The Stacks project

Proof. Let $\zeta \in K$ be a primitive $n$th root of $1$. Let $\sigma $ be a generator of $\text{Gal}(L/K)$. Consider $\sigma : L \to L$ as a $K$-linear operator. Note that $\sigma ^ n - 1 = 0$ as a linear operator. Applying linear independence of characters (Lemma 9.13.1), we see that there cannot be a polynomial over $K$ of degree $< n$ annihilating $\sigma $. Hence the minimal polynomial of $\sigma $ as a linear operator is $x^ n - 1$. Since $\zeta $ is a root of $x^ n - 1$ by linear algebra there is a $0 \neq z \in L$ such that $\sigma (z) = \zeta z$. This $z$ satisfies $z^ n \in K$ because $\sigma (z^ n) = (\zeta z)^ n = z^ n$. Moreover, we see that $z, \sigma (z), \ldots , \sigma ^{n - 1}(z) = z, \zeta z, \ldots \zeta ^{n - 1} z$ are pairwise distinct which guarantees that $z$ generates $L$ over $K$. Hence $L = K[z]$ as required. $\square$

Proof. The polynomial $x^ p - 1$ splits completely over $K(\zeta )$ as its roots are $1, \zeta , \zeta ^2, \ldots , \zeta ^{p - 1}$. Hence $K(\zeta )/K$ is a splitting field and hence normal. The extension is separable as $x^ p - 1$ is a separable polynomial. Thus the extension is Galois. Any automorphism of $K(\zeta )$ over $K$ sends $\zeta $ to $\zeta ^ i$ for some $1 \leq i \leq p - 1$. Thus the Galois group is a subgroup of $(\mathbf{Z}/p\mathbf{Z})^*$. $\square$

Proof. Observe that for $d | e$ the subfield $K(\alpha ^ d)$ has $[K(\alpha ^ d) : K] = e/d$ and $[L : K(\alpha ^ d)] = d$ and that both extensions $K(\alpha ^ d)/K$ and $L/K(\alpha ^ d)$ are extensions as in the lemma.

We will use induction on the pair of integers $([L : L'], [L' : K])$ ordered lexicographically. Let $p$ be a prime number dividing $e$ and set $d = e/p$. If $K(\alpha ^ d)$ is contained in $L'$, then we win by induction, because then it suffices to prove the lemma for $L/L'/K(\alpha ^ d)$. If not, then $[L'(\alpha ^ d) : L'] = p$ and by induction hypothesis we have $L'(\alpha ^ d) = K(\alpha ^ i)$ for some $i | d$. If $i \not= 1$ we are done by induction. Thus we may assume that $[L : L'] = p$.

If $e$ is not a power of $p$, then we can do this trick again with a second prime number and we win. Thus we may assume $e$ is a power of $p$.

If the characteristic of $K$ is $p$ and $e$ is a $p$th power, then $L/K$ is purely inseparable. Hence $L/L'$ is purely inseparable of degree $p$ and hence $\alpha ^ p \in L'$. Thus $L' = K(\alpha ^ p)$ and this case is done.

The final case is where $e$ is a power of $p$, the characteristic of $K$ is not $p$, $L/L'$ is a degree $p$ extension, and $L = L'(\alpha ^{e/p})$. Claim: this can only happen if $e = p$ and $L' = K$. The claim finishes the proof.

First, we prove the claim when $K$ contains a primitive $p$th root of unity $\zeta $. In this case the degree $p$ extension $K(\alpha ^{e/p})/K$ is Galois with Galois group generated by the automorphism $\alpha ^{e/p} \mapsto \zeta \alpha ^{e/p}$. On the other hand, since $L$ is generated by $\alpha ^{e/p}$ and $L'$ we see that the map

is an isomorphism of $K$-algebras (look at dimensions). Thus $L$ has an automorphism $\sigma $ of order $p$ over $K$ sending $\alpha ^{e/p}$ to $\zeta \alpha ^{e/p}$. Then $\sigma (\alpha ) = \zeta ' \alpha $ for some $e$th root of unity $\zeta '$ (as $\alpha ^ e$ is in $K$). Then on the one hand $(\zeta ')^{e/p} = \zeta $ and on the other hand $\zeta '$ has to be a $p$th root of $1$ as $\sigma $ has order $p$. Thus $e/p = 1$ and the claim has been shown.

Finally, suppose that $K$ does not contain a primitive $p$th root of $1$. Choose a primitive $p$th root $\zeta $ in some algebraic closure $\overline{L}$ of $L$. Consider the diagram

By Lemma 9.24.2 the vertical extensions have degree prime to $p$. Hence $[L(\zeta ) : K(\zeta )]$ is divisible by $e$. On the other hand, $L(\zeta )$ is generated by $\alpha $ over $K(\zeta )$ and hence $[L(\zeta ) : K(\zeta )] \leq e$. Thus $[L(\zeta ) : K(\zeta )] = e$. Similarly we have $[K(\alpha ^{e/p}, \zeta ) : K(\zeta )] = p$ and $[L(\zeta ) : L'(\zeta )] = p$. Thus the fields $K(\zeta ), L'(\zeta ), L(\zeta )$ and the element $\alpha $ fall into the case discussed in the previous paragraph we conclude $e = p$ as desired. $\square$