Uniqueness of values of automorphisms on a primitive root

Question

Let $L$ be a field and let $\zeta \in L$ be a primitive root of $n$-th degree. Let $\alpha$ and $\beta$ be automorphisms belonging to $Aut(L)$. Is it possible that $\alpha(\zeta) \ne \beta(\zeta)$? It strongly seems to me that the equality $\alpha(\zeta) = \beta(\zeta)$ must hold. Here is my reasoning behind this statement:
as $\alpha$ is automomorphism, it must be that $\alpha(1) = 1$. Since $\zeta$ is a primitive root of $n$-th degree, we have that $\zeta^n = 1$. Hence, we have that $\alpha(\zeta^n) = 1$. Similarly $\beta(\zeta^n) = 1$. So we have $$\alpha(\zeta^n) = \beta(\zeta^n)$$ Thus: $$(\alpha(\zeta^n))^{1-n} = (\beta(\zeta^n))^{1-n}$$ and finally we obtain: $$\alpha(\zeta) = \beta(\zeta)$$

Answer 1

You've made an algebra error in the final step. In fact, there are many field automorphisms that act nontrivially on primitive roots of unity: the cyclotomic field $F = \mathbb{Q}(\zeta_n)$ is a Galois extension of degree $\varphi(n) = \lvert (\mathbb{Z}/n\mathbb{Z})^\times \rvert$ (where $\varphi$ is the Euler phi function), and has $\varphi(n)$ automorphisms. Given $d \in \mathbb{Z}$ with $\gcd(d, n) = 1$, the assignment $\zeta_n \mapsto \zeta_n^d$ extends to a field automorphism of $F$, and all automorphisms are of this form. (In particular, the only automorphism of $F$ that fixes the generator $\zeta_n$ is the identity automorphism. This is a general phenomenon: if an automorphism fixes the generators of a field extension, then it's the identity automorphism.)

More generally, one can show using Galois theory that the only algebraic numbers that are fixed by every field automorphism are rational numbers.

Answer 2

The following answer is thanks to Roland and Daniel Hast, I write it here for documentation purposes. The reasoning that I presented is seriously flawed. First of all, there is a mistake because $(\alpha(\zeta)^n)^{1-n} \neq \alpha(\zeta)$, but $(\alpha(\zeta^n)^{1-n} = (\alpha(\zeta^{n \cdot 1-n \cdot n})) = (\alpha(\zeta^{n-n^2}))$. Second thing is that the idea itself is wrong. While it may seem for a while that we could rescue the situation by writing that $$\alpha(\zeta^n) = \beta(\zeta^n)$$ implies $$(\alpha(\zeta^n))^{1/n} = (\beta(\zeta^n))^{1/n}$$ and hence $$\alpha(\zeta) = \beta(\zeta),$$ that is plainly wrong. It is due to the fact that a given field element can have many (possibly different) roots which makes it impossible to deduce the above equality. As a concrete example let us take the field $C$ - we have $(i^2)^{\frac{1}{2}} = 1^\frac{1}{2}= \{1, -1\}$ but if we (incorrectly) try $(i^2)^\frac{1}{2} = i^{2\cdot \frac{1}{2}}$ then we get farther $= i^1 = i$.
The formula $(a^b)^\frac{1}{n} = a^\frac{b}{n}$ that works in $R$ if some common conventions are preserved does not work in fields in general.