# A tower of normal extensions is normal

I am reading Kaplansky's notes on Galois theory. He defines a normal extension as follows:

Let $M$ be any field and $K$ any subfield. $M$ is normal over $K$ if for any $u\in M$ but not in $K$, there exists an automorphism of $M$ leaving every element of $K$ fixed but actually moving $U$.

As a side note, this definition does not appear on Wikipedia. Could someone verify it is equivalent to the normal definitions?

He sets the following exercise:

Problem: Let $K\subset L \subset M$ be fields with $L$ normal over $K$ and $M$ normal over $L$. Suppose any automorphism of $L/K$ can be extended to $M$. Prove that $M$ is normal over $K$.

My solution attempt is in the answers. I also wonder how any automorphism of $L/K$ can be extended to $M$.

-

The concept of a normal extension is invented to capture the behaviour that polynomials that ought to split actually do into linear factors.

The extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ does not behave in this way...the polynomial $x^3 - 2$ has the generator $\sqrt[3]{2}$ as a root so we would like to hope that the other roots also lie in this field, but the other roots do not. This is a problem as far as the Galois group is concerned...there are "missing roots".

The elements of the Galois group permute roots of polynomials that are "lying in the field" so that in normal extensions, there are no "missing roots", the Galois group really has maximum choice as to where to send a given root (as long as no roots are repeated but that is what separability captures).

Now your definition is equivalent to $K$ being the fixed field of Gal$(M/K)$...which is a property of normal extensions (when working with subfields of $\mathbb{C}$). So yes the definition is equivalent.

-
I have fixed this now...the definition actually IS equivalent. – fretty Jun 9 '12 at 9:36
I deleted my previous comment, which makes no sense after the correction. – Mark Bennet Jun 9 '12 at 10:13

Fix an element $u\in M$ but $u\notin K$. We want to show there is an automorphism of $M$ that fixes $K$ but moves $u$.

Suppose $u\notin L$. By hypothesis, there is an automorphism of $K$ that moves $u$ but fixes $L$, and this also fixes $K$, so we are done.

Suppose $u\in L$. Then we can find an automorphism of $L$ that fixes $K$ but moves $u$. By hypothesis, this automorphism can be extended to $M$, so it works, and we are done.

As noted in the problem, I do not know how the extension to $M$ can be carried out. Could someone indicate how this can be done?

-