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This is a question from Lang's ANT, Thm 6, ch.IV, $\S2$.

It states that every quadratic extension of $\mathbb{Q}$ is contained in a cyclotomic extension and that it's a direct consequence of the following result:

Let $\zeta_n$ be a primitive $n$-th root of unity for $n$ odd and $$S=\sum_\nu\left(\frac{\nu}{n}\right)\zeta^\nu_n$$ the sum being taken over non-zero residue classes mod $n$. Then $$S^2=\left(\frac{-1}{n}\right)n$$

Here, I guess the brackets denote the Jacobi symbol though Lang doesn't make it clear.

Anyway, assuming this, I understand that if the quadratic extension $K$ looks like $\mathbb{Q}(\sqrt{n})$ where n is a positive square-free odd integer then $K\subset \mathbb{Q}(\zeta_n)$.

Question: how does the result help us if either $n$ is a negative or an even (square-free) integer? Do we need some other results on cyclotomic extensions?

Many thanks in advance.

Mark Rodriguez
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    Yes, it's the Jacobi symbol. For negative $n$, consider $\mathbb{Q}(\zeta_{4\lvert n\rvert})$, which contains $\sqrt{\lvert n\rvert}$ and $i$, and for $m = 2n$, consider $\mathbb{Q}(\zeta_{8n})$, which also contains $\sqrt{2}$. – Daniel Fischer Aug 24 '13 at 21:55
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    @Daniel: that looks like a complete answer to me. Would you mind posting it? – Qiaochu Yuan Aug 24 '13 at 23:17

1 Answers1

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The symbol $\left(\frac{\nu}{n}\right)$ is indeed the Jacobi symbol. Note that since

$$S^2 = \left(\frac{-1}{n}\right)n,$$

it might be the case that $\mathbb{Q}(\zeta_n)$ contains $\sqrt{-n}$ and not $\sqrt{n}$.

But, since $\mathbb{Q}(\zeta_{4n})$ contains $\mathbb{Q}(\zeta_n)$ and $i$, the cyclotomic extension $\mathbb{Q}(\zeta_{4n})$ contains both $\sqrt{n}$ and $\sqrt{-n}$. That also covers the case of negative odd squarefree $n$, $\mathbb{Q}(\zeta_{4\lvert n\rvert})$ does the trick.

For even $n$, note that $$\zeta_8 = \frac{1+i}{\sqrt{2}}$$

and hence $\mathbb{Q}(\zeta_8)$ contains $\sqrt{2} = \zeta_8 + \zeta_8^{-1}$. Then, for squarefree $n = 2m$, the cyclotomic extension $\mathbb{Q}(\zeta_{8\lvert m\rvert})$ contains $\sqrt{n}$.

Daniel Fischer
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