As it is well-known we have $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\cong(\mathbb Z/n\mathbb Z)^\ast$. An approach often taken by textbooks is to first establish the irreducibility and degree of the cyclotomic polynomials $\Phi_n$, then using this fact in order to obtain the isomorphism. This usually goes along the lines of defining an embedding
$$
\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\to(\mathbb Z/n\mathbb Z)^\ast,\,(\zeta_n\mapsto\zeta_n^c)\mapsto\overline c
$$
combined with $|\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)|=\deg(\Phi_n)=\varphi(n)$.
This approach strikes me as… impractical? Below I present a different ansatz which I find much simpler but which for some reason textbooks seem to avoid entirely.
Show that $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\cong(\mathbb Z/n\mathbb Z)^\ast$.
Proof. Since $K=\mathbb Q(\zeta_n)$ is a simple extension, the action of any $\mathbb Q$-automorphism is completely determined by its action on $\zeta_n$. Let $\sigma$ be such a $\mathbb Q$-automorphism of $K$. As
$$
\operatorname{ord}(\sigma(\zeta_n))=\operatorname{ord}(\zeta_n)=n
$$
all conjugates have to be primitive $n$-th roots of unity.
We claim that $\zeta_n^c$ is a primitive $n$-th root of unity iff $(c,n)=1$. Indeed, if $(c,n)=d>1$ then we find $k,k'$ such that $c=dk$ and $n=dk'$, hence
$$
(\zeta_n^c)^{k'}=(\zeta_n^n)^k=1
$$
with $k'<n$. Conversely, if $o=\operatorname{ord}(\zeta_n^c)\le n$ we have
$$
(\zeta_n^c)^o=\zeta_n^{co}=1
$$
implying that $n\mid co$. However, since $(c,n)=1$ we deduce that $n\mid o$ and therefore $o=n$ as desired.
This leaves us with at most $\varphi(n)$ possible $\mathbb Q$-automorphism, as they have to map $\zeta_n$ to $\zeta_n^c$ for $(c,n)=1$, and any such choice yields a different $\mathbb Q$-automorphism. Hence $|\operatorname{Gal}(K/\mathbb Q)|=\varphi(n)$. Now the natural embedding
$$
\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\to(\mathbb Z/n\mathbb Z)^\ast,\,(\zeta_n\mapsto\zeta_n^c)\mapsto\overline c
$$
yields the result. $\square$
I see two potential problems with this proof:
- It requires some rudimentary knowledge on elemetary number theory. The arguments to establish that $\zeta_n^c$ for $(c,n)=1$ are the primitive $n$-th roots of unity are based on some non-algebraic knowledge. I see how this could discourage their usage when there are purely algebraic proofs avaiable (which even are applications of earlier presented material on rings/fields).
- Taking powers and, especially, interchanging powers of complex numbers is a problematic thing. Frankly speaking, I am note completely sure if what I have done is correct or relies on a wrong principle (I know that $(a^b)^c\ne a^{bc}$ in general for $a,b,c\in\mathbb C$; not sure if this matters here)
Does someone have some insight regarding this matter? Is my proof correct? or at least savable without too drastic changes?
Thanks in advance!
Note: Once it is established that $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\cong(\mathbb Z/n\mathbb Z)^\ast$ we get the irreducibility of the cyclotomic polynomials for free by simple degree considerations.
Best Answer
There is a problem with this proof as written. Say that $(m,n) = 1$, and that $\zeta$ is a primitive $n$th root of unity. It is certainly also the case that $\zeta^m$ is a primitive $n$th root of unity. But why does that mean there is an automorphism sending $\zeta \rightarrow \zeta^m$? That is, the condition that $(m,n)=1$ is certainly necessary but not (obviously) sufficient. (So the problematic part is "and any such choice yields a ... $\mathbf{Q}$-automorphism".) To prove this you need to show that $\zeta^m$ is conjugate to $\zeta$ for $(m,n) = 1$, which is (more or less) equivalent to showing that $\Phi_n(X)$ is an irreducible polynomial.
In otherwords, your argument as written shows that there is an injection from the Galois group to $(\mathbf{Z}/n \mathbf{Z})^{\times}$. It actually shows that if $F$ is any field (let's say of characteristic prime to $n$) that there is an injection
$$\mathrm{Gal}(F(\zeta_n)/F) \rightarrow (\mathbf{Z}/n \mathbf{Z})^{\times}.$$
But that certainly won't be surjective for all $F$.