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Suppose I have $K = \mathbb{Q}(\theta)$ and let $f$ be the minimal polynomial of $f$ over $\mathbb{Q}$. Suppose $f$ has degree $n$ so that the degree of $K$ over $\mathbb{Q}$ is $n$. Suppose further that $K/\mathbb{Q}$ is a Galois extension. Let $G = \{ \sigma_{(j)} : j = 1, ..., n \}$ be the Galois group of $k /\mathbb{Q}$.

Does it then follow that every $\sigma_{(j)}(\theta)$ is distinct for each $j$?

Does this also imply that for any $\beta \in K \backslash {\mathbb{Q}}$, $\sigma_{(j)}(\beta)$ is distinct for each $j$?

Thank you very much!

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    $\begingroup$ Hint : for your first one, notice that an element in $K$ is of the form $P(\theta)$, with $P(X) \in \Bbb Q[X]$. $\endgroup$ – Watson Feb 6 '16 at 17:51
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    $\begingroup$ For part two consider $K=\Bbb{Q}(\sqrt2,\sqrt3)$, $\beta=\sqrt2$. $\endgroup$ – Jyrki Lahtonen Feb 6 '16 at 18:04
  • $\begingroup$ Thank you! I got the second part but I am not still quite getting the first part... $\endgroup$ – Johnny T. Feb 6 '16 at 18:50
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Hint 1 :

  • For your first question, notice that an element in $K$ is of the form $P(\theta)$, with $P(X) \in \Bbb Q[X]$ and that for any $g \in G$, $g(P(\theta)) = P(g(\theta))$.

Answer 1 :

Therefore, if $\sigma_j(\theta) = \sigma_k(\theta)$ for some $j,k$, then you can conclude that for any $x=P(\theta) \in K$, $$\sigma_j(x) = \sigma_j(P(\theta)) = P(\sigma_j(\theta))=P(\sigma_k(\theta))=\sigma_k(P(\theta))=\sigma_k(x),$$ so that $\sigma_j = \sigma_k$. Hence the answer is "yes".

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