# 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

## 1 Answer

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.