# Normal extension and effect of field automorphisms

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.

• Isn't that the definition of a $k$-automorphism? – Elliot G Mar 23 at 6:16
• 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 $E_2 \ne E$ – reuns Mar 23 at 6:24
• 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 – reuns Mar 23 at 6:30

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.