Galois, normal and separable extensions

Question

Theorem: Every finite extension, normal and separable is a Galois extension.

Is the theorem equivalent to: $\mathbb K:\mathbb F$ is Galois $\iff \mathbb K:\mathbb F$ is normal & $\mathbb K:\mathbb F$ is separable ?

thus, $\mathbb K:\mathbb F$ is not a Galois extension $\iff \mathbb K:\mathbb F$ is not normal or $\mathbb K:\mathbb F$ is not separable, but only one holds, either normal or separable.

Answer 1

For a finite extension $K/k$ the following are equivalent:

1. $\require{begingroup}\begingroup\DeclareMathOperator{\Aut}{Aut}|\Aut(K/k)| = [K:k]$
2. $k$ is the fixed field of $\Aut(K/k)$
3. $K/k$ is normal and separable
4. $K$ is the splitting field of a finite number of separable polynomials over $k$
5. $K$ is the splitting field of a single separable polynomial over $k$
6. There is a one to one correspondence between subgroups of $\Aut(K/k)$ and intermediate extensions of $K/k$

For a possibly infinite algebraic extension $K/k$ the following are equivalent:

1. $k$ is the fixed field of $\Aut(K/k)$
2. $K/k$ is normal and separable
3. $K$ is the splitting field of a possibly infinite family of separable polynomials over $k$
4. There is a one to one correspondence between closed subgroups of $\Aut(K/k)$ and intermediate extensions of $K/k$

$\endgroup$