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What is the image of $\zeta_3$ under the non-identity embedding of $\mathbb{Q}(\zeta_3)$ in $\mathbb{C}$?

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$\zeta_3$ is...? –  Mariano Suárez-Alvarez Mar 1 '11 at 23:03
    
Cube root of unity $\frac{-1+\sqrt{-3}}{2}$. –  Jason Smith Mar 1 '11 at 23:16

3 Answers 3

up vote 7 down vote accepted

If $\zeta_3$ is a primitive cubic root of $1$, then its image under the non-identity embedding is $\zeta_3^{-1}$.

Indeed, there is an endomorphism $\sigma:\mathbb Q(\zeta_3^{-1})\to\mathbb Q(\zeta_3^{-1})$ such that $\sigma(\zeta_3)=\zeta_3^{-1}$, which in fact generates the Galois group of the field over $\mathbb Q$. The non-trivial embedding is then the composition of $\sigma$ with the trivial embedding.

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So, $\sigma(a+b\zeta_3)=a+b\sigma(\zeta_3)=a+b\zeta_3^{-1}$, for $a+b\zeta \in \mathbb{Z}[\zeta_3]$? –  Jason Smith Mar 1 '11 at 23:27

If you define an algebraic extension of $\Bbb Q$ as $K={\Bbb Q}[x]/(f(x))$ where $f(x)$ is an irreducible polynomial of degree $d$, you obtain the $d$ embeddings of $K$ by sending $x\mapsto\alpha$ where $\alpha$ is a complex root of $f(x)$ (they always exist, thanks to Gauss).

In the case of ${\Bbb Q}(\zeta_3)$, this is just ${\Bbb Q}[x]/(x^2-x+1)$ because $x^2-x+1$ is the quadratic irreducible factor of $x^3-1$. Thus you obtain two complex embeddings sending $$ x=\zeta_3\mapsto\alpha^\pm=\frac12(1\pm\sqrt{-3}). $$ If you decide to identify $\zeta_3=\alpha^{+}=(1+\sqrt{-3})/2$, a moment of thought (in the form of a short computation) will convince you that the "other" root, i.e. the other embedding, $\alpha^{-}$ is just $\zeta_3^2$. Mind that the two roots (and so the two embeddings) are switched by complex conjugation.

The identities $\zeta_3^2=\overline{\zeta}_3=\zeta_3^{-1}$ are also clear observing that $1$, $\alpha^{+}$ and $\alpha^{-}$ are the three complex roots of $x^3-1$, so that their product must be $1$.

Adrian's answer gives the generalization to the $p$-th cyclotomic field

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If you have a cyclotomic field $K = \mathbb{Q}(\zeta)$ where $\zeta := e^{2 \pi i /p}$ with $p$ an odd prime number, then it can be easily seen that its $p-1$ embeddings in $\mathbb{C}$ are given by $\sigma _i : K \hookrightarrow \mathbb{C}$ with $\sigma _i (\zeta) = \zeta ^{i}$ for $1 \leq i \leq p-1$. This is just a consequence of the fact that the minimum polynomial for $\zeta$ over $\mathbb{Q}$ is $$f(t) = \frac{t^p - 1}{t - 1} = t^{p-1} + t^{p-2} + \cdots + t + 1$$

and $f(t)$ factors as $$f(t) = (t - \zeta)(t - \zeta ^{2}) \cdots (t - \zeta ^{p-1})$$

So in general the image of $\zeta$ under the non-trivial embeddings of $K$ will be the powers $\zeta \mapsto \zeta^{i}$ for $2 \leq i \leq p-1$, which in your case for $p = 3$ will only correspond to $\zeta_{3} \mapsto \zeta_{3}^{2}$ but in this case $\zeta_{3}^{2} = \zeta_{3}^{-1}$ as in Mariano's answer.

And about the question you posed in your comment to Mariano's answer, yes, if $a + b \zeta_{3} \in \mathbb{Z}[\zeta_{3}]$ then $\sigma_{2}(a + b \zeta_{3}) = a + b \zeta_{3}^{2} = a + b \zeta_{3}^{-1}$.

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