Lemma 9.24.1 (Kummer extensions). Let $L/K$ be a Galois extension of fields whose Galois group is $\mathbf{Z}/n\mathbf{Z}$. Assume moreover that the characteristic of $K$ is prime to $n$ and that $K$ contains a primitive $n$th root of $1$. Then $L = K[z]$ with $z^ n \in K$.
9.24 Kummer extensions
Let $K$ be a field. Let $n \geq 2$ be an integer such that $K$ contains a primitive $n$th root of $1$. Let $a \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root $b$ of the equation $x^ n = a$. Then $L/K$ is Galois. If $G = \text{Gal}(L/K)$ is the Galois group, then the map
\[ G \longrightarrow \mu _ n(K),\quad \sigma \longmapsto \sigma (b)/b \]
is an injective homomorphism of groups. In particular, $G$ is cyclic of order dividing $n$ as a subgroup of the cyclic group $\mu _ n(K)$. Kummer theory gives a converse.
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. $\square$
Lemma 9.24.2. Let $K$ be a field with algebraic closure $\overline{K}$. Let $p$ be a prime different from the characteristic of $K$. Let $\zeta \in \overline{K}$ be a primitive $p$th root of $1$. Then $K(\zeta )/K$ is a Galois extension of degree dividing $p - 1$.
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 $(\mathbf{Z}/p\mathbf{Z})^*$. $\square$
Lemma 9.24.3. Let $K$ be a field. Let $L/K$ be a finite extension of degree $e$ which is generated by an element $\alpha $ with $a = \alpha ^ e \in K$. If every $e$th root of unity in $L$ is contained in $K$, then any sub extension $L/L'/K$ is generated by $\alpha ^ d$ for some $d | e$.
Proof. Observe that for $d | e$ the subfield $K(\alpha ^ d)$ has $[K(\alpha ^ d) : K] = e/d$ and $[L : K(\alpha ^ d)] = d$. Let $L/L'/K$ be a subextension. Say $d = [L : L']$. If $\alpha ^ d \in L'$, then we have $L' = K(\alpha ^ d)$ for degree reasons. Let $P \in L'[x]$ be the minimal polynomial of $\alpha $ over $L'$. Then $P$ divides $x^ e - a$ and $P$ has degree $d$. Let us write
\[ x^ e - a = \prod \nolimits _{i = 1, \ldots , e} (x - \zeta _ i \alpha ) \]
in a splitting field of $x^ e - a$ over $L$. The $\zeta _ i$ are $e$th roots of unity and after renumbering we have
\[ P = \prod \nolimits _{i = 1, \ldots , d} (x - \zeta _ i \alpha ) \]
The constant term of $P$ is equal to
\[ c = (\prod \nolimits _{i = 1, \ldots , d} \zeta _ i) \alpha ^ d \]
and is in $L' \subset L$. Since $\alpha \in L$ this implies that $\zeta = \prod _{i = 1, \ldots , d} \zeta _ i$ is in $L$ and hence in $K$ by our assumption. Thus $\alpha ^ d = \zeta ^{-1}c \in L'$ and we conclude. $\square$
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