Finding a normal field extension of degree 7

[MathNugget]

I am supposed to find an extension $k \subset K$ 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 $F_7$ (it should be the double line F, but I can't find the latex code for it). (the idea is that the field $F_7$ 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 $k \subset K$ is normal if every irreducible polynomial in $k[X]$ that has a root in K, has all its roots in K.
Let's suppose the degree of the field extension is finite. then $K=k(\alpha, \beta, ...)$, 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 $k=\mathbb{Q}$ and $K=\mathbb{Q}(\theta_1,...\theta_7)$, with $\theta_k=\sqrt[7]{2}(cos(\frac{360k}{7})+isin(\frac{360k}{7}))$, $k=0,...6$ (the roots of $x^7=2$).

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

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

(it should be the double line F, but I can't find the latex code for it)

Like $\mathbb F$ ? For that one can use \mathbb

 

[MathNugget]

Like $\mathbb F$ ? For that one can use \mathbb

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.

 

[blamocur]

I am quite rusty on field extensions, but here is my question regarding your approach:
Consider extension $\mathbb L$ of $\mathbb Q$ defined by the polynomial $x^7-1 = 0$. Am I right to believe that the degree of $\mathbb L$ over $\mathbb Q$ is 7? And that $\mathbb Q \subset \mathbb L \subset \mathbb K$ and $\mathbb L \neq \mathbb K$ ? Then this would mean that the degree of $\mathbb K$ must be greater than 7, wouldn't it?

 

[MathNugget]

I am quite rusty on field extensions, but here is my question regarding your approach:
Consider extension $\mathbb L$ of $\mathbb Q$ defined by the polynomial $x^7-1 = 0$. Am I right to believe that the degree of $\mathbb L$ over $\mathbb Q$ is 7? And that $\mathbb Q \subset \mathbb L \subset \mathbb K$ and $\mathbb L \neq \mathbb K$ ? Then this would mean that the degree of $\mathbb K$ must be greater than 7, wouldn't it?

I am not convinced that besides 1, the roots of $x^7-1=0$ are in the field $\mathbb{K}$ I tried to create there, but that seems so painful to prove...
say $cos(\frac{360}{7})+isin(\frac{360}{7})=\sqrt[7]{2}[a_1(cos(\frac{360}{7})+isin(\frac{360}{7})) +a_2(cos(\frac{360\cdot 2}{7})+isin(\frac{360\cdot 2}{7})) + ... +a_7(cos(\frac{360\cdot 7}{7})+isin(\frac{360\cdot 7}{7}))]+t$, with $a_1,...a_7, t$ 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 $\mathbb{Q}$

 

[blamocur]

$x_1 = 2^{1/7} \in \mathbb K \wedge x_2 = 2^{1/7} e^{i\pi/7} \in\mathbb K \Longrightarrow x_1/x_2 = e^{i\pi/7} \in \mathbb K$

 

[MathNugget]

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...

 

[blamocur]

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...

I am confused myself I am guessing that my post #4 might be wrong and $\mathbb L = \mathbb K$, but at this point I don't know how to get there.

 

[MathNugget]

I am confused myself I am guessing that my post #4 might be wrong and $\mathbb L = \mathbb K$, but at this point I don't know how to get there.

Sounds like we aren't passing the Galois theory exam on Tuesday .

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.

 

[fresh_42]

Degree seven means that the Galois group is $\mathbb{Z}_7,$ 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 $\mathbb{Q} \subseteq \mathbb{Q}[x]/(x^7+2)$?

 

[MathNugget]

Degree seven means that the Galois group is $\mathbb{Z}_7,$ 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 $\mathbb{Q} \subseteq \mathbb{Q}[x]/(x^7+2)$?

So $\mathbb{Q}$ and the polynomials in $\mathbb{Q}[x]$ 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...

 

[blamocur]

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

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

Does this make sense?

Good luck with the exam.

 

[MathNugget]

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

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

Does this make sense?

Good luck with the exam.

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.

 

[blamocur]

Looks like I'm back to where I started.

Looks like we both learned something useful

 

[fresh_42]

So $\mathbb{Q}$ and the polynomials in $\mathbb{Q}[x]$ 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...

Yes. $\mathbb{Q}[x]/(x^7+2)$ is the quotient of a polynomial ring, an integral domain, factored by a prime ideal, since the polynomial is irreducible over $\mathbb{Q}$. This makes it a field. As it contains $\mathbb{Q}$ 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 $2\pi / 7$ of the circle of radius two, a cyclic group of order seven.

 

[MathNugget]

Yes. $\mathbb{Q}[x]/(x^7+2)$ is the quotient of a polynomial ring, an integral domain, factored by a prime ideal, since the polynomial is irreducible over $\mathbb{Q}$. This makes it a field. As it contains $\mathbb{Q}$ 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 $2\pi / 7$ of the circle of radius two, a cyclic group of order seven.

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 $K[\alpha]$, looking for the monic minimal polynomial etc etc, which doesn't work here. What is the strategy in this case?

 

[blamocur]

Here is my summary (to be taken with a grain of salt) :
Field $\mathbb K$ in your post is called splitting field of $f(x) = x^7-2$. 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 $n!$ (which is 5040 in your case ) BTW/PS: did you look at $f(x) = x^7-1$ ?

 

[fresh_42]

I'm sorry, I guess I have (once again) got confused between $\mathbb{Z}_n$ and $\mathbb{Z}_n^*$. Sorry.

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

I thought, I could cheat with $x^7+2=0$ but this doesn't work, it's six-dimensional.

 

[MathNugget]

Here is my summary (to be taken with a grain of salt) :
Field $\mathbb K$ in your post is called splitting field of $f(x) = x^7-2$. 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 $n!$ (which is 5040 in your case ) BTW/PS: did you look at $f(x) = x^7-1$ ?

I thought $f(x) = x^7-1=(x-1)(x^6+x^5+...+x+1)$ so it supposedly would have degree 6 at most...but right now, I am not sure anymore.

 

[MathNugget]

I'm sorry, I guess I have (once again) got confused between $\mathbb{Z}_n$ and $\mathbb{Z}_n^*$. Sorry.

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

I thought, I could cheat with $x^7+2=0$ but this doesn't work, it's six-dimensional.

If this is bad, the exam also had the same question but with order 14 . Pure madness, I guess I'll have to find an easier exercise