If I have some Galois number field $K$, a sequence of elements $(x_n)_{n=1}^{\infty}\subset K$ converging to some $z\in K$, and an automorphism $\sigma$ of K, when does $\lim\limits_{n\to\infty}(\sigma(x_n)) = \sigma(z)$? Assuming the sequence $(\sigma(x_n))_{n=1}^{\infty}$ converges to something in $K$, that is.

For some sequences it obviously works, for example taking $K=\mathbb{Q}(\sqrt{2})$ and $x_n=\frac{1}{(\sqrt{2})^n}$, which is null and therefore fixed by any automorphism. For a more general field $K=\mathbb{Q}(\theta)$, writing $x_n=\sum_{k=1}^{m}\alpha_{k,n}\theta_k$, with $\alpha_{k,n}\in\mathbb{Q}$ and $\theta_k$ the Galois conjugates of $\theta$, the limit of $x_n$ is just $\sum_{k=1}^{m}\lim\limits_{n\to\infty}(a_{k,n})\theta_k$, provided these limits exist.

I'm looking for any kind of strengthening of this to other, more general sequences, as well as any books or papers that might deal with analysis of this kind in number fields.

I know it's not true in all cases, for example $K=\mathbb{Q}(\sqrt{2})$, with $x_n=a_n\sqrt{2}$ and $a_n$ a sequence of rational numbers converging to $\sqrt{2}$. Here the conjugation map takes the limit from $2$ to $-2$.

Is there an in-between point between these two cases, where the idea is salvageable?

  • $\begingroup$ What topology on $K$ do you take? Are you fixing some embedding $K\hookrightarrow\mathbb C$ and considering the induced topology? If so, the only continuous automorphisms of $K$ (so that your condition holds for all sequences $x_n$) are identity and complex conjugation. $\endgroup$ – Wojowu Mar 17 at 10:31
  • $\begingroup$ Yes, I'm looking for convergence with respect to the standard topology on $\mathbb{C}$. Since your answer covers all sequences and just lets you use identity and complex conjugation, are there conditions on the sequence that could make it sequentially continuous for a certain automorphism that isn't either of these? $\endgroup$ – ArchiePi Mar 17 at 10:50

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