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Let $\mathbb{Q}(\alpha)$ be a number field, and suppose that $[\mathbb{Q}(\alpha) : \mathbb{Q} ] = m$. Then there are precisely $m$ different embeddings of $\mathbb{Q}(\alpha)$ into $\mathbb{C}$, essentially by sending $\alpha$ to each of its $m$ conjugates.

On the other hand, some of the images of these embeddings can coincide, for instance if $\alpha = i$ (in which case the other conjugate is $-i$, and $\mathbb{Q}(i) = \mathbb{Q}(-i)$).

QUESTION: is there a criterion (for instance in terms of the Galois group of the extension) which tells us precisely how many different images we obtain ?

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    $\begingroup$ Shouldn't it just be something like $[K:\mathbb Q]/\#\mathrm{Aut}(K/\mathbb Q)$? $\endgroup$
    – Ariel Weiss
    Jun 1, 2022 at 13:15
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    $\begingroup$ Indeed. And for $K=\mathbb Q(\alpha)$, the size of $\mathrm{Aut}(K/\mathbb Q)$ equals the number of conjugates of $\alpha$ inside $\mathbb Q(\alpha)$. $\endgroup$
    – Emil Jeřábek
    Jun 1, 2022 at 13:20
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    $\begingroup$ Or in terms of the Galois group, the centralizer of the Galois group in the symmetric group. $\endgroup$
    – Will Sawin
    Jun 1, 2022 at 13:56
  • $\begingroup$ @EmilJeřábek My comment was supposed to go on the end of your comment. "The size of $\operatorname{Aut}(K/\mathbb Q)$ equals $\dots$ or, in terms of the Galois group, $\dots$" $\endgroup$
    – Will Sawin
    Jun 1, 2022 at 14:20
  • $\begingroup$ @WillSawin Oh I see, all right. $\endgroup$
    – Emil Jeřábek
    Jun 2, 2022 at 17:27

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