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I have the following question: Let $K|\mathbb{Q}$ be a numberfield and $x \in K$ with $N_{K|\mathbb{Q}}(x) = 1$. Is it true that $x$ or $-x$ is a totally positive element i.e. $\sigma(x) > 0$ or $\sigma(-x) > 0$ for all real embeddings $\sigma$ from $K$ to $\mathbb{C}$? For example let $K|\mathbb{Q}$ be a galois extension. Then by Hilbert 90 there exists an element $a \in K$ and an embedding $\mu$ such that $x = \mu(a)a^{-1}$. Now we can use the product formula of the Norm i.e. $N_{K|\mathbb{Q}}(x) = \prod\limits_{\sigma \in Gal(K|\mathbb{Q})}\sigma(\mu(a)a^{-1})$. Ok, nice. But where do we go from here? And what's about the non-galois case?

Thanks for your help.

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    $\begingroup$ Something is strange/missing in the formulation of the question. Consider $K = \mathbb{Q}(i)$ and $x=i$. $\endgroup$
    – user14972
    Dec 23, 2016 at 21:55
  • $\begingroup$ @Hurkyl no, that's fine, there are no real embeddings so it is vacuously true in this case. $\endgroup$ Dec 23, 2016 at 22:00
  • $\begingroup$ @Adam: Which is part of why it's weird, since "totally positive" usually implies "totally real"; i.e. all of the embeddings are real. $\endgroup$
    – user14972
    Dec 23, 2016 at 22:03
  • $\begingroup$ @Hurkyl I agree the question is non-standard, but he explained his terminology unambiguously is what I meant. $\endgroup$ Dec 23, 2016 at 22:04
  • $\begingroup$ @Adam: IMO, it's weird enough that it deserves comment, and since the OP didn't it's fairly likely to be a mistake. Which is why I prompted clarification. $\endgroup$
    – user14972
    Dec 23, 2016 at 22:05

1 Answer 1

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Let $f(x) = x^3 - 4x - 1$. If $\alpha$ is a root of $f(x)$, then $K = \mathbb{Q}(\alpha)$ is a totally real number field and is the splitting field of $f$. We can see from $f$ that $\alpha$ has norm $1$; however, $f$ has one positive and two negative roots, thus neither $\alpha$ nor $-\alpha$ is totally positive.

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  • $\begingroup$ I've already guessed. Thank you very much! $\endgroup$
    – Club-Mate
    Dec 23, 2016 at 22:20

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