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I have to solve an exercise that requires to find all extensions $\hat{\sigma}:\mathbb{Q}\left(\sqrt{2},\xi\right)\rightarrow\mathbb{Q}\left(\sqrt{2},\xi\right)$ (with $\xi$ being a third root of the unity) of the field morphism

$$ \sigma:\mathbb{Q}\left(\sqrt{2}\right)\rightarrow\mathbb{Q}\left(\sqrt{2}\right),\ a+b\sqrt{2}\mapsto a-b\sqrt{2}. $$ What I don't understand is the following:

  • Shouldn't extensions of morphisms preserve the range of mapping, i.e. should we be looking at extensions $\hat{\sigma}:\mathbb{Q}\left(\sqrt{2},\xi\right)\rightarrow\mathbb{Q}\left(\sqrt{2}\right)$ of the above morphism ?

  • Are all the possible extensions the mappings $\hat{\sigma}_{1},\hat{\sigma}_{2}:\mathbb{Q}\left(\sqrt{2},\xi\right)\rightarrow\mathbb{Q}\left(\sqrt{2}\right)$ given by \begin{eqnarray*} & \hat{\sigma}_{1}:\sqrt{2}\mapsto-\sqrt{2},\ \xi\mapsto\xi\\ & \hat{\sigma}_{2}:\sqrt{2}\mapsto-\sqrt{2},\ \xi\mapsto\xi^{2}\ ? \end{eqnarray*}

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What you want are maps $\hat{\sigma}:\mathbb{Q}(\sqrt{2},\zeta_{3})\to\mathbb{Q}(\sqrt{2},\zeta_{3})$ s.t $\hat{\sigma}|_{\mathbb{Q}(\sqrt{2})}=\sigma$ – Belgi Jan 30 '13 at 16:02
up vote 2 down vote accepted

You wish to extend $\sigma$ to $\mathbb{Q}(\sqrt{2},\zeta_{3})$. Since you know where $\hat{\sigma}$ maps $\mathbb{Q}(\sqrt{2})$ and since the extension $$\mathbb{Q}(\sqrt{2},\zeta_{3})/\mathbb{Q}(\sqrt{2})$$ is generated by $\zeta_{3}$all that remains is to determine $\hat{\sigma}$ of $\zeta_{3}$.

You know that $\hat{\sigma}$ needs to map $\zeta_{3}$to a conjugate of his, there are $\phi(3)=2$ different options and you have found both of them.

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Ok, understood, thanks. Though I wonder: What is $\phi$ ? – user47580 Jan 30 '13 at 16:25
@user47580 - – Belgi Jan 30 '13 at 17:13
I'm glad that helps! – Belgi Jan 30 '13 at 17:13

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