Uniqueness of automorphisms on root

Question

The following is from A. Białynicki-Birula, "Algebra" (the translation is mine).

Example 3. Let $p$ be a prime number. Let us consider the splitting field $L$ of the polynomial $\frac{x^p - 1}{x - 1}$ over $Q$. The polynomial $f$ is (see p. 214) irreducible in $Q[x]$. The roots of this polynomial are primitive roots of $p$-th degree. Let $\zeta \in L$ be a fixed primitive root of $p$-th degree. Then $L_{f_p} = \{\zeta, \zeta^2, ..., \zeta^{p-1}\}$. For every $i \in N_+$, $1 \le i \le p-1$ there is exactly one automorphism $\alpha \in Aut(L \backslash K)$ such that $\alpha(\zeta) = \zeta^i$. Indeed, the existence of such an automorphism is implied by the theorem 11.3. Besides that, if $\alpha(\zeta) = \beta(\zeta)$ for $\alpha, \beta \in Aut(L \backslash K)$, then for every natural number $j, 1 \leq j \leq p-1$ we have $$\alpha(\zeta^j) = (\alpha(\zeta))^j = (\beta(\zeta))^j = \beta(\zeta^j),$$ hence $$\alpha|L_{f_p} = \beta|L_{f_p}.$$ This - by the final remark of the \$5, example 7, implies that $\alpha = \beta$.
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Example 4. Let us assume that $n$ is a natural number and that the field $K$ contains $n$ different $n$-th degree roots of unity (thus it contains - by theorem VI.7.9 - a primitive $n$-th degree root). Let $a \in K$; let us consider the polynomial $f=x^n-a \in K[x]$ and let $L$ be the extension of splitting field of this polynomial over $K$. Let us denote by $\sqrt[n]a$ whichever root of polynomial $f$ in the field $L$. Then $$K(\sqrt[n]a) = L \text{ and } L_f = \{\sqrt[n]a, \zeta \sqrt[n]a, \zeta ^2 \sqrt[n]a, ..., \zeta^{n-1} \sqrt[n]a\}$$ where $\zeta$ is any fixed $n$-th degree primitive root. Applying the reasoning similar to that in the example 3 one can easily show that for every integer number $i $ there exists at most one such an automorphism $\alpha \in Aut(L \backslash K)$ that $\alpha(\sqrt[n]a) = \zeta^i \sqrt[n]a$.

The final remark of the \$5, example 7 says (a bit roughly speaking, but I think that for these needs it is sufficient) that if the restrictions of $\alpha$ and $\beta$ to the set of roots of the polynomial are the same, then $\alpha = \beta$.

Now, how in the case of example 4 the fact that $\alpha (\sqrt[n]a) = \beta (\sqrt[n]a)$ would imply that $\alpha(\zeta^j \sqrt[n]a) = \beta(\zeta^j \sqrt[n]a)$? In the case of example 3, the situation was relatively easy because we had $\alpha(\zeta) = \beta(\zeta)$. Here it seems that $\alpha(\zeta) = \beta(\zeta)$ would also be immediately useful - then we would have $$\alpha(\zeta^j \sqrt[n]a) = \alpha(\zeta)^j \alpha (\sqrt[n]a) = \beta(\zeta)^j \beta (\sqrt[n]a) = \beta(\zeta^j \sqrt[n]a).$$ I tried some other things like raising the whole expression in the parentheses to the $j$-th power, but all these attempts were in vain.

Answer 1

The key is that both $x_1:=\root n\of a$ as well as $x_2:=\zeta\root n\of a$ are roots of the polynomial $x^n-a$. So if the restrictions of $\alpha$ and $\beta$ to the set of roots are equal, then we have, in addition to $\alpha(x_1)=\beta(x_1)$ also $$ \alpha(\zeta)=\alpha(\frac{x_2}{x_1})=\frac{\alpha(x_2)}{\alpha(x_1)}= \frac{\beta(x_2)}{\beta(x_1)}=\beta(\frac{x_2}{x_1})=\beta(\zeta). $$ All because we were also given that $\alpha(x_2)=\beta(x_2)$.

In other words, we can write $\zeta$ in terms of $x_1$ and $x_2$. So if we know where both $x_1$ and $x_2$ are mapped, the image of $\zeta$ is also determined.