Definition 1. An extension $E$ of a field $k$ is called normal extension if
(i) $E$ is algebaric extension of $k$.
(ii) Every irreducible polynomial over $k$, which has a root in $E$, has all its roots (considered in $\overline{k}$) lie in $E$.
Definition: If $E$ is an extension of a field $k$, then by an $k$-automorphism of $E$, we mean a ring automorphism from $E$ to $E$ which is $k$-linear map of the $k$-vector space $E$, i.e. which is identity on $k$.
Following is a result from Basic Algebra , Cohn P. M.
Corollary 11.4.9 An algebraic extension $E$ of $k$ is normal if and only if for any field $\Omega$ containing $E$, any $k$-automorphism of $\Omega$ takes $E$ into $E$.
Question: In the corollary, can we replace a $k$-automorphism by a field automorphism of $\Omega$ which takes $k$ to $k$?
In other words, I want to see whether following is correct:
An algebraic extension $E$ of $k$ is normal if and only if for any field $\Omega$ containing $E$, any field automorphism $\sigma:\Omega\rightarrow\Omega$ which takes $k$ to itself, also takes $E$ to itself.
While the stronger condition of course implies normality, it is strictly stronger. That is, the “if” holds, but not the “only if”.
You may know that any extension of degree $2$ is normal. Consider $k=\mathbb{Q}(\sqrt{2})$ and $E=\mathbb{Q}(\sqrt{\alpha})$, with $\alpha=\sqrt{2}$ and $E\subseteq\mathbb{R}$ (that is, pick a real root). Take $\Omega$ to be the splitting field of $x^4-2$ over $\mathbb{Q}$, which contains $E$. Then $[E:k]=2$, so $E$ is normal over $k$. Now consider the automorphism of $k$ that maps $\sqrt{2}$ to $-\sqrt{2}$. This can be extended to $\Omega$. This automorphism sends $E=k(\sqrt{\alpha})$ to $k(\sqrt{-\alpha})$, which is not contained in $\mathbb{R}$, hence cannot equal $E$.
This is essentially because normality is not preserved in towers: $L/E$ normal and $E/k$ normal does not imply $L/k$ is normal.