Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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*}

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
    
Ok, understood, thanks. Though I wonder: What is $\phi$ ? –  user47580 Jan 30 '13 at 16:25
    
@user47580 - en.wikipedia.org/wiki/Euler%27s_totient_function –  Belgi Jan 30 '13 at 17:13
    
I'm glad that helps! –  Belgi Jan 30 '13 at 17:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.