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Let $F ⊆ K$ be finite fields. Show that $\operatorname{Stab}_{\operatorname{Gal}(K/F)}(z)=\{1\}$ for at least one $z\in K$.

My attempt
Firstly $z\notin F$, since any element of $F$ is fixed by $\operatorname{Gal}(K/F)$.

$\operatorname{Stab}_{\operatorname{Gal}(K/F)}(z)=\{1\}$ is equivalent to that $f(z)$ for all $f\in\operatorname{Gal}(K/F)$ are distinct.

Existence of $z$ is equivalent to that $\bigcup_{f\in\operatorname{Gal}(K/F)\setminus\{1\}}\text{Fix}(f)$ is a proper subset of $K$.


I'm a beginner to Galois theory, could you help me proving this?

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    $\begingroup$ The answer below is great. This also follows from the weaker assertion that $F = \Bbb F_p(\alpha)$ for some $\alpha$ (a "primitive element"). You can prove this in the case of finite fields by a nice counting argument using the fact that subfields of $F$ have order a power of $p$ and there is at most one subfield of each cardinality. (Suppose no element of $F$ generates $F$. Then every element of $F$ lies in a proper subfield...) $\endgroup$ Commented Oct 11, 2023 at 23:21
  • $\begingroup$ Primitive element theorem $\endgroup$
    – hbghlyj
    Commented Oct 12, 2023 at 12:43

1 Answer 1

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$K^\times$ is cyclic, let $a$ be a generator. If some $f \in \mathrm{Gal}(K/F)$ fixes $a$, then it fixes all of $K$. (Do you see why?) By Galois theory, this implies that $f=1$ is the identity.

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