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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.

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  • $\begingroup$ Isn't that the definition of a $k$-automorphism? $\endgroup$ – Elliot G Mar 23 at 6:16
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    $\begingroup$ Every irreducible polynomial $\in k[x]$ which has a root in E splits completely in E. If $E/k$ is not normal then there is $a \in E, b \in \overline{k},\not \in E$ such that $a,b$ are roots of the same irreducible polynomial so there is a field isomorphism $k(a) \to k(b)$ which can be extended to a field isomorphism $E \to E_2 \subset \overline{k}$ fixing $k$ such that $E_2 \ne E$ $\endgroup$ – reuns Mar 23 at 6:24
  • $\begingroup$ A $k$-(iso)morphism is an (iso)morphism of $k$-algebra, that is $\sigma(ua) = u \sigma(a)$ for $u \in k$, which is the case iff it fixes $k$, that's the definitions $\endgroup$ – reuns Mar 23 at 6:30
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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.

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