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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.

Peter_Pan
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  • I think the issue is that you are assuming that the Galois group is a cyclic group of order $n$ in the last few lines your proof. In which case, the result is trivial. –  May 07 '20 at 04:44
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    @saltandpepper: Certainly not that, given that $\mathbb{Z}_n^{\times}$ is not of order $n$, and almost never cyclic. I do not see that assumptions. Could you pinpoint where you see it? – Arturo Magidin May 07 '20 at 04:52
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    I cannot tell for sure, but it is possible that those "more complicated" proofs include, in one form or another, a proof for the fact that $$\Phi_n(x)=\prod_{t\in\Bbb{Z}_n^\times}(x-\zeta^t)$$ is irreducible over $\Bbb{Q}$. This is somewhat non-trivial (though well known) when $n$ is not a prime. (when it is the textbook application of Eisenstein). You do realize that should $\Phi_n$ be reducible, you could not use all the possible $\zeta^t$:s as automorphic images of $\zeta$, don't you? – Jyrki Lahtonen May 07 '20 at 06:08
  • If the irreducibility of $\Phi_n(x)$ over $\Bbb{Q}$ has been covered in an earlier Lemma/Proposition, then your argument is fine. Otherwise that's exactly the catch. You need to have the minimal polynomial of an element before you get control of its automorphic images. – Jyrki Lahtonen May 07 '20 at 06:16
  • See here for a local proof. – Jyrki Lahtonen May 07 '20 at 06:17
  • Could you elaborate on why $\Phi_n$ needs to be irreducible? @JyrkiLahtonen – Peter_Pan May 07 '20 at 06:19

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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 $g(x)$.

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 $K$: $$ (x^4+x^3+x^2+x+1)=(x^2+\frac{1+\sqrt{5}}2x+1)(x^2+\frac{1-\sqrt5}2x+1). $$

Jyrki Lahtonen
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  • Instead of proving that $\Phi_n$ is irreducible over $\mathbb{Q}$, can we instead find a basis for $\mathbb{Q}(\zeta)$? A possible basis is ${\zeta, \cdots, \zeta^{n-1} }$. And since $$\zeta+ \cdots + \zeta^{n-1} = -1,$$ then we need $1 \leq i \leq n-1$, in order for $1$ to be a linear combination of the $\zeta^i$'s. (i.e., we can't have a basis with less elements). So then $$[\mathbb{Q}(\zeta); \mathbb{Q}] =n.$$ – Peter_Pan May 07 '20 at 16:04
  • This means that there are $n$ automorphisms (using the fact that splitting fields are Galois), so all the possible values for $t$ appear in the image of $\phi$ (the isomorphism I defined in the question). Or am I implicitly using the gact that $\Phi_n$ is irreducible in this argument? – Peter_Pan May 07 '20 at 16:04
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    @Jess Yes, you are implicitly using irreducibility of $\Phi_n(x)$ when deducing linear independence of powers of $\zeta$. The minimal polynomial of $\zeta$ having degree $d$ is equivalent to $1,\zeta,\ldots,\zeta^{d-1}$ being linearly independent. – Jyrki Lahtonen May 07 '20 at 16:25
  • How does $x^n-1$ irreducible over $\mathbb{Q}$ work with the proof that the splitting field of $x^n-1$ has degree $\varphi(n)$ (where $\varphi(n)$ is Euler's Totient Function)? Isn't $\varphi(n) < n$ when $n$ is not prime, and $x^n-1$ irreducible over $\mathbb{Q} \implies $ the degree of the splitting field is $n$. https://math.stackexchange.com/questions/2231355/prove-that-the-splitting-field-of-xn-1-has-degree-phin-over-mathbbq – Peter_Pan May 07 '20 at 18:51
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    @Jess $x^n-1$ is most emphatically not irreducible over $\Bbb{Q}$. We have the factorization $$x^n-1=\prod_{d\mid n}\Phi_d(x)$$ with the cyclotomic polynomials $\Phi_d(x)$ as above. $\Phi_n(x)$ is the minimal polynomial of $\zeta$ (and has degree $\phi(n)$). Even when $n=p$ is a prime, we have $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots+x+1=\frac{x^p-1}{x-1}.$$ After all, $\phi(p)=p-1$. – Jyrki Lahtonen May 07 '20 at 19:10
  • Oh I see, I missed that detail. Thanks! – Peter_Pan May 07 '20 at 19:39