# Converse to Prop. V.6.11 from Lang's *Algebra*: $E/k$ normal $\impliedby E^{\operatorname{Aut}(E/k)}/k$ purely inseparable?

In Lang's Algebra, he proves in Proposition 6.11 of Chapter V (page 251, third edition) the decomposition of a normal extension into a tower of a purely inseparable extension followed by a separable extension. In particular, he shows that if $$E/k$$ is normal, then $$E^{\operatorname{Aut}(E/k)}/k$$ is purely inseparable.

I came to think about the converse of the proposition:

Is it true that if $$E^{\operatorname{Aut}(E/k)}/k$$ is purely inseparable, then $$E/k$$ is normal?

I tried to prove this by proving the contrapositive, and most parts of the proof went on easy but I had a problem with just one part.

To prove the contrapositive, I started with the hypothesis that $$E/k$$ is not normal. Let $$L/k$$ be the smallest normal extension containing $$E$$ and let $$k_0=L^{\operatorname{Aut}(L/k)}$$. If $$k_0\subset E$$, since $$\operatorname{Aut}(E/k)=\operatorname{Aut}(E/k_0)$$, using Galois correspondence on the subset $$\{\sigma \in Gal(L/k_0):\text{restriction of }\sigma \text{ to }E\text{ is an automorphism}\}$$ easily yields $$E^{\operatorname{Aut}(E/k)}\supsetneq k_0$$ and thus that $$E$$ is not purely inseparable over k.

The problem I'm having is that I'm not sure about $$k_0 \subset E$$. That is,

Given an algebraic extension $$E/k$$, if $$L/k$$ is the smallest normal extension containing $$E$$, is it always true that $$L^{\operatorname{Aut}(L/k)} \subset E$$?

It seems like a trivial thing but maybe I missed something very basic?

It is not always true that $$L^{\operatorname{Aut}(L/k)} \subset E$$ (counterexample given at the bottom), but it is true in your scenario.

I also feel that the application of the Galois correspondence you mentioned is happening in the case where $$[E:k] < \infty$$. Perhaps there are some more checks to be done in the case where $$[E:k] = \infty$$, but I am not sure.

Here is how I would prove the result you're after:

Let $$E/k$$ be a finite extension. Let $$G = \operatorname{Aut}(E/k)$$ and let $$k_0 = E^G$$. Let $$I$$ be the compositum of all subfields $$F$$ of $$E$$ such that $$F \supset k$$ and $$F$$ is purely inseparable over $$k$$. Note that $$I$$ is purely inseparable over $$k$$, and in fact, $$I$$ consists of all elements $$\alpha$$ in $$E$$ that are purely inseparable over $$k$$.

First, we show that $$I \subset E^G$$.
Let $$\alpha \in I$$ and let $$\sigma \in G$$. We want to show that $$\sigma(\alpha) = \alpha$$. Since $$\alpha$$ is purely inseparable over $$k$$, $$\operatorname{Irr}(\alpha,k,X)$$ has only one distinct root, namely $$\alpha$$. Since $$\sigma(\alpha)$$ is also a root of $$\operatorname{Irr}(\alpha,k,X)$$, it must be that $$\sigma(\alpha) = \alpha$$. Thus, $$\alpha \in E^G$$.

Second, we remark that $$E/E^G$$ is separable.
This is proved in Proposition 6.11 of Chapter V in Lang's Algebra (pages 251-252, third edition), so we skip the proof here.

Third, we introduce the hypothesis that $$E^G \subset I$$.
But, we proved that $$I \subset E^G$$, so we have that $$E^G = I$$. Hence, we have a chain of extensions $$E \supset E^G \supset k$$ such that $$E^G/k$$ is purely inseparable and $$E/E^G$$ is separable. Hence, by Proposition A in this note by Joseph Lipman, there exists a separable extension $$K/E$$ such that $$K/k$$ is normal. Moreover, if $$H = \operatorname{Aut}(K/k)$$, then $$K^H = I$$.

To show that $$E/k$$ is normal, it suffices to show that $$E/I$$ is normal.
This is because $$E/I$$ normal and $$I/k$$ purely inseparable together imply that $$E/k$$ is normal (see here or here).

Now, since $$E/E^G = E/I$$ is separable, showing that $$E/I$$ is normal is equivalent to showing that $$E/I$$ is Galois. Since $$E/k$$ is a finite extension, this is equivalent to showing that the fixed field of $$\operatorname{Aut}(E/I)$$ equals $$I$$. To do this, we will show that $$\operatorname{Aut}(E/I)$$ equals $$G = \operatorname{Aut}(E/k)$$. Then, the fixed field of $$\operatorname{Aut}(E/I)$$ will be $$E^G = I$$, proving the result.

Since $$\operatorname{Aut}(E/k) \supset \operatorname{Aut}(E/I)$$, it suffices to show that $$\operatorname{Aut}(E/k) \subset \operatorname{Aut}(E/I)$$. So, let $$\sigma \in G$$. We need to show that $$\sigma \in \operatorname{Aut}(E/I)$$, that is, that $$\sigma$$ fixes every element of $$I$$. Extend $$\sigma$$ to an embedding of $$K$$ in an algebraic closure of $$K$$, also denoted $$\sigma$$. Since $$K/k$$ is normal, $$\sigma$$ is an automorphism of $$K$$ over $$k$$. Hence, $$\sigma \in H = \operatorname{Aut}(K/k)$$ and we have shown that $$K^H = I$$. Thus, $$\sigma$$ fixes every element of $$I$$, as was to be shown.

Notice that we have used the facts used in Lipman's note to prove that the fixed field of $$\operatorname{Aut}(K/k)$$ is a subfield of $$E$$, which is where you were stuck. In the same note, Lipman gives an example of an extension of fields for which this does not happen: let $$\Bbb{F}_2$$ be the field of two elements, let $$Y$$, $$Z$$ be indeterminates, let $$k=\Bbb{F}_2(Y, Z)$$, and let $$E=k(x)$$, $$x$$ being a root of $$f(X) = X^4 + YX^2 + Z = 0.$$ Then it turns out that there is no normal extension $$K$$ of $$k$$ containing $$E$$ such that $$K^{\operatorname{Aut}(K/k)} \subset E$$, based on the facts proved in the note.