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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 $i\notin \mathbb{Q}(\zeta_p)$.

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

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+1 This is really clever. – Derek Allums Jun 24 '12 at 18:31

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