4
$\begingroup$

While reading this solution set, I came across the following statement.

In a Galois extension $K / F$, the number of Galois conjugates of any $\alpha \in K$ is equal to:

\begin{equation*} \left| \text{Gal}(K/F) : \text{Gal}(K/F(\alpha)) \right| \end{equation*} and I am a bit perplexed where this number comes from.

After thinking about this question for a bit, I realized that there is a "backdoor" approach to figuring this out. We know that the minimal polynomial for $\alpha$, i.e. $m_{\alpha}(x)$, is equal to the squarefree part of the polynomial \begin{equation*} \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma(\alpha)) \end{equation*} which has degree equal to the number of distinct Galois conjugates $\sigma(\alpha)$. We also know that $\text{deg}(m_{\alpha}(x))$ is equal to $[F(\alpha): F]$. But this is exactly what the index of $\text{Gal}(K/F)$ in $\text{Gal}(K/F(\alpha))$ is equal to, since: \begin{align*} \left| \text{Gal}(K/F) : \text{Gal}\big(K/F(\alpha) \big) \right| & = \frac{ [K:F]}{[K:F(\alpha)]} \\[0.65em] & = [F(\alpha):F] \end{align*}

But I'm wondering if it is possible to see more "directly" that the number of (distinct) Galois conjugates of $\alpha$ is the number \begin{equation*} \left| \text{Gal}(K/F) : \text{Gal}\big(K/F(\alpha) \big) \right| \end{equation*}

$\endgroup$
6
  • $\begingroup$ The number of conjugates of $\;\alpha\in K\;$ ( over $\;F\;$ ) is the number of different roots of the minimal polynomial of $\;\alpha\;$ in $\;F[x]\;$ , which is $\;[F(\alpha):F]=\;$ the degree of the polynomial mentioned above. $\endgroup$
    – DonAntonio
    May 22, 2017 at 22:16
  • 1
    $\begingroup$ @DonAntonio That is more or less the OP's argument in the post. $\endgroup$
    – Kenny Wong
    May 22, 2017 at 22:18
  • $\begingroup$ @SamY. Why do you think that ${\rm Gal}(K/F(\alpha))$ is a normal subgroup of ${\rm Gal}(K/F)$? This is true if and only if $F(\alpha)$ is a normal extension of $F$; it is not true in general. [By the way, you also wrote ${\rm Gal}(K/F(\alpha))$ and ${\rm Gal}(K/F)$ the wrong way round in your post.] $\endgroup$
    – Kenny Wong
    May 22, 2017 at 22:19
  • $\begingroup$ @KennyWong Indeed. These posts that write "I don't understand this or that..." and then later it happens to be they actually do perplex me. I many times address the OP's very question and don't read the whole thing. $\endgroup$
    – DonAntonio
    May 22, 2017 at 22:20
  • $\begingroup$ @KennyWong You are right. I forgot that the symbol $|G:H|$ makes sense even when $H$ is not normal. When I wrote that question, I was thinking you needed $H$ to be a normal subgroup of $G$...but you are right - $|G:H|$ is just the number of left cosets of $H$ in $G$. $\endgroup$
    – Sam Y.
    May 22, 2017 at 22:25

1 Answer 1

3
$\begingroup$

$\newcommand{\Gal}{\text{Gal}}$ Sometimes when $|A| = |B|$, it is because there is really an underlying bijection between $A$ and $B$.

Since $|\Gal(K/F) : \Gal(K/F(\alpha))|$ is the number of cosets of $\Gal(K/F(\alpha))$ in $\Gal(K/F)$, and this is the same as the number of Galois conjugates of $\alpha$, we might ask if we can establish a bijection between cosets of $\Gal(K/F(\alpha))$ in $\Gal(K/F)$ and conjugates of $\alpha$.

This is true, and we can see it as follows: Let $S$ be the set of conjugates of $\alpha$. $G$ acts on $S$ transitively, and the stabilizer of $\alpha \in S$ is the set of Galois elements fixing $\alpha$, equivalently, fixing $F(\alpha)$, i.e. $\Gal(K/F(\alpha))$, which establishes that $S$ and $\Gal(K/F)/\Gal(K/F(\alpha))$ are isomorphic as $\Gal(K/F)$-sets.

If you only care about the numbers, then you can use the Orbit-Stabilizer theorem to conclude that $|\Gal(K/F)| = |S||\Gal(K/F(\alpha))|$.

Edit: If you'd rather have something more concrete, the bijection between cosets and conjugates is given by $\sigma\Gal(K/F(\alpha)) \mapsto \sigma\alpha$. You can check this is a bijection in the same way you would check the stuff about group actions above.

$\endgroup$
2
  • 1
    $\begingroup$ If $\sigma\alpha = \sigma'\alpha$ then $\sigma^{-1}\sigma\alpha = \alpha$ and thus $\sigma^{-1}\sigma'$ fixes alpha and is therefore in $\text{Gal}(K/F(\alpha))$, which means that $\sigma$ and $\sigma'$ are in the same coset. $\endgroup$
    – user263190
    May 23, 2017 at 2:23
  • $\begingroup$ I think this plus the last 3 lines should be your answer (and that $Gal(K/F(\alpha))$ is naturally a subgroup of $Gal(K/F)$) $\endgroup$
    – reuns
    May 23, 2017 at 2:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .