5
$\begingroup$

This question is an extension to the question in math.stackexchange.com/questions/759230/subfield-of-the-galois-group-of-x5-1

It seems the discussion in that topic is dead and I still have a major question.

For the $\Phi_5(x)$ the complex roots are $\{\zeta, \zeta^2, \zeta^3, \zeta^4 \}$ for $\zeta:=e^{\frac{2\pi i}5}$. In his response Don Antonio defined $\;\omega\in Gal(\Bbb Q(\zeta)/\Bbb Q)\;$ to be complex conjugation.

Don went on to explain:

for$\;z\in\Bbb C\;\;,\;\;\overline z=z^{-1}\iff |z|=1\;$ , and thus:

$$\omega(\zeta+\zeta^{-1})=\omega(\zeta)+\omega(\zeta)^{-1}=\zeta^{-1}+\zeta\implies \zeta+\zeta^{-1}\in\Bbb R$$

end of quote

Now Let $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{id, \rho, \rho^2, \rho^3\}$ where $\rho(\zeta) = \zeta^2$ and $\rho^2(\zeta) = \zeta^3$ and $\rho^3(\zeta) = \zeta^4$ and so on.

Could someone write this complex conjugation $\omega$ in terms of the automorphism $\rho$

I tried all kinds of conjugations and cannot find the one that will send $\zeta \rightarrow \zeta^4, \zeta^2 \rightarrow \zeta^3$

$\endgroup$
4
  • $\begingroup$ because $\mathbb{Q}(\zeta)$ is a splitting field. $\endgroup$
    – Quester
    Commented Apr 23, 2014 at 23:38
  • 2
    $\begingroup$ Ahem, if $\rho(\zeta)=\zeta^2$, then $\rho^2(\zeta)=\zeta^4$, not $\zeta^3$. $\endgroup$
    – ccorn
    Commented Apr 23, 2014 at 23:39
  • $\begingroup$ Better yet; if $\rho^3(\zeta)=\zeta^4$ then $$\rho^3(\zeta^2)=(\rho^3(\zeta))^2=(\zeta^4)^2=\zeta^8=\zeta^3.$$ So it seems that $\omega=\rho^3$. $\endgroup$
    – Servaes
    Commented Apr 23, 2014 at 23:42
  • $\begingroup$ Hey thank you all, especially Servaes. I now know the mistake I was making. $\endgroup$
    – Quester
    Commented Apr 24, 2014 at 0:16

2 Answers 2

6
$\begingroup$

The Galois group for any cyclotomic field $\mathbb{Q}(\zeta_n)$ will be the multiplicative group $\mathbb{Z}_n^{\times}$. When $n$ is prime, as is the case here, then $\mathbb{Z}_n^{\times} \cong \mathbb{Z}_{n-1}$, which is cyclic. (The latter fact can be proven using the structure theorem for finitely generated abelian groups.)

Therefore, we simply need to find a generator for the Galois group. To do this, note that every $\mathbb{Q}$-automorphism of this field is determined by its action on $\zeta$, and further, $\zeta \mapsto \zeta^k$ always defines a legitimate automorphism (since $\text{Gal}(f)$ acts transitively on the roots of $f$ when $f$ is irreducible). Therefore, I simply experiment as follows:

Does $\phi$ such that $\phi(\zeta) = \zeta^2$ generate the group? If not, what about $\phi(\zeta) = \zeta^3$? And I continue until I have found my generator - it'll end up being $\zeta \mapsto \zeta^k$ for a $k$ that generates $\mathbb{Z}_n^\times$.

Once you've found the generator $\phi$, the Galois group is simply $\{ \text{id}, \phi$, $\phi^2$, $\phi^3\}$. Noting that $\mathbb{Z}_4$ has one subgroup isomorphic to $\mathbb{Z}_2$, you will get a subgroup $\{ \text{id}, \phi^2\}$ that sends $\zeta \mapsto \zeta^4$ and $\zeta^2 \mapsto \zeta^3$ when it acts on the roots.

Long story short, I think this is simpler than trying to think about the automorphisms in terms of complex conjugation, etc.

$\endgroup$
1
  • 2
    $\begingroup$ I like this answer (+1), but the issue seems to be that the OP has an inconsistent definition of his $\rho$. Should be made explicit I think. $\endgroup$
    – ccorn
    Commented Apr 23, 2014 at 23:55
4
$\begingroup$

If $\rho^3(\zeta)=\zeta^4$ then $$\rho^3(\zeta^2)=(\rho^3(\zeta))^2=(\zeta^4)^2=\zeta^8=\zeta^3,$$ which shows that $\omega=\rho^3$. Your notation is rather misleading though; if $\rho(\zeta)=\zeta^2$ then $$\rho^2(\zeta)=\rho(\zeta^2)=(\rho(\zeta))^2=(\zeta^2)^2=\zeta^4.$$ You may want to consider renaming the elements of the Galois group.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .