1

We know by standard theorems that $Gal(Q(\zeta_n)/Q)\approx (Z_n)^*$. Thus the order of the extension is $\phi(n)$.

Now if $n=p$ then we know that $(Z_p)^*\approx Z_{p-1}$ and that given a subgroup of $Gal(Q(\zeta_p)/Q)$ H, the fixed field of it is $$Q(\Sigma_{\sigma \in H}\sigma(\zeta_p))$$ which is easily computable. Are there similar "theorems" for $n=p^n$ or $n=2p^n$ when $(Z_n)^*$ is cyclic?

Furthermore, I was wondering, when can we determine for which $a,b$ do these occur: $Q(\zeta_n^a)=Q(\zeta_n^b)$ or $Q(\zeta_n^a)\subseteq Q(\zeta_n^b)$.

It is clear that when $gcd(a,n)=1$ then $Q(\zeta_n^a)=Q(\zeta_n)$ that is because then $\zeta_n^a$ is the generator of the group $<\zeta_n>$. Can we extend this reasoning to other cases? For example, $Q(\zeta_n^a)=Q(\zeta_n^b)$ $\iff$ they generate the same subgroup in $<\zeta_n>$? Also $Q(\zeta_n^a)\subseteq Q(\zeta_n^b)$ $\iff$ $<\zeta_n^a>\subseteq \zeta_n^b>$?

For my last question, consider ,$Q(\zeta_n)$ the degree of the extension is $\phi(n)$. Now let us say we find all the automorphisms and so the Galois group elements. Any element is of the form $Q(\zeta_n)=\{a_1+a_2\zeta_n+...+a_{\phi(n)-1}\zeta_n^{\phi(n)-1}\}$. Thus everything is determined by where we map $\zeta_n$. Now when we apply some automorphism on an arbitary element $a_1+a_2\zeta_n+...+a_{\phi(n)}\zeta_n^{\phi(n)}$ then the powers of $\zeta_n$ can exceed $\phi(n)$, how would we now find the decomposition of $\zeta_n^a$ in terms of the basis (we can assume $a<n$ as we can always reduce mod $n$.)? So that we can see what becomes fixed and what does not?

Sorfosh
  • 3,680
  • Note in general that if $\gcd(a,n) = d$, then $Q(\zeta_n^a) = Q(\zeta_{n/d})$. – arkeet Dec 10 '19 at 00:57
  • You want a normal basis of $\Bbb{Q}(\zeta_{p^2})$ and $\Bbb{Q}(\zeta_{p^k})$ (it will give one for $\Bbb{Q}(\zeta_{n})$) – reuns Dec 10 '19 at 02:13
  • @reuns Can you elaborate on that? I don't know what a normal basis is. – Sorfosh Dec 10 '19 at 03:26
  • @arkeet that seems equivalent to the statement i made about the subgroups – Sorfosh Dec 10 '19 at 03:27
  • The answer to your last question is that you make an euclidean division by the minimal polynomial - this is how you proceed for any primitive extension, not just a cyclotomic one. And by the way, the sum needn't go up to degree $\phi(n)$, degree $\phi(n)-1$ suffices. – Ewan Delanoy Dec 12 '19 at 15:00
  • The equivalence you propose is also true. Indeed, by arkeet's comment $Q(\zeta_n^a) \subseteq Q(\zeta_n^b)$ iff $Q(\zeta_A) \subseteq Q(\zeta_B)$ (where $A=\frac{n}{\gcd(n,a)},B=\frac{n}{\gcd(n,b)}$), iff $Q(\zeta_A) \cap Q(\zeta_B) = Q(\zeta_A)$ , iff $Q(\zeta_{lcm(A,B)}) =Q(\zeta_A)$ (see this question ), iff either $A=lcm(A,B)$ or $A$ is odd and $lcm(A,B)=2A$ (see this question). – Ewan Delanoy Dec 12 '19 at 15:35
  • Hi, I study cyclotomic extensions of number fields, and I've never seen that the fixed field is $$Q(\Sigma_{\sigma \in H}\sigma(\zeta_p))$$ Can you give a reference for this? I could probably read the proof and see how it would generalize to other fields (and maybe even publish it in the arXiv :) ) – Alex Dec 18 '19 at 09:00
  • @Alex I don't know a reference of this, and I cannot find a proof now, but I believe this is correct for prime $p$ while incorrect for general prime-power $p$. For the latter, counterexample occurs already at $p = 9$. For the former, I verified it for all $p < 150$. I'm quite surprised that you study cyclotomic extensions but have never seen such a claim, either as a known result or as a conjecture (I tend to believe that it's a known result, though). – WhatsUp Dec 19 '19 at 00:23
  • @Alex It is dummet and foote. – Sorfosh Dec 19 '19 at 01:53
  • 1
    @Sorfosh I see, the proof is on page 597-598 in the third edition (available online https://www.academia.edu/13527708/Abstract_algebra_Dummit_and_Foote). Actually not that difficult. Thanks. – WhatsUp Dec 19 '19 at 02:49
  • That's why I asked to read it. I think I started my studies after D&F, and that's the reason I missed it. But in general, things that I never had to utilize I naturally wouldn't know, even if the proof is in D&F. – Alex Dec 19 '19 at 08:08

1 Answers1

1

For your third question, you may simply divide $X^a$ by the cyclotomic polynomial $\Phi_n(X)$.

Example: $n = 9$.

We have $\Phi_9(X) = x^6 + x^3 + 1$. Therefore: $$\zeta^6 = -\zeta^3 - 1;$$ $$\zeta^7 = -\zeta^4 - \zeta;$$ $$\zeta^8 = -\zeta^5 - \zeta^2.$$ In general, for $a \geq \phi(n)$, the polynomials expressing $\zeta^a$ in terms of $\zeta$ are quite complicated.

When $n$ is squarefree, it's much easier to work with the normal basis $(\zeta^r)_{r\in (\Bbb Z/n\Bbb Z)^\times}$.

In terms of the normal basis, it's quite easy to tell when an element is fixed by an automorphism.

Let's say the automorphism is $\sigma_k:\zeta\mapsto \zeta^k$, with $k\in(\Bbb Z/n\Bbb Z)^\times$. The element $k$ generates a subgroup $\langle k\rangle$ of $(\Bbb Z/n\Bbb Z)^\times$.

Now if $x = \sum_{r \in(\Bbb Z/n\Bbb Z)^\times}a_r\zeta^r$ is an element of $\Bbb Q(\zeta)$, then the element $\sigma_k(x)$ is simply $\sum\limits_{r \in(\Bbb Z/n\Bbb Z)^\times}a_r\zeta^{kr} = \sum\limits_{r \in(\Bbb Z/n\Bbb Z)^\times}a_{rk^{-1}}\zeta^r$.

Since the elements $(\zeta^r)_{r\in(\Bbb Z/n\Bbb Z)^\times}$ form a basis, we have $\sigma_k(x) = x$ if and only if $a_r = a_{rk^{-1}}$ for all $r$.

This is again equivalent to saying that there is a function $a:(\Bbb Z/n\Bbb Z)^\times/\langle k\rangle \rightarrow \Bbb Q$ such that $a_r = a(r\mod \langle k\rangle)$.

In other words, an element of $\Bbb Q(\zeta)$ is fixed by $\sigma_k$ if and only if it is a $\Bbb Q$-linear combination of elements of the form $\zeta^r + \zeta^{rk} + \zeta^{rk^2} + \dotsc + \zeta^{rk^{m - 1}}$, where $r \in (\Bbb Z/n\Bbb Z)^\times$ and $m$ is the order of $k$ in $(\Bbb Z/n\Bbb Z)^\times$.

WhatsUp
  • 22,614