# Does this lemma still holds when replacing 'Galois' by ‘normal’?

All of the field extensions involved are assumed to be finite.

This lemma is from Isaacs' Algebra, serving as a key step toward the establishment of Galois' criterion of solvability. The proof of the original lemma is as shown in the pictures.

The original lemma: Suppose $$F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$$, where for $$1 \leq i \leq r$$, each of the extensions $$F_{i-1} \subset F_i$$ is Galois with an abelian Galois group. Suppose $$F \subseteq E \subseteq L$$ and $$E$$ is Galois over $$F$$. Then $$\operatorname{Gal}(E/F)$$ is solvable.

It seems to me that the lemma holds still if one replace all 'Galois' in the original version by 'normal'. Then it becomes like this:

The modified lemma: Suppose $$F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$$, where for $$1 \leq i \leq r$$, each of the extensions $$F_{i-1} \subset F_i$$ is normal with an abelian Galois group. Suppose $$F \subseteq E \subseteq L$$ and $$E$$ is normal over $$F$$. Then $$\operatorname{Gal}(E/F)$$ is solvable.

But I am not sure if the modified lemma am right. I also write down a proof for the version lemma assuming normality. If it will not take a lot of time, please verify the following proof.

Proof: We work by induction on $$r$$. Let $$E_1=\langle F_1, E\rangle$$. The fact that $$E_1$$ is normal over $$F$$ follows from normality of $$F \subseteq E$$ and $$F \subseteq F_1$$. Thus, $$F_1 \subseteq E_1$$ is also normal. We have a tower of length $$r-1$$ of normal extensions from $$F_1$$ to $$L$$, and by the inductive hypothesis one concludes $$\operatorname{Gal}(E_1/F_1)$$ is solvable.

Consider the tower $$F \subset F_1 \subset E_1$$. Since $$F \subset F_1$$ and $$F \subset E_1$$ are normal, $$\operatorname{Gal}(E_1/F_1) \lhd \operatorname{Gal}(E_1/F)$$, and $$\operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/F_1) \cong \operatorname{Gal}(F_1/F)$$. Given both $$\operatorname{Gal}(E_1/F_1)$$ and $$\operatorname{Gal}(F_1/F)$$ are solvable, one knows $$\operatorname{Gal}(E_1/F)$$ is solvable.

Next consider the tower $$F \subset E \subset E_1$$. Again we have two normal extensions $$F \subset E$$ and $$F \subset E_1$$, which follows $$\operatorname{Gal}(E_1/E) \lhd \operatorname{Gal}(E_1/F)$$, and $$\operatorname{Gal}(E/F) \cong \operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/E)$$. One deduce $$\operatorname{Gal}(E/F)$$ is solvable as a quotient of a solvable group.

• Please do not rely on pictures of text. Jan 4 at 22:18
• It will help the readability of your question if you quoted the lemma first. Jan 4 at 22:19
• Prove that if $E/F/K$ is a finite tower in characteristic $p$ and $E/F$ is normal then taking $n$ large enough such that $p^n \nmid [E:K]$, $K(E^{p^n})/K$ is separable and $K(E^{p^n})/K(F^{p^n})$ is Galois with $Aut(K(E^{p^n})/K(F^{p^n}))=Aut(E/F)$. Jan 5 at 1:52
• @Shaun I have edited the descriptions on your advice, trying to clarify my question.
– zyy
Jan 5 at 11:38
• It means that replacing each $F_j$ by $F_0(F_j^{p^n})$ then everything becomes separable, the normal extensions become Galois, and their automorphism groups are left unchanged. Jan 5 at 11:57

I try to give a proof based on the sketch given by reuns. His sketch of proof is quoted as follows (reuns states that $$Aut(K(E^{p^n})/K(E^{p^n})) = Aut(E/F)$$, but I cannot prove the equality) :

Prove that if $$E/F/K$$ is a finite tower in characteristic $$p$$ and $$E/F$$ is normal then taking $$n$$ large enough such that $$p^n∤[E:K]$$, $$K(E^{p^n})/K$$ is separable and $$K(E^{p^n})/K(F^{p^n})$$ is Galois with $$Aut(K(E^{p^n})/K(F^{p^n})) \cong Aut(E/F)$$.

Proof: Let $$S = \{\gamma \in E \mid \gamma$$ is separable over $$K \}$$, i.d. the maximal separable extension of $$K$$ in $$E$$.

We first prove that $$K(E^{p^n}) = S$$. To do this we must show that $$K(E^{p^n})$$ is separable over $$K$$ and $$E$$ is purely inseparable over $$K(E^{p^n})$$. $$[E:S]$$ is a power of $$p$$, say $$p^m$$, where $$m \lt n$$. Each $$\alpha \in E$$ has minimal polynomial over $$S$$ of the form $$f(X)=X^{p^l}-a$$ for some element $$a\in S$$ and some integer $$l\leq m\lt n$$. That is, $$\alpha ^{p^l} = a \in S$$. So $$\alpha ^{p^n} \in S$$, $$K(E^{p^n}) \subseteq S$$, hence separable over $$K$$. On the other hand, for each $$\beta\in E$$, $$\beta^{p^n} \in K(E^{p^n})$$, therefore $$E/K(E^{p^n})$$ is purely inseparable. By the uniqueness of intermediate field that is separable over $$K$$ and over which $$E$$ is purely inseparable, we conclude that $$S=K(E^{p^n})$$.

Similarly we can show that $$K(F^{p^n})$$ is the maximal separable extension of $$K$$ if $$F$$.

By the first part of proof, we know that $$K(E^{p^n})/K$$ is separable. Since $$K \subseteq K(F^{p^n}) \subseteq K(E^{p^n})$$, $$K(E^{p^n})/K(F^{p^n})$$ is finite separable. We shall show that $$K(E^{p^n})$$ is normal over $$K(F^{p^n})$$. $$E$$ is a splitting field over $$F$$ for some polynomial $$g(X) = \Pi (X-\beta_i)^{e_i}$$, where $$\beta_i \in E$$ are distinct. Let $$h(X) = \Pi (X - \beta_i^{p^n})^{e_i}$$, then $$h(X^{p^n}) = g^{p^n}(X) \in F^{p^n}[X]$$. So $$h(X) \in F^{p^n}[X]$$, and $$E^{p^n} = F^{p^n}(\beta_i^{p^n})$$ is a splitting field of $$h(X)$$ over $$K(F^{p^n})$$, hence Galois over $$K(F^{p^n})$$.

Finally we need to show $$Aut(K(E^{p^n})/K(F^{p^n})) \cong Aut(E/F)$$. Now that $$K(E^{p^n})$$ is Galois over $$F$$, we can define a homomorphism $$\rho: Aut(E/F) \to Aut(K(E^{p^n})/K(F^{p^n}))$$ to be the restriction of $$\sigma \in Aut(E/F)$$ to $$K(E^{p^n})$$. Each $$\sigma \in \ker \rho$$ fixes $$K(E^{p^n}) = S$$, and by the properties of such intermediate field $$S$$, $$\sigma$$ must fix $$E$$. It follows that $$\rho$$ is injective. For the same reason, any $$\tau$$ that fixes $$K(F^{p^n})$$ must fix $$F$$, therefore each $$\sigma \in Aut(K(E^{p^n})/K(F^{p^n}))$$ extends to some $$\tau \in Aut(E)$$ that fixed $$F$$, which follows the surjectivity of $$\rho$$.