Considering all the following in the context of Galois theory.

I believe, given say the primitive $9^{th}$ root of unity, that this will have as its minimum polynomial , the cyclotomic polynomial


Clearly this has 6 roots, and I know it's a Galois extension so the order of the Galois group is 6. I also know that the group is in fact $C_6$ for the extension $\Bbb Q(w)/\Bbb Q$.

More generally I believe it can be said that for any primitive $n^{th}$ root of unity the extension field over $\Bbb Q$ corresponds to a Galois group which is isomorphic to the group of units mod n.

My question is why exactly ?

I can see from the example that I gave there that the minimum polynomial has powers of $w$ which correspond to the units (i.e. elements with multiplicative inverses) in $\Bbb Z_9$, but beyond this I'm quite foggy. I feel it's something like associating $w$ with $1\in \Bbb Z_9$ , $w^2$ with $2\in \Bbb Z_9$, but I don't really understand the mathematical connection. Could anyone elaborate on this point for me ?

  • $\begingroup$ I mean, $\omega^a\cdot\omega^b = \omega^{a+b}$? So the connection to the cyclic group isn't surprising imo. Maybe I misunderstand the question. $\endgroup$ – Don Thousand Apr 19 at 20:55
  • $\begingroup$ @DonThousand I had been asking about why it is isomorphic to the multiplicative group of units $\endgroup$ – bhapi Apr 19 at 21:06

Let $\zeta$ be a primitive $n$-th root of unity, and $K=\Bbb Q(\zeta)$. Then $K/\Bbb Q$ is a normal extension, since all primitive $n$-th roots of unity lie within it. So $K$ is a Galois extension of $\Bbb Q$.

Consider an automorphism $\sigma$ of $K$. It is determined by the value $\sigma(\zeta)$. But $\sigma(\zeta)^n=\sigma(\zeta^n)=\sigma(1)=1$, so $\sigma$ is an $n$-th root of unity. But for $0<m<n$, $\zeta^m\ne1$ and so $\sigma(\zeta)^m=\sigma(\zeta^m)\ne1$, that is, $\sigma(\zeta)$ is a primitive $n$-th root of unity. Thus $\sigma(\zeta)=\zeta^k$ where $\gcd(k,n)=1$. Thus $k$ is a unit in the ring $\Bbb Z_n$.

The deep part of of this is proving that if $\gcd(k,n)=1$, there really is an automorphism $\sigma_k$ with $\sigma_k(\zeta)=\zeta^k$. This is essentially the irreducibility of the cyclotomic polynomials over $\Bbb Q$. If $k\equiv k'\pmod n$, then $\zeta^k=\zeta^{k'}$ and $\sigma_k=\sigma_{k'}$, so we the Galois group consists of the $\sigma_k$ with $\gcd(k,n)=1$, that is it corresponds to the group of units of $\Bbb Z_n$.

This correspondence is a group isomorphism. Consider $\sigma_k\circ\sigma_l$. Then $$\sigma_k\circ\sigma_l(\zeta)=\sigma_k(\zeta^l)=\sigma_k(\zeta)^l =(\zeta^k)^l=\zeta^{kl}=\sigma_{kl}(\zeta).$$ This proves that $k\mapsto\sigma_k$ is a group homomorphism from the unit group of $\Bbb Z_n$ to the Galois groups, and it's an isomorphism as it's a bijection.


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