Prove that $G(\mathbb{Q}(\zeta)/\mathbb{Q}) $ and $\mathbb{Z}_n^\times$ are Isomorphic

Question

Prove that $G(\mathbb{Q}(\zeta)/\mathbb{Q}) \rightarrow \mathbb{Z}_n^\times$ is an isomorphism.

$\mathbb{Q}(\zeta)$ is the cyclotomic extension of $\mathbb{Q}$ ($\zeta$ is a root of $x^n - 1$). $\mathbb{Z}_n^\times$ is the group of units of $\mathbb{Z}_n$.

Let $T := \{ s \in \mathbb{Z}_n \mid \, \big( \, s,n \big) = 1 \}$. I have previously proved that $T = \mathbb{Z}_n^\times$.

$\sigma \in G(\mathbb{Q}(\zeta)/\mathbb{Q})$ is defined by $\sigma(\zeta)$, which necessary has the same order as $\zeta$. Thus, $\sigma(\zeta) = \zeta^t$, where $(t,n)=1$.

Now, define $\phi: G(\mathbb{Q}(\zeta)/\mathbb{Q}) \rightarrow \mathbb{Z}_n^\times$ by $\phi(\sigma) = \sigma^t \mapsto t \in\mathbb{Z}_n^\times$.

This is clearly an injective group homomorphism.

Now, recall $\sigma \in G(\mathbb{Q}(\zeta)/\mathbb{Q}$ is defined by $\sigma(\zeta) = \zeta^t$, where $(t,n)=1$. So $t \in \mathbb{Z}$ s.t. $(t,n) = 1 \Leftrightarrow \zeta^t$ has order $n \Leftrightarrow \exists \ \sigma \in G(\mathbb{Q}(\zeta) $ s.t. $\sigma(\zeta) = \zeta^t$. It follows that $\# G(\mathbb{Q}(\zeta) = \# T =\# \mathbb{Z}_n^\times$.

Since $n$ is finite, then $\phi$ injective implies that $\phi$ is also surjective.

Is this a valid argument? I think I am missing something because all the other proofs that I have found for the surjectivity of this map are more complicated.

Answer 1

Those "more complicated" proofs include, in one form or another, a proof for the fact that the cyclotomic polynomial $$\Phi_n(x)=\prod_{t\in\Bbb{Z}_n^\times}(x-\zeta^t)$$ is irreducible over $\Bbb{Q}$.

This is absolutely essential. If we had a non-trivial factorization $\Phi_n(x)=g(x)h(x)$ with $g(\zeta)=0$ and $g(x)\in\Bbb{Q}[x]$, then $\sigma(\zeta)$ must also be a zero of $g(x)$, limiting the choice of $t\in\Bbb{Z}_n^\times$. More precisely, if $$g(x)=x^d+g_1x^{d-1}+\cdots+g_{d-1}x+g_d\in\Bbb{Q}[x]$$ with $g_i\in\Bbb{Q}$ and $d<\phi(n)=\deg\Phi_n(x)$, then $$0=\zeta^d+g_1\zeta^{d-1}+\cdots+g_{d-1}\zeta+g_d$$ and applying $\sigma$ to that gives $$0=\sigma(\zeta)^d+g_1\sigma(\zeta)^{d-1}+\cdots+g_{d-1}\sigma(\zeta)+g_d$$ implying that $\sigma(\zeta)$ also has to be a zero of $g(x)$.

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Also consider the following. What would change if look for $K$-automorphisms of $\Bbb{Q}(\zeta)$ instead of $\Bbb{Q}$-automorphisms as here? Here $K$ is some intermediate field. The number of automorphisms must drop as $K$ grows (all the way down to $1$ when $K=\Bbb{Q}(\zeta)$). For example, when $n=5$ (a prime number!) it is easy to figure out that $\zeta+\zeta^{-1}=2\cos(2\pi/5)=(-1+\sqrt5)/2$ generates the intermediate field $K=\Bbb{Q}(\sqrt5)$. Any $K$-automorphism $\sigma$ of $\Bbb{Q}(\zeta)$ must map $\sqrt5$ to itself, implying that only the choices $\sigma(\zeta)=\zeta^{\pm1}$ are available.

Note that this is reflected in the factorization of $\Phi_5(x)=x^4+x^3+x^2+x+1$, irreducible over $\Bbb{Q}$ but reducible over $K$: $$ (x^4+x^3+x^2+x+1)=(x^2+\frac{1+\sqrt{5}}2x+1)(x^2+\frac{1-\sqrt5}2x+1). $$