Galois Theory HT25, Kummer extensions

$K/F$ is a Kummer extension if:

• $x^n - 1$ splits in $F$
• $n$ and $\text{char}(F)$ are coprime
• $K/F$ is a splitting field for $x^n - a$ for some $a \in F$

Suppose:

• $K/F$ is a Kummer extension, i.e.
  • $x^n - 1$ splits in $F$
  • $n$ and $\text{char}(F)$ are coprime
  • $K/F$ is a splitting field for $x^n - a$ for some $a \in F$

Then:

• $K/F$ is Galois
• There is a unique group homomorphism $\text{Gal}(K/F) \hookrightarrow \mu _ n(K)$
• (therefore $\text{Gal}(K/F)$ is isomorphic to a subgroup of $\mathbb Z / n \mathbb Z$)

Because $\varepsilon \ne 1$ and $p$ is prime, $\varepsilon$ generates a cyclic group of order $p$ in $L^\times$, hence

\[t^p - a = \prod^{p-1}_{i = 0} (t - \varepsilon^i \gamma)\]

splits completely in $L$.

We may assume $a \ne 0$, since otherwise $L = F$ trivially. Hence $t^p - a$ has distinct roots and is therefore separable.

Its roots generate $M$ since $\gamma$ is a root and $M = L(\gamma)$. Hence $L$ is Galois over $F$.

To show it’s abelian, pick some $\sigma \in \text{Gal}(L/F)$. Since

\[\sigma(\gamma)^p = a\]

It follows that $\sigma(\gamma) = \varepsilon^{\psi(\sigma)} \gamma$ for some

\[\psi(\sigma) \in \\{0, 1, \ldots, p-1\\}\]

Since

\[\begin{aligned} (\sigma \circ \tau)(\gamma) &= \sigma(\varepsilon^{\psi(\tau)} \gamma) \\\\ &= \varepsilon^{\psi(\sigma) + \psi(\tau)} \alpha \\\\ &= (\tau \circ \sigma)(\gamma) \end{aligned}\]

It follows that $\sigma \circ \tau = \tau \circ \sigma$ (since $L = F(\gamma)$), and therefore $\text{Gal}(M/F)$ is abelian.

• $K/F$ is a Galois extension
• For some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^n = 1$ (note the root of unity is in $F$)
• $\text{char}(F)$ and $n$ are coprime
• $\text{Gal}(K/F)$ is a cyclic group of order $n$

Then $K/F$ is Kummer, i.e.

• $K/F$ is a Kummer extension, i.e.
  • $x^n - 1$ splits in $F$
  • $n$ and $\text{char}(F)$ are coprime
  • $K/F$ is a splitting field for $x^n - a$ for some $a \in F$
• …and thus in particular $F = K(\alpha)$ for some root $\alpha$ of $a$

Suppose:

• $K/F$ is a Galois extension
• $[K : F] = p$ where $p$ prime
• For some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^p = 1$

Then there exists $u \in K$ such that $u^p \in F$ and $K = F(u)$.

(i.e. they are simple and radical, this is a special case of a more general theorem; you also obtain a simple field extension $K = F(u)$ if $F$ contains an $n$-th root of unity, $\gcd(n, \text{char}(F)) = 1$ and $\text{Gal}(K/F) \cong \mathbb Z / n \mathbb Z$, although the proof is slightly more involved)

Let $G = \text{Gal}(K/F)$. Since $K/F$ is Galois, $ \vert G \vert = [K : F] = p$ and hence $G = \langle \sigma \rangle$ is a cyclic group of order $p$ where $\sigma : K \to K$ is an $F$-linear map satisfying $\sigma^p = 1$. The minimal polynomial of this map divides

\[t^p - 1 = \prod^{p-1}_{i = 0} (t-\varepsilon^i)\]

and therefore splits completely over $F$.

Hence $\sigma$ is diagonalisable since its minimal polynomial has no repeated roots. Since $\sigma \ne 1$, some eigenvalue $\sigma$ is a nontrivial $p$-th root of unity. Without loss of generality, $\varepsilon$ is an eigenvalue of $\sigma$ (we just have that for some $\varepsilon \in F$ where $\varepsilon \ne 1$, we have $\varepsilon^p = 1$).

Let $u \in K$ be a corresponding eigenvector, so

\[\sigma(u) = \varepsilon u\]

Since $\sigma(u) \ne u$, $u \notin F$ (otherwise it would be fixed). Therefore $[F(u) : F] > 1$, but this must divide $[K : F] = p$ by the tower law and since $p$ is prime, it follows that $[F(u) : F] = p$ therefore and $K = F(u)$.

To see $u^p \in F$, note that

\[\sigma(u^p) = \sigma(u)^p = (\varepsilon u)^p = \varepsilon^p u^p = u^p\]

so $u^p \in K^G = F$.