Let $f_{i}$ be different automorphisms of field $\mathbb{K}$. Does there exist an $x \in \mathbb{K}$ such that $f_{i}(x)$ are pairwise distinct?

Question

Let $f_{1},f_{2},\ldots,f_{n}$ be different automorphisms of field $\mathbb{K}$ . What I want to ask is:

Does there exist an element $x \in \mathbb{K}$ such that $f_{1}(x),f_{2}(x),\ldots,f_{n}(x)$ are pairwise distinct?

Answer 1

This is true in the characteristic $0$ case at least.

For any two automorphisms of $\mathbb{K}$, the set of points where they are equal is a subfield of $\mathbb{K}$. Thus a counterexample to the claim gives us an expression of $\mathbb{K}$ as the union of finitely many proper subfields, namely the $\binom{n}{2}$ fields $\{ x|f_i(x)=f_j(x)\}$. Now if $\mathbb{K}$ has characteristic $0$, then it is a vector space over $\mathbb{Q}$, and these subfields are vector subspaces. But a vector space over an infinite field can never be a finite union of proper subspaces.

The case when $\mathbb{K}$ is finite is easy: any finite field is generated by a single element, and different automorphisms must differ on that element. I don't know if the claim is true for infinite fields of positive characteristic.

Answer 2

Let's take the case where $K$ is a finite, normal extension of the rationals. Then by the Primitive Element Theorem, there is an element $\alpha$ of $K$ such that $K={\bf Q}(\alpha)$. If $f_i(\alpha)=f_j(\alpha)$ then $f_i=f_j$ since $\alpha$ generates $K$ over $\bf Q$. So $\alpha$ will do as your $x$.

Now you can try to loosen some of the hypotheses to see how much of the argument still works and whether you can plug the holes.

EDIT: Note that if one asks about $f_1,f_2,\dots$ (instead of $f_1,f_2,\dots,f_n$), then there's generally no such element $x$. E.g., if $K={\bf Q}(\sqrt2,\sqrt3,\sqrt5,\dots)$, and $f_i(\sqrt{p_j})$ is $-\sqrt{p_j}$ for $i=j$, $\sqrt{p_j}$ otherwise, then every $x$ in $K$ is fixed by all but finitely many of the $f_i$.