I have following question: Let finite field and $ G = \varprojlim Gal(\bar{\mathbb{F}_p}|\mathbb{F}_p)$ equiped with canonical profinite topology indeced by inverse system $ (\phi _{i,j}, Gal(\mathbb{F}_{p^i},\mathbb{F}_p)$. My question is how to see that for every $ a \in \bar{\mathbb{F}_p}$ every $Stab_G(a) = \{g \in G | g(a) =a \}$ is open in G.
My idea:
According to finite Galois theory $a^G$ is finite, more precisely thats is composed of roots of the minimal polynom of a. Because the roots are algebraic we can find a finite field extension $ \mathbb{F}_{p^i}$ which includes all roots. Futhermore $Stab_G(a)_{\mathbb{F}_{p^i}} := \{g \in Gal(\mathbb{F}_{p^i},\mathbb{F}_p) | g(a) =a \}$ is open in $ Gal(\mathbb{F}_{p^i},\mathbb{F}_p) $ since it has the discrete topology. So so the preimage of $Stab_G(a)_{\mathbb{F}_{p^i}}$ unter the projection map $\phi_i: Gal(\bar{\mathbb{F}_p}|\mathbb{F}_p) \to Gal(\mathbb{F}_{p^i},\mathbb{F}_p)$ is open. So if it would hold that $Stab_G(a) = \phi^{-1}Stab_G(a)_{\mathbb{F}_{p^i}}$ the proof whould be finished. But is this equality correct?
If we have that $ \mathbb F_p(a) $ is the finite field of cardinality $ p^n $ for some $ n $, we have that
$$ \textrm{Stab}_G(a) = \left( \prod_{2 \leq i < n} \textrm{Gal}(\mathbb F_{p^i}/\mathbb F_p) \times \{ \textrm{id} \} \times \prod_{i > n} \textrm{Gal}(\mathbb F_{p^i}/\mathbb F_p) \right) \cap G $$
which is open since $ G $, as an inverse limit, inherits the subspace topology from the product; and the factors are discrete groups.