p-Extensions and Galois Closures (14.4.5)

Dummit and Foote Abstract Algebra, section 14.4, exercise 5:

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. Such an extension is called a $p$-extension.

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

Proof: Since $L/K$ and $K/F$ are Galois of $p$-power degree, we have $L/F$ is separable and finite (and of $p$-power degree), thus simple. Let $L=F(α)$. We note that the Galois closure of $L$ must contain all the roots of $m_α(x)$ and also that the splitting field of $m_α(x)$ over $F$ is Galois, hence the latter is precisely the closure. Since $K/F$ is Galois, by 14.4.4 we see that in particular $m_α(x)∈F[x]$ splits over $K$ into a product of $n$ irreducibles of same degree $d$, and since $dn=p^a$ (the degree of $L/F$ and $m_α(x)$) we must have $d$ is a $p$-power thus $K(β)/K$ is of $p$-power degree for any root $β$ of $m_α(x)$.

Since the isomorphism $φ~:~K(α)→K(β)$ from mapping $α↦β$ induces an isomorphism $\text{Aut}(K(α)/K)≅\text{Aut}(K(β)/K)$ given by $σ↦φσφ^{-1}$, we see $K(β)/K$ is Galois. By observing degrees in Proposition 14.4.21 we see that composites of $p$-extensions are themselves $p$-extensions ($\text{Gal}(K_1/F)×\text{Gal}(K_2/F)$ is a $p$-group, and so too are its subgroups), hence it follows that the composite of the extensions $K(β_i)$ for roots $β_i$ of $m_α(x)$ (i.e. the splitting field of $m_α(x)$ over $F$, the Galois closure of $L$ over $F$) is in fact of $p$-power degree over $K$ and a $p$-extension of $F$.$~\square$