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.

My main question is the title: for an odd prime $p$, denote a primitive $p^{\text{th}}$ root of unity by $\zeta_p$. Is it true that $i$ is not contained in the cyclotomic extension $\mathbb{Q}(\zeta_p)$? If this is true, is the following proof correct, and if it is not true, where does the following proof break down:

Recall that the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{\pm p})$ where there is a $"+"$ is $p\equiv 1 \mod 4$ and a $"-"$ if $p\equiv 3 \mod 4$ (Source: exercise 11 of section 14.7 of Dummit and Foote). Assume to the contrary that $i \in \mathbb{Q}(\zeta_p)$. Then $\mathbb{Q}(i)$ is a quadratic extension of $\mathbb{Q}$ of degree 2 contained in $\mathbb{Q}(\zeta_p)$. But this yields an immediate contradiction since of course $\mathbb{Q}(\sqrt{\pm p}) \neq \mathbb{Q}(i)$. So $i\notin \mathbb{Q}(\zeta_p)$.

share|improve this question
    
looks good to me –  Scaramouche Jun 24 '12 at 17:29
    
I think your proof looks fine.. –  Cocopuffs Jun 24 '12 at 17:41

1 Answer 1

up vote 7 down vote accepted

Another proof: if $i\in{\mathbb Q}(\zeta_p)$ then ${\mathbb Q}(i\zeta_p) \subseteq {\mathbb Q}(\zeta_p)$. But $i\zeta_p$ is a primitive $(4p)$th root of unity, so we have a field of degree $\phi(4p)=2(p-1)$ over $\mathbb Q$ contained in a field of degree $\phi(p)=p-1$ over $\mathbb Q$, a contradiction.

share|improve this answer
    
+1 This is really clever. –  Derek Allums Jun 24 '12 at 18:31

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.