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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?

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Hi, maths! Your last name seems familiar: haven't we met met before in some friendly circle? –  Georges Elencwajg Aug 16 '12 at 11:46
    
I'd imagine the problem to be clearer if more restrictions were made on the field K. For example, whether K is Galois/algebraic/finite over its base field (Q or F_p) –  Michael Luo Aug 16 '12 at 12:01
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Dear @Michael, the problem is perfectly clear and very interesting as stated. Your suggestion would only make it easier and/or less interesting. –  Georges Elencwajg Aug 16 '12 at 12:51
    
Equivalently: Given $n$ automorphisms $f_1, \ldots, f_n$ different from the identity, prove that there is an element $x$ in $K$ such that $f_i(x) \neq x$ for all $f_i$. –  Mikko Korhonen Aug 16 '12 at 19:31
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2 Answers 2

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.

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+1 for this really ingenious answer, Chris: congratulations! –  Georges Elencwajg Aug 16 '12 at 14:40
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It is true in general that no field is a union of finitely many proper subfields. See this MO question –  Mikko Korhonen Aug 16 '12 at 17:20
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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 $f_i$.

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The same argument can be used if we know that the group $G$ generated by the $f_i$ is finite, because then the extension $\mathbb{K}^G\subseteq\mathbb{K}$ is finite Galois. –  Hagen Aug 16 '12 at 15:02
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