Finding a normal field extension of degree 7

[MathNugget]

I am back with some insight:
I think the problem isn't manageable with the idea $\mathbb{K}=\mathbb{k}[\alpha, \beta...]$
Here's a property that I think is correct: $[\mathbb{\mathbb{k}[\alpha, \beta]}:\mathbb{k}]=[\mathbb{\mathbb{k}[\alpha]}:\mathbb{k}][\mathbb{\mathbb{k}[\alpha, \beta]}:\mathbb{\mathbb{k}[\alpha]}]$

Given all those degrees are integers, and 7 is a prime, there must be a (single) thing adjoined to k...

So, I think we can rewrite the issue as this:

Is there a polynomial $f(x)=(x-\alpha)(x-(a\alpha+b))(x-(c\alpha+d))(x-(e\alpha+f))...$ (there should be 7 brackets, and a, b, c, d, ... are rational numbers)
Thatis irreducible in $\mathbb{Q}[X]$, and more importantly, is a part of $\mathbb{Q}[X]$...

I'll attempt it if we replace 7 with 3 (sounds more reasonable, and might give some useful observations...

 

[MathNugget]

What I got so far: at least 1 root is real, so then all must be real. From 1st Viete relation, in the case of trying to find a degree 3 normal extension with the polynomial above, I get c=-1-a, but I cannot seem to write d using a and b... the other 2 Viete relations aren't helping at all

 

[blamocur]

I am back with some insight:
I think the problem isn't manageable with the idea $\mathbb{K}=\mathbb{k}[\alpha, \beta...]$
Here's a property that I think is correct: $[\mathbb{\mathbb{k}[\alpha, \beta]}:\mathbb{k}]=[\mathbb{\mathbb{k}[\alpha]}:\mathbb{k}][\mathbb{\mathbb{k}[\alpha, \beta]}:\mathbb{\mathbb{k}[\alpha]}]$

Given all those degrees are integers, and 7 is a prime, there must be a (single) thing adjoined to k...

So, I think we can rewrite the issue as this:

Is there a polynomial $f(x)=(x-\alpha)(x-(a\alpha+b))(x-(c\alpha+d))(x-(e\alpha+f))...$ (there should be 7 brackets, and a, b, c, d, ... are rational numbers)
Thatis irreducible in $\mathbb{Q}[X]$, and more importantly, is a part of $\mathbb{Q}[X]$...

I'll attempt it if we replace 7 with 3 (sounds more reasonable, and might give some useful observations...

$\alpha, a\alpha+b, c\alpha+d, ...$ are linearly dependent, which means that the field's extension degree is 2.

 

[MathNugget]

$\alpha, a\alpha+b, c\alpha+d, ...$ are linearly dependent, which means that the field's extension degree is 2.

Is that so? I thought the degree would be the degree of the minimal polynomial in $\mathbb{Q}[X]$ with the root $\alpha$, which isn't necessarily 2...

Them being linearly dependent just means $\mathbb{Q}$ [one of the roots] = $\mathbb{Q}$ [any number of those linearly dependent roots], I suppose

 

[blamocur]

Is that so? I thought the degree would be the degree of the minimal polynomial in $\mathbb{Q}[X]$ with the root $\alpha$, which isn't necessarily 2...

Them being linearly dependent just means $\mathbb{Q}$ [one of the roots] = $\mathbb{Q}$ [any number of those linearly dependent roots], I suppose

You are right and I am wrong: typically the basis consists of $1, \alpha, \alpha^2, ..., \alpha^{n-1}$ for extensions of degree 'n'. But I don't know what your approach with linear forms of $\alpha$ can gain.

What about $\mathbb Q[2^{1/7}]$? Isn't its degree equal to 7 with a basis $\{2^{k/7}\}$ for $0\leq k \leq 6$ ?

 

[MathNugget]

You are right and I am wrong: typically the basis consists of $1, \alpha, \alpha^2, ..., \alpha^{n-1}$ for extensions of degree 'n'. But I don't know what your approach with linear forms of $\alpha$ can gain.

What about $\mathbb Q[2^{1/7}]$? Isn't its degree equal to 7 with a basis $\{2^{k/7}\}$ for $0\leq k \leq 6$ ?

From what I remember from yesterday, I thought we said that $\mathbb{Q}[2^{1/7}]$ is not a normal extension.

Normal extension - Wikipedia

en.wikipedia.org

The polynomial $g(x)=x^7-2$ is in $\mathbb{Q}[X]$, has a root in $\mathbb{Q}[2^{1/7}]$, but not the other roots.

The smallest normal extension we can get by using the 7th root of 2 (or whatever is called) would be (I suppose): $\mathbb{Q}[2^{1/7}, \zeta_7]$, with $\zeta_7$ one of the roots (not 1) of $x^7=1$. And the degree of that (the extension) would be 7x6=42.

 

[blamocur]

From what I remember from yesterday, I thought we said that $\mathbb{Q}[2^{1/7}]$ is not a normal extension.

Normal extension - Wikipedia

en.wikipedia.org

The polynomial $g(x)=x^7-2$ is in $\mathbb{Q}[X]$, has a root in $\mathbb{Q}[2^{1/7}]$, but not the other roots.

The smallest normal extension we can get by using the 7th root of 2 (or whatever is called) would be (I suppose): $\mathbb{Q}[2^{1/7}, \zeta_7]$, with $\zeta_7$ one of the roots (not 1) of $x^7=1$. And the degree of that (the extension) would be 7x6=42.

True. I did not claim the extension is normal, just 7th degree.

 

[MathNugget]

True. I did not claim the extension is normal, just 7th degree.

You are very much correct, then. I apologize .

You are right and I am wrong: typically the basis consists of $1, \alpha, \alpha^2, ..., \alpha^{n-1}$ for extensions of degree 'n'. But I don't know what your approach with linear forms of $\alpha$ can gain.

Well, I just realized something in my long argument was wrong, so I deleted it. I was expanding on the (simpler) case of a normal field extension of degree 3.

I was trying to prove such a construction impossible, by playing around with the 2nd degree polynomial I should get with 2 of those roots (without loss of generality, I can suppose they're $\alpha$ and $a\alpha+b$). Given these 2 roots are not rational, they'd have to be conjugates , in the sense that if one is $x+y\sqrt{z}$, with x, y, z real, z-square free, then the other must be $x-y\sqrt{z}$. I was getting (with faulty logic ) to the dream of proving a=-1 (with the linear dependence argument, because the coefficients would be rational, and the sqrt irrational), and then the 3rd root would fall apart and be rational, giving the final contradiction...

Maybe I'll find more inspiration tomorrow, although opening a different problem and trying to solve it starts seeming like a better idea to pass that exam. After checking someone else's notes, I believe I misremembered things, and the extension asked for at the exam was not actually supposed to be normal . Maybe I just invented a painful exercise.

 

[MathNugget]

I think I can still further restrict that weird 2nd degree polynomial to get some results out of it, maybe I can prove it to have rational coefficients...but at this rate, I'm better off pretending to be half blind at the exam, and writing an 'abnormal' extension even if I am asked for a normal one... Let's start on a clean sheet with a (simpler, hopefully) problem... taking 10 mins break, then opening a thread about radical extensions

 

[blamocur]

I was expanding on the (simpler) case of a normal field extension of degree 3.

I like this idea. Looking forward to seeing some results, positive or negative.

 

[MathNugget]

I don't know, it feels like I am just going in circles, I keep re-evaluating what I have and what I want to prove, hoping to find an easy contradiction that would show that finding a degree 3 normal extension is impossible, and maybe infer from that information that the same happens for degree 7 (and maybe all primes)...
I guess I'll give this problem a break.

so far:
I am looking for a polynomial $f(x)=x^3+sx^2+tx+v \in \mathbb{Q}[X]$, with roots $\alpha, a\alpha+b, c\alpha+d \notin \mathbb{Q}$.
I am able to prove:
$\alpha^2, \alpha^3 \notin \mathbb{Q}$ (same for other 2 roots)
c=-1-a

I tried going with Viete's 2nd relation (sum of products of 2 roots), and with $f(\alpha)-f(a\alpha+b)=0$, but I get stuck. I think it would be miraculous if I could prove $x\alpha^2+y\alpha \notin \mathbb{Q}, \forall x, y \in \mathbb{Q}$, but that is probably going to remain a dream...
Addition: $x\alpha^2+y\alpha \notin \mathbb{Q}, \forall x, y \in \mathbb{Q}$ is equivalent to $x\alpha^2+\alpha \notin \mathbb{Q}, \forall x \in \mathbb{Q}$

 

[blamocur]

BTW, there is a Wikipedia page on cubic fields. I tried to read it, but it is too much for me.

 

[MathNugget]

BTW, there is a Wikipedia page on cubic fields. I tried to read it, but it is too much for me.

Well, that's horrifying. From the looks of it, $x^3+x^2-2x-1$ is a polynomial like the one I was looking for, so there is a normal field extension of degree 3...

So it is highly likely there is one for degree 7 too

Jokes aside, thank you very much. I guess we can give up on this problem, it's way more difficult (since it likely has an answer).

 

[blamocur]

You got me interested enough to ask Sympy for the roots of your polynomial and post them below. They all look like complex numbers but evaluate to real ones (1.24697960371747, -1.80193773580484, -0.445041867912629). They also look hairy enough for me to give up on this topic
$$\theta_1 = - \frac{1}{3} + \frac{\sqrt[3]{2} \cdot 7^{\frac{2}{3}}}{3 \sqrt[3]{1 + 3 \sqrt{3} i}} + \frac{\sqrt[3]{28} \sqrt[3]{1 + 3 \sqrt{3} i}}{6}$$$$\theta_2 = \frac{7^{\frac{2}{3}} \left(- \frac{2 \sqrt[3]{2}}{3} + \frac{\left(-4 + 2^{\frac{2}{3}} \left(-1 + \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}\right) \left(1 - \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}}{84}\right)}{\left(1 - \sqrt{3} i\right) \sqrt[3]{1 + 3 \sqrt{3} i}}$$$$\theta_3 = - \frac{7^{\frac{2}{3}} \left(\frac{2 \sqrt[3]{2}}{3} + \frac{\left(1 + \sqrt{3} i\right) \left(4 - 2^{\frac{2}{3}} \left(-1 - \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}\right) \sqrt[3]{7 + 21 \sqrt{3} i}}{84}\right)}{\left(1 + \sqrt{3} i\right) \sqrt[3]{1 + 3 \sqrt{3} i}}$$

 

[MathNugget]

You got me interested enough to ask Sympy for the roots of your polynomial and post them below. They all look like complex numbers but evaluate to real ones (1.24697960371747, -1.80193773580484, -0.445041867912629). They also look hairy enough for me to give up on this topic
$$\theta_1 = - \frac{1}{3} + \frac{\sqrt[3]{2} \cdot 7^{\frac{2}{3}}}{3 \sqrt[3]{1 + 3 \sqrt{3} i}} + \frac{\sqrt[3]{28} \sqrt[3]{1 + 3 \sqrt{3} i}}{6}$$$$\theta_2 = \frac{7^{\frac{2}{3}} \left(- \frac{2 \sqrt[3]{2}}{3} + \frac{\left(-4 + 2^{\frac{2}{3}} \left(-1 + \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}\right) \left(1 - \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}}{84}\right)}{\left(1 - \sqrt{3} i\right) \sqrt[3]{1 + 3 \sqrt{3} i}}$$$$\theta_3 = - \frac{7^{\frac{2}{3}} \left(\frac{2 \sqrt[3]{2}}{3} + \frac{\left(1 + \sqrt{3} i\right) \left(4 - 2^{\frac{2}{3}} \left(-1 - \sqrt{3} i\right) \sqrt[3]{7 + 21 \sqrt{3} i}\right) \sqrt[3]{7 + 21 \sqrt{3} i}}{84}\right)}{\left(1 + \sqrt{3} i\right) \sqrt[3]{1 + 3 \sqrt{3} i}}$$

True, it looks horrible, and it doesn't look like we can find the $a, b \in \mathbb{Q}$ such that $a\theta_1+b=\theta_2$. I guess this needs a different approach... but I am quitting on it for now too.

 

[MathNugget]

Well, it looks like there is some result saying in $k$ a field of characteristic p, $x^p-b$ is irreducible, if $b \in k/k^p$. (meaning, I suppose, that the field extension between k and the splitting field of that polynomial isn't k). I got taught it as "abel's lemma", but can't seem to find it on the internet.

So there's an $a \notin k$, $(x-a)^p=x^p-b$ irreducible in k, and therefore a normal extension... Now the problem about this is proving that the extension has degree p and not less.

There's also a result here https://en.wikipedia.org/wiki/Normal_extension saying how we can get a normal extension of degree p(p-1). (it's exactly the style of extension we talked about here when trying to get a normal extension of degree p).

I suppose I have to read my courses a little more, the results are probably around there...although not very easy to find, grasp or use...

 

[blamocur]

I suppose I have to read my courses a little more, the results are probably around there...although not very easy to find, grasp or use...

Sounds like a good idea to me I've noticed in the past that banging my head, even in vain, at some problems helped me understand both the meaning of and the reason for some of the theorems.

Good luck, and feel free to share any insights you get.