Notes - Galois Theory HT25, Kummer extensions

$L/F$ is a cyclotomic field extension of $F$ if $L = F(\varepsilon)$ where $\varepsilon \in L$ and $\varepsilon^p = 1$ for some prime $p$.

If $\varepsilon = 1$, $L = F$ and the result is trivial. So we may assume that $\varepsilon \ne 1$.

Since $p$ is a prime, $\varepsilon$ generates a cyclic group of order $p$ in $L^\times$.

Hence (@todo) $t^p - 1$ splits completely in $L$ as

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

So $L$ is a splitting field of $t^p - 1$ over $F$. Since $t^p - 1$ has no repeated roots, it is separable.

So $L/F$ is Galois.

To show commutativity, fix some $\sigma \in \text{Gal}(L/F)$. Since $\sigma(\varepsilon)^p = \sigma(1) = 1$, it follows $\sigma(\varepsilon) \in \{1, \varepsilon, \ldots, \varepsilon^{p-1}\}$ (@todo).

Let $\chi : \text{Gal}(L/F) \to \mathbb N$ be the function such that

\[\sigma(\varepsilon) = \varepsilon^{\chi(\sigma)}\]

for any $\sigma$. For $\sigma, \tau \in \text{Gal}(L/F)$,

\[\begin{aligned} (\sigma \circ \tau)(\varepsilon) &= \sigma(\varepsilon^{\chi(\tau)}) \\\\ &= (\varepsilon^{\chi(\sigma)})^{\chi(\tau)} \\\\ &= \varepsilon^{\chi(\sigma) \chi(\tau)} \\\\ &= (\tau \circ \sigma)(\varepsilon) \end{aligned}\]

Since $L = F(\varepsilon)$, it follows that $\sigma \circ \tau = \tau \circ \sigma$ for all $\sigma, \tau$ and therefore $\text{Gal}(L/F)$ is abelian.

$L/F$ is a Kummer extension if $L$ contains an element $\varepsilon$ such that $\varepsilon^p = 1$ and $\varepsilon \ne 1$ (i.e. a $p$-th root of unity) and it also contains an element $\gamma$ such that $\gamma^p \in F$ (i.e. a $p$-th root).

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.

Let $G = \text{Gal}(K/F)$.

Since $K/F$ is Galois, $ \vert G \vert = [K : F] = p$. Hence $G = \langle \sigma \rangle$ is a cyclic group of order $p$.

$\sigma : K \to K$ is an $F$-linear map satisfying $\sigma^p = 1$, and hence 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 $\sigma$, viewed as a linear map, 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 (@todo), $\varepsilon$ is an eigenvalue of $\sigma$.

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

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

Since $\sigma$ is also a ring homomorphism,

\[\begin{aligned} \sigma(u^p) &= \sigma(u)^p \\\\ &= (\varepsilon u)^p \\\\ &= \varepsilon^p u^p \\\\ &= u^p \end{aligned}\]

so $u^p \in K^G = F$. Since $\sigma(u) \ne u$, $u \notin F$ (otherwise it would be fixed.) Therefore

\[[F(u) : F] > 1\]

But this divides $[K : F] = p$ by the tower law. Since $p$ is prime, it follows that

\[[F(u) : F] = p\]

and $K = F(u)$.