Finding Gal (K=Q[ζ15]/QK=\mathbb{Q}[\zeta_{15}] / \mathbb{Q} (up to an isomorphism)

[MathNugget]

$K=\mathbb{Q}[\zeta_{15}]$
Given $\zeta_{15}$ is the fundamental root of the polynomial $f(x)=x^{15}-1$, that is, $\zeta_{15}=cos(\frac{360}{15})+isin(\frac{360}{15})$.

The Galois groups is made out of automorphisms from K to K, that fix the points of $\mathbb{Q}$. I observe 1 is the only rational root, and $\phi(1)=1$. Furthermore, I notice $\zeta_{15}^5, \zeta_{15}^{10}$ have order 3, so the authomorphism can only fix them or switch these 2.
Then I notice something similar for $\zeta_{15}^{3}, \zeta_{15}^{6}, \zeta_{15}^{9}, \zeta_{15})^{12}$, they have order 5, so they should switch around too. I also realize that these 2 'chains' are independent, as they only would only 'connect' on multiples of 15, which are 1 (hope this makes sense).

But then, how would $\zeta_1$ act? Since $\phi(\zeta_1^3)=\phi(\zeta_1)^3$, if I, for example, permute $\zeta_{15}^{3}$ and $\zeta_{15}^{6}$, I'd have to do the same for $\zeta_1$ and $\zeta_1^2$, and then check for everything to work...

Any tips? Worth mentioning that (if I could give a complete proof), I could just state to which group this is isomorphic, for example $\mathbb{Z}_3 \times \mathbb{Z}_5$ (I like to think it is this, because of the order 3 and 5 subgroups, but who knows?)

Also I'd like to thank everyone helping on this forum, you're all great.

 

[blamocur]

If you look at where $\zeta_{15}$ can be mapped by an automorphism you'll see that it is not $\mathbb Z_3 \times \mathbb Z_5$. Since every root of $f(x)$ must be mapped to its other root every automorphism must map $\zeta_{15}$ to $\zeta^m_{15}$ for some $m\neq 0$. What is the set of all possible $m$'s ?

 

[MathNugget]

If you look at where $\zeta_{15}$ can be mapped by an automorphism you'll see that it is not $\mathbb Z_3 \times \mathbb Z_5$. Since every root of $f(x)$ must be mapped to its other root every automorphism must map $\zeta_{15}$ to $\zeta^m_{15}$ for some $m\neq 0$. What is the set of all possible $m$'s ?

Galois group - Wikipedia

en.wikipedia.org

apparently all m's relatively prime to n. So, Euler's totient function, and I do $\phi(15)=15 \times \frac{4}{5} \times \frac{2}{3}=8$. There's 8 possible m's...m=1, 2, 4, 7, 8, 11, 13, 14.

So it's all about not sending $\zeta_{15}$$$$$ to a member of a smaller subgroup... or the relation between all the zetas breaks, and it's no longer an automorphism...

How do we find the group it is isomorphic to, though?

 

[MathNugget]

I looked through some courses, apparently it is isomorphic to the (sub)group of invertibles of $\mathbb{Z}_{15}$, which is made out of those 8 numbers I mentioned earlier...1, 2, 4, 7, 8, 11, 13, 14

 

[blamocur]

I looked through some courses, apparently it is isomorphic to the (sub)group of invertibles of $\mathbb{Z}_{15}$, which is made out of those 8 numbers I mentioned earlier...1, 2, 4, 7, 8, 11, 13, 14

True, but you want to be careful with the terms used: invertibles of the $\mathbb Z_{15}$ ring are not a subgroup of the $\mathbb Z_{15}$ (additive) group.

 

[blamocur]

Galois group - Wikipedia

en.wikipedia.org

apparently all m's relatively prime to n. So, Euler's totient function, and I do $\phi(15)=15 \times \frac{4}{5} \times \frac{2}{3}=8$. There's 8 possible m's...m=1, 2, 4, 7, 8, 11, 13, 14.

So it's all about not sending $\zeta_{15}$$$$$ to a member of a smaller subgroup... or the relation between all the zetas breaks, and it's no longer an automorphism...

How do we find the group it is isomorphic to, though?

What is the product of automorphisms corresponding to $m_1$ and $m_2$ from the set you found?

 

[MathNugget]

What is the product of automorphisms corresponding to $m_1$ and $m_2$ from the set you found?

Not a clue how to find that. Just got back from exam, I think I'll be looking at these polynomials for a loong time...

so $\phi_1(\zeta_9)=\zeta_9, \phi_2(\zeta_9)=\zeta_9^2$. If I multiply them...they're no longer an automorphism, so the group's wrong?

 

[blamocur]

Not a clue how to find that. Just got back from exam, I think I'll be looking at these polynomials for a loong time...

so $\phi_1(\zeta_9)=\zeta_9, \phi_2(\zeta_9)=\zeta_9^2$. If I multiply them...they're no longer an automorphism, so the group's wrong?

Using plain $\zeta$ for $\zeta_{15}$ here:

If $A_m(\zeta) = \zeta^m$ and $A_k(\zeta) = \zeta^k$ what is $A_m (A_k (\zeta))$ ? E.g. $A_2 (A_4 (\zeta))$? [imath]A_4 \(A_4 (\zeta))[/imath]?

 

[blamocur]

so ϕ1(ζ9)=ζ9,ϕ2(ζ9)=ζ92\phi_1(\zeta_9)=\zeta_9, \phi_2(\zeta_9)=\zeta_9^2ϕ1(ζ9)=ζ9,ϕ2(ζ9)=ζ92. If I multiply them...they're no longer an automorphism, so the group's wrong?

Can't you see that $\phi_1$ is the identity element of the group?

 

[blamocur]

Not a clue how to find that. Just got back from exam, I think I'll be looking at these polynomials for a loong time...

I wouldn't be so pessimistic. Just because finding normal extensions degree 7 and 3 of Q turned out to be so difficult does not mean the time was wasted. Quite opposite, I'd expect much better understanding of the material on your part after seeing how the problems might be real hard without a bunch of useful theorems and constructs. I've seen you providing useful insights which I've often missed, which makes me confident that you can learn this stuff eventually.

Here are some more wisdom morsels: Getting stuck with a difficult problem is normal and an important part of learning. Getting stuck for too long, on the other hand, can be counter-productive. Figuring out when it becomes too long comes with experience.

 

[MathNugget]

What is the product of automorphisms corresponding to $m_1$ and $m_2$ from the set you found?

Noo, when I read product I forgot everything I knew, and I did basic multiplication... Forgot automorphisms are a group with composition, and it's easier to see these things as permutations (although I don't know how to write them in latex).

so $\phi_1(\phi_2(\zeta))=\zeta^2$, or $1 \rightarrow 2 \rightarrow 2$, if we see them as permutations...

Using plain $\zeta$ for $\zeta_{15}$ here:

If $A_m(\zeta) = \zeta^m$ and $A_k(\zeta) = \zeta^k$ what is $A_m (A_k (\zeta))$ ? E.g. $A_2 (A_4 (\zeta))$? [imath]A_4 \(A_4 (\zeta))[/imath]?

$A_m(A_k(\zeta))=A_m(\zeta^k)=\zeta^{mk}$, (mk can be reduced mod 15)...

Here are some more wisdom morsels: Getting stuck with a difficult problem is normal and an important part of learning. Getting stuck for too long, on the other hand, can be counter-productive. Figuring out when it becomes too long comes with experience.

True, but I am flying from 1 exam to another, so there's little time left. Sadly it is more about finding an easy way to finish than learning right now.

 

[blamocur]

mk can be reduced mod 15

"... must be reduced..."?
One way to describe this group is the multiplicative group of the $\mathbb Z_{15}$ ring often written as $\mathbb Z_{15}^*$

Sadly it is more about finding an easy way to finish than learning right now.

Sadly, I can only help you with learning. But I do wish you good luck with you exams!

 

[blamocur]

if we see them as permutations...

Any finite group is a subgroup of a permutation group.

 

[MathNugget]

"... must be reduced..."?

Is that so? I think you can still write $\zeta^{104}$, just like writing a $720\degree$ angle. It won't be fundamentally wrong (like in the case of writing $x=16$ when $x \in \mathbb{Z}_{15}$).

Sadly, I can only help you with learning. But I do wish you good luck with you exams!

Thank you. It's alright, I didn't mean I want to cheat just that I am focusing on practical exercises, rather than understanding concepts through curious research. So there is still some learning going on, it's just a heavily directed minimal learning. Also the time pressure means for the next 2 weeks I'll be hopping from subject to subject. If things 'go south', I'll be tormenting this forum all the way to September, every day.

 

[blamocur]

Is that so? I think you can still write ζ104\zeta^{104}ζ104, just like writing a 720°720\degree720° angle. It won't be fundamentally wrong

It won't be fundamentally wrong but pretty misleading. Using different names/expressions for the same element ($\zeta^{104}\equiv \zeta^{14}$) makes it more difficult to figure out the structure of the group.

 

[MathNugget]

You're right.

True, but you want to be careful with the terms used: invertibles of the $\mathbb Z_{15}$ ring are not a subgroup of the $\mathbb Z_{15}$ (additive) group.

Regarding this, I am still not sure how finding the isomorphism works.
I know the Galois group has 8 elements:

apparently all m's relatively prime to n. So, Euler's totient function, and I do $\phi(15)=15 \times \frac{4}{5} \times \frac{2}{3}=8$. There's 8 possible m's...m=1, 2, 4, 7, 8, 11, 13, 14.

Would it be $(\mathbb{Z}_8, +)$? If yes, can I (also) find a grup with multiplication?

I remember finding the order of the elements (or whatever it is called, the minimal number of times to form a loop), was a way to figure out which of the groups with n elements is isomorphic to another group with n elements.
Here, for example, $\phi_2(\zeta)=\zeta^2$ I think has order 8...$\zeta^{2 \times 8}=\zeta$? I think I didn't remember this correctly, as looking for $7 \times ? \equiv 1 mod 15$ gives 13, but I think the order of elements needed to divide the order of the group...

 

[blamocur]

Would it be (Z8,+)(\mathbb{Z}_8, +)(Z8,+)? If yes, can I (also) find a grup with multiplication?

No, it isn't isomorphic to $\mathbb Z_8$.

ϕ2(ζ)=ζ2 I think has order 8.

No, the order of $\phi_2$ is not 8. It is true that $\phi_2^8 = 1$, but 8 is not the minimal number $m$ for which $\phi_2^m = 1$ -- can you figure out the minimum? How about $\phi_4$. BTW, if the group were isomorphic to $\mathbb Z_{8}$ it would have to have an element with order 8 -- can you find such element?

 

[MathNugget]

No, it isn't isomorphic to $\mathbb Z_8$.

No, the order of $\phi_2$ is not 8. It is true that $\phi_2^8 = 1$, but 8 is not the minimal number $m$ for which $\phi_2^m = 1$ -- can you figure out the minimum? How about $\phi_4$. BTW, if the group were isomorphic to $\mathbb Z_{8}$ it would have to have an element with order 8 -- can you find such element?

Alright, looks like I missed the shot...m was 4? $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$

I looked through some courses, apparently it is isomorphic to the (sub)group of invertibles of $\mathbb{Z}_{15}$, which is made out of those 8 numbers I mentioned earlier...1, 2, 4, 7, 8, 11, 13, 14

$\phi_x$ | permutations:
4 | $1 \rightarrow 4\rightarrow 1$
7 | $1 \rightarrow 7 \rightarrow 4 \rightarrow 13 \rightarrow 1$
8 | $1 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1$
11 | $1 \rightarrow 11 \rightarrow 1$
13 | $1 \rightarrow 13 \rightarrow 4 \rightarrow 7 \rightarrow 1$
14 | $1 \rightarrow 14 \rightarrow 1$

Consider this attempt to make a table. Would this be the answer? all elements seem to have order 2, or 4...

 

[MathNugget]

Would this be $\mathbb{Z}_2 \times \mathbb{Z}_4, +$?

 

[blamocur]

Good job! I agree that it is $\mathbb Z_2 \times \mathbb Z_4$. If you get bored you can try figuring out which elements from $\mathbb Z_{15}^*$ map to (1,0) and (0,1) in $\mathbb Z_2 \times \mathbb Z_4$.