Let $K/F$ be a Galois extension with $G=\text{Gal}(K/F)$.
$\mathcal{I}=\lbrace E\text{ }|\text{ }E/F \text{ is Galois and }[E:F]<\infty\rbrace$
$\mathcal{N}=\lbrace N\text{ }|\text{ } N=\text{Gal}(K/E) \text{ for some } E\in\mathcal{I}\rbrace$
The Krull topology over $G$ is defined as follows. The set
$B=\lbrace \sigma_iN_i\text{ }|\text{ }\sigma_i\in G \text{ and }N_i\in\mathcal{N}\rbrace$ is the base for the Krull topology.
I am trying to prove that the product map $f:G\times G\rightarrow G$ defined as $(g,h)\mapsto gh$ and the inverse map $h:G\rightarrow G$ defined as $g\mapsto g^{-1}$ are continuous.
My idea:
Since $B$ is a base for the topology, it is enough to show that $f^{-1}(\sigma_iN_i)$ and $h^{-1}(\sigma_iN_i)$ is open for all $\sigma_i\in G$ and $N_i\in\mathcal{N}$. What to do after this ?
I think I have a solution. Please correct me if I am wrong.
Consider any $\sigma,\tau\in G$. To prove that $f$ is continuous, we have to show that given any neighborhood $W$ of $\sigma\tau$, we have to find a neighborhood $U$ of $\sigma$ and a neighborhood $V$ of $\tau$ such that $f(U\times V)\subseteq W$ (This is just the definition of continuity).
We have $f(U\times V)=UV$ (This is by the definition of $f$). Any arbitrary neighborhood W of $\sigma\tau$ contains a set of the form $(\sigma\tau) N$ for some $N\in\mathcal{N}$( for any $g\in G$ the set $\lbrace gN\text{ }|\text{ }N\in\mathcal{N}\rbrace$ is a neighborhood base of $g$). Consider the neighborhoods $U=\sigma N$ of $\sigma$ and $V=\tau N$ of $\tau$. We have $$UV=\sigma N\tau N=(\sigma\tau)N\subseteq W$$
The step $\sigma N\tau N=(\sigma\tau)N$, follows from the fact that $N$ is normal in $G$.