# Why are the Galois groups that correspond to extensions which adjoin primitive roots of unity given by the group of units mod n

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

$$\Psi_9=(x-w)(x-w^2)(x-w^4)(x-w^5)(x-w^7)(x-w^8)$$.

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 ?

• 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. – Don Thousand Apr 19 at 20:55
• @DonThousand I had been asking about why it is isomorphic to the multiplicative group of units – 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 $$0, $$\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.