# union of fix points of $f\in\operatorname{Gal}(K/F)\setminus\{1\}$ is a proper subset of $K$

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?

• 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...) Commented Oct 11, 2023 at 23:21
• Primitive element theorem Commented Oct 12, 2023 at 12:43

$$K^\times$$ 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.