus one as roots on the unit circle and are therefore not suited. So how about [imath] \mathbb{Q} \subseteq \mathbb{Q}[x]/(x^7+2) [/imath]?
 
fresh_42 said:
Degree seven means that the Galois group is [imath] \mathbb{Z}_7, [/imath] i.e. cyclic. The usual candidates have either minus one or plus one as roots on the unit circle and are therefore not suited. So how about [imath] \mathbb{Q} \subseteq \mathbb{Q}[x]/(x^7+2) [/imath]?
Click to expand...

So [imath] \mathbb{Q} [/imath] and the polynomials in [imath]\mathbb{Q}[x][/imath] with degree < 7? I don't think I understood...

Or you're saying the splitting field of that polynomial? Because I was trying something similar, but it seems that boat sank...
 
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Here is my understanding after looking at Wikipedia pages on algebraic elements and algebraic extensions: if [imath]\alpha[/imath] is a root of irreducible polynomial [imath]P[/imath] of degree [imath]n[/imath] then [imath]\mathbb Q[\alpha][/imath] is an extension of [imath]\mathbb Q[/imath] of degree [imath]n[/imath]. But it is not necessarily true that [imath]\mathbb Q[\alpha_1, \alpha_2, ..., \alpha_n][/imath] (where [imath]\alpha_k[/imath] are roots of [imath]P[/imath]) has degree [imath]n[/imath]. In fact [imath]\mathbb Q[\alpha_i][/imath] and [imath]\mathbb Q[\alpha_j][/imath] can be different fields when [imath]i\neq j[/imath], although they are probably isomorphic.

In your case I think [imath]\mathbb Q[2^{1/7}][/imath] is an extension of degree 7 even though it does not contain, say, [imath]2^{1/7}e^{2\pi i / 7}[/imath].

Does this make sense?

Good luck with the exam.
 
blamocur said:
Here is my understanding after looking at Wikipedia pages on algebraic elements and algebraic extensions: if [imath]\alpha[/imath] is a root of irreducible polynomial [imath]P[/imath] of degree [imath]n[/imath] then [imath]\mathbb Q[\alpha][/imath] is an extension of [imath]\mathbb Q[/imath] of degree [imath]n[/imath]. But it is not necessarily true that [imath]\mathbb Q[\alpha_1, \alpha_2, ..., \alpha_n][/imath] (where [imath]\alpha_k[/imath] are roots of [imath]P[/imath]) has degree [imath]n[/imath]. In fact [imath]\mathbb Q[\alpha_i][/imath] and [imath]\mathbb Q[\alpha_j][/imath] can be different fields when [imath]i\neq j[/imath], although they are probably isomorphic.

In your case I think [imath]\mathbb Q[2^{1/7}][/imath] is an extension of degree 7 even though it does not contain, say, [imath]2^{1/7}e^{2\pi i / 7}[/imath].

Does this make sense?

Good luck with the exam.
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Thank you... I guess that was it. I was thinking that containing a root means it had to contain all of them (which is precisely the definition of a normal extension). I made the mistake of trying to find a normal extension, by supposing all extensions are normal...
Looks like I'm back to where I started.
 
MathNugget said:
So [imath] \mathbb{Q} [/imath] and the polynomials in [imath]\mathbb{Q}[x][/imath] with degree < 7? I don't think I understood...

Or you're saying the splitting field of that polynomial? Because I was trying something similar, but it seems that boat sank...
Click to expand...

Yes. [imath] \mathbb{Q}[x]/(x^7+2) [/imath] is the quotient of a polynomial ring, an integral domain, factored by a prime ideal, since the polynomial is irreducible over [imath] \mathbb{Q} [/imath]. This makes it a field. As it contains [imath] \mathbb{Q} [/imath] it is a field extension of the rational numbers of dimension seven. The polynomial has only simple roots, and the automorphisms between the roots are generated by the rotations of angle [imath] 2\pi / 7 [/imath] of the circle of radius two, a cyclic group of order seven.
 
fresh_42 said:
Yes. [imath] \mathbb{Q}[x]/(x^7+2) [/imath] is the quotient of a polynomial ring, an integral domain, factored by a prime ideal, since the polynomial is irreducible over [imath] \mathbb{Q} [/imath]. This makes it a field. As it contains [imath] \mathbb{Q} [/imath] it is a field extension of the rational numbers of dimension seven. The polynomial has only simple roots, and the automorphisms between the roots are generated by the rotations of angle [imath] 2\pi / 7 [/imath] of the circle of radius two, a cyclic group of order seven.
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I am confused as to how it's proven that the dimension of the field extension is 7... I am used to fields of the form [imath]K[\alpha][/imath], looking for the monic minimal polynomial etc etc, which doesn't work here. What is the strategy in this case?
 
Here is my summary (to be taken with a grain of salt) :
Field [imath]\mathbb K[/imath] in your post is called splitting field of [imath]f(x) = x^7-2[/imath]. It is normal, like all splitting fields, but seems to have degree larger than 7. I don't know what the actual degree is, but one section of Wikipedia page on splitting fields mentions that splitting fields can have degrees no larger than [imath]n![/imath] (which is 5040 in your case :)) BTW/PS: did you look at [imath]f(x) = x^7-1[/imath] ?
 
I'm sorry, I guess I have (once again) got confused between [imath] \mathbb{Z}_n [/imath] and [imath] \mathbb{Z}_n^* [/imath]. Sorry.

A normal extension of [imath] \mathbb{Q} [/imath] of order seven would have the Galois group [imath] \mathbb{Z}_7 [/imath] because there is no other group of order 7. It would furthermore have to be a cyclotomic field extension, say with the polynomial [imath] x^n -1.[/imath] Finally, there is no such number [imath] n [/imath] so that [imath] \varphi (n)=7. [/imath]

I thought, I could cheat with [imath] x^7+2=0 [/imath] but this doesn't work, it's six-dimensional.
 
blamocur said:
Here is my summary (to be taken with a grain of salt) :
Field [imath]\mathbb K[/imath] in your post is called splitting field of [imath]f(x) = x^7-2[/imath]. It is normal, like all splitting fields, but seems to have degree larger than 7. I don't know what the actual degree is, but one section of Wikipedia page on splitting fields mentions that splitting fields can have degrees no larger than [imath]n![/imath] (which is 5040 in your case :)) BTW/PS: did you look at [imath]f(x) = x^7-1[/imath] ?
Click to expand...
I thought [imath]f(x) = x^7-1=(x-1)(x^6+x^5+...+x+1)[/imath] so it supposedly would have degree 6 at most...but right now, I am not sure anymore.
 
fresh_42 said:
I'm sorry, I guess I have (once again) got confused between [imath] \mathbb{Z}_n [/imath] and [imath] \mathbb{Z}_n^* [/imath]. Sorry.

A normal extension of [imath] \mathbb{Q} [/imath] of order seven would have the Galois group [imath] \mathbb{Z}_7 [/imath] because there is no other group of order 7. It would furthermore have to be a cyclotomic field extension, say with the polynomial [imath] x^n -1.[/imath] Finally, there is no such number [imath] n [/imath] so that [imath] \varphi (n)=7. [/imath]

I thought, I could cheat with [imath] x^7+2=0 [/imath] but this doesn't work, it's six-dimensional.
Click to expand...
If this is bad, the exam also had the same question but with order 14 :D . Pure madness, I guess I'll have to find an easier exercise
 
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