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I'm working on a problem in Dummit & Foote and I'm quite stumped. The problem reads:

Let $p$ be a prime and let $F$ be a field. Let $K$ be a Galois extension of $F$ whose Galois group is a $p$-group (i.e., the degree $[K:F]$ is a power of $p$). Such an extension is called a $p$-extension (note that $p$-extensions are Galois by definition).

Let $L$ be a $p$-extension of $K$. Prove that the Galois closure of $L$ over $F$ is a $p$-extension of $F$.

This is what I've done so far:

Using the tower law we can readily show that $L$ is a $p$ extension of $F$ so we have $[L:F]=p^{\ell}$ for some integer $\ell$. Then if $M$ is the Galois closure of $L$ over $F$ then $$[M:F]=[M:L][L:F]$$ and therefore $[M:F]=p^{\ell}n$ for some integer $n$ that is not divisible by $p$. So $[M:L]=n$.

From here it seems like I want to show that either $n=1$ or that $n$ is in fact a power of $p$. I just don't see how to proceed. I've considered using the Sylow Theorems, but I'm not sure how that would really work. I also realize that this statement depends on $K$ being Galois over $F$ but I can't figure out how to take advantage of that.

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I don't want to spoil it for you, but if you have some self-restraint, there's a brief solution to this problem on this old CIT coursepage. – Buble Apr 20 '12 at 2:56
Using the Galois correspondence, this problem translates into a group theory problem. Let $1 < A \lhd B \lhd C$ be groups such that $B/A$ and $C/B$ are finite $p$-groups for some prime $p$. Then the core of $A$ in $C$ (that is, the intersection of the $C$-conjugates of $A$) has index a power of $p$ in $B$. This is true because the core is the intersection of finitely many normal subgroups of $B$, each having index a power of $p$ in $B$. – Derek Holt Apr 20 '12 at 12:36
@DerekHolt, You should post your comment as an answer. – JSchlather Jan 25 '13 at 7:00

First, $L/F$ is a finite and separable extension, thus by the primitive element theorem $L = F(\alpha)$ for some $\alpha \in L$. Let $m_\alpha \in F[X]$ be the minimal polynomial of $\alpha$. Since $K/F$ is Galois, over $K$, $m_\alpha$ splits into irreducible factors of the same degree, $d$ say. In Dummit-Foote this last statement is Exercise 14.4.4, the exercise right before the one in the question. Since $d$ divides $p^\ell = [L:F]$, it is a power of $p$ itself. This proves

Claim 1. For any root $\beta \in M$ of $m_\alpha$, $K(\beta)/K$ has degree a power of $p$.

Now the Galois closure $M$ is the composite field of the $K(\beta_i)$ where $\beta_i$ runs through the roots of $m_\alpha$. Reason: The composite field is the splitting field of $m_\alpha$, so it is Galois and contains $M$. On the other hand $M$ contains the splitting field of $m_\alpha$ because it contains the root $\alpha$.

Claim 2. $K(\beta)/K$ is Galois for every root $\beta$ of $m_\alpha$.

Proof. $L/K = K(\alpha)/K$ is Galois. Let $\phi : K(\alpha) \to K(\beta)$ be the isomorphism induced by $\alpha \mapsto \beta$ and the identity on $K$. Then $\text{Aut}(K(\alpha)/K) \cong \text{Aut}(K(\beta)/K)$ via $\sigma \mapsto \phi\sigma\phi^{-1}$.

Finally, the Galois group of $M$ (the composite of the $K(\beta)$) over $K$, is a subgroup of the direct product of the Galois groups of $K(\beta)/K$, which are all $p$-groups by Claim 1. Therefore it is a p-group by Dummit-Foote Proposition 14.12. Therefore $[M:K]$ is a power of $p$ and since $[K:F]$ is a power of $p$ and the same follows for $[M:F]$ by the tower formula.

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Why is $L/F = K(\alpha)/K$ Galois? Thanks! – TheNicanova Feb 12 at 20:09
@TheNicanova It is $L/K$, not $L/F$ that is equal to $K(\alpha)/K$. Now, $m_\alpha$ splits over $K$ into some irreducibles, so $K(\alpha)/K$ is the splitting field of a separable polynomial. – Thomas Feb 13 at 13:57

We denote by $G(K/F)$ the Galois group of a Galois extension $K/F$. We need the following lemma.

Lemma Let $F$ be a field, $\Omega$ its algebraic closure. Let $K_1$ and $K_2$ be subfields of $\Omega$ each of which is a Galois extension of $F$. Then $K_1K_2$ is a Galois extension of $F$ and $G(K_1K_2/F)$ is isomorphic to a subgroup of $G(K_1/F)\times G(K_2/F)$.

Proof. Let $\sigma$ be an automorphism of $\Omega/F$. Then $\sigma(K_1K_2) = \sigma(K_1)\sigma(K_2) = K_1K_2$. Hence $K_1K_2/F$ is a normal extension. Clearly it is a separable extension. Hence $K_1K_2/F$ is a Galois extension. We define a homomorphism $\psi\colon G(K_1K_2/F) \rightarrow G(K_1/F)\times G(K_2/F)$ by $\psi(\sigma) = (\sigma\mid K_1, \sigma\mid K_2)$. Since it is clearly injective, we are done.

Now let $F$ be a field, $\Omega$ its algebraic closure. Let $F \subset K \subset L \subset \Omega$ be a tower of fields such that $K/F$ and $L/K$ are $p$-extensions. Let $M/F$ be the Galois closure of $L/F$ in $\Omega$. Then $M = \sigma_1(L)\cdots \sigma_n(L)$, where $\sigma_i(L), i = 1,\cdots, n$ are all the distinct conjugates of $L$ over $F$ in $\Omega$. Since each $\sigma_i(L)$ is a Galois extension over $K$, $G(M/K)$ is isomorphic to a subgroup of $G(\sigma_1(L)/K)\times \cdots \times G(\sigma_n(L)/K)$ by the lemma. Since each $G(\sigma_i(L)/K)$ is a $p$-group, $G(M/K)$ is a $p$-group. Since $K/F$ is a $p$-extension, $(M : F)$ is a power of $p$. Hence $M/F$ is a $p$-extension and we are done.

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This proof is wrong. There are elements $\sigma_i$ in the Galois group of $M/F$ that don't fix $K$. Therefore the Lemma is not applicable. Also this proof has not used that $K/F$ is Galois. This is necessary since otherwise $p=3$, $F=\mathbb{Q}$, $K=L=\mathbb{Q}(\sqrt[3]{2})$ would have a Galois closure with a 3-group as Galois group. In reality it has $S_6$. – Thomas Dec 16 '15 at 13:37

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