When is $\sqrt n$ in $\Bbb Q[\omega_m]$?

Question

Given a positive integer $n$ and a primitive $m^{th}$ root of unity $\omega_m$ over $\Bbb Q$, how could one determine if $\sqrt{n}$ lies in $\Bbb Q[\omega_m]$?

In the case of $n=p>0$ being an odd prime, the question is known by some algebraic number theory. Indeed primes ramified in $\Bbb Q\left[\sqrt p\right]$ are $p$ and if $p\equiv 3$ modulo $4$, also $2$. If $p\equiv 1$ modulo $4$, the inclusion $\Bbb Q\left[\sqrt p\right]\subset\Bbb Q\left[\omega_m\right]$ requires $p$ to also ramify in $\Bbb Q[\omega_m]$, which happens iff $p|m$. If $p\equiv 3$ modulo $4$, then $2$ also need to ramify in $\Bbb Q[\omega_m]$, forcing $4p|m$. Finally $\sqrt 2\in\Bbb Q[\omega_m]$ iff $8|m$. Conversely one verifies $p\equiv 1$ modulo $4$ implies $\sqrt{p}\in\Bbb Q[\omega_p]$ and that $p\equiv 3$ modulo $4$ implies $\sqrt{p}\in\Bbb Q[\omega_{4p}]$. The following criteria summarize the situation:

Theorem 1. Let $p>0$ be a positive prime. If $p\equiv 1$ modulo $4$, then $\sqrt p\in\Bbb Q[\omega_m]$ iff $p|m$. If $p\equiv 3$ modulo $4$, then $\sqrt p\in\Bbb Q[\omega_m]$ iff $4p|m$. Finally $\sqrt 2\in\Bbb Q[\omega_m]$ iff $8|m$.

Are there similar criteria to determine if $\sqrt{n}$ lies in a given cyclotomic field $\Bbb Q[\omega_m]$ for integer $n>0$? From the prime $p$ case a necessary condition would be $n|m$, but this is clearly not sufficient.

Answer 1

Recall that $\mathrm{Gal}(\mathbb Q^{ab}/\mathbb Q)\simeq\widehat{\mathbb Z}^\times$ which sends $\mathrm{Frob}_p$ to $[p]$. Now the homomorphism $\varphi\colon\mathrm{Gal}(\mathbb Q^{ab}/\mathbb Q)\to\{\pm1\}$ sends $\mathrm{Frob}_p$ to $\big(\frac np\big)$. Thus your question is equivalent to computing the minimal $m$ such that $\varphi(1+m\widehat{\mathbb Z})=1$. In other words, your question is equivalent to computing the kernel of $\varphi$.

For example, when $n=6$ then by quadratic reciprocity $$\big(\frac 6p\big)=\big(\frac2p\big)\big(\frac3p\big)=(-1)^{\frac{(p-1)(p-3)}8}\big(\frac p3\big),$$ so $$\ker(\varphi)=\{a\in\widehat{\mathbb Z}^\times:a\equiv1\ \mathrm{mod}\ 3,\ a\equiv1,3\ \mathrm{mod}\ 8\}.$$ Thus $\varphi(1+m\widehat{\mathbb Z})=1$ is equivalent to $24|m$. Thus $\sqrt6\in\mathbb Q(\zeta_m)$ is equivalent to $24|m$.

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The following is the complete characterization, which follows readily from the above and quadratic reciprocity. Let $n$ be square-free. Then $\sqrt n\in\mathbb Q(\zeta_m)$ if and only if:

• if $n$ is even then $4n|m$
• if $n\equiv1\pmod4$ then $n|m$
• if $n\equiv3\pmod4$ then $4n|m$.

Answer 2

The other solution uses class field theory. I will avoid that here.

Your question is the same as asking when $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$.

By replacing $n$ by is biggest squarefree factor, we can assume $n$ is squarefree. And surely you are not interested in $n$ being a square, so we take $n$ to be a squarefree integer other than $1$. When $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$, primes that ramify in the quadratic field also ramify in the cyclotomic field.

When $n$ is squarefree and not $1$, which primes ramify in $\mathbf Q(\sqrt{n})$? An odd prime ramifies in $\mathbf Q(\sqrt{n})$ if and only if it divides $n$, while $2$ ramifies in $\mathbf Q(\sqrt{n})$ if and only if $n \equiv 2, 3 \bmod 4$.

Which primes ramify in $\mathbf Q(\omega_m)$? When $p \nmid m$, $p$ is unramified in $\mathbf Q(\omega_m)$ since $x^m - 1 \bmod p$ is separable. When $p \mid m$ and $p > 2$, $p$ is ramified in $\mathbf Q(\omega_p)$, so $p$ is ramified in the larger field $\mathbf Q(\omega_m)$. When $m$ is divisible by $2$ just once, $2$ is unramified in $\mathbf Q(\omega_m)$ since this field equals $\mathbf Q(\omega_{m/2})$ and $m/2$ is odd. When $4 \mid m$, $2$ is ramified in $\mathbf Q(i)$ and thus also in the larger field $\mathbf Q(\omega_m)$.

By looking at ramified primes, if $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$ then every odd prime factor of $n$ is a factor of $m$, so when $n$ is odd we have $n \mid m$ (because $n$ is squarefree). When $n \equiv 3 \bmod 4$, also $2$ ramifies in $\mathbf Q(\sqrt{n})$, so $4 \mid m$, and thus $4n \mid m$.

I am going to look only at odd $n$ for simplicity. A necessary condition to have $\mathbf Q(\sqrt{n}) \subset\mathbf Q(\omega_m)$ is then $n \mid m$ when $n \equiv 1 \bmod 4$ and $4n \mid m$ when $n \equiv 3 \bmod 4$.

Now we want to prove the converse: if $n$ is odd with $n \mid m$ when $n \equiv 1 \bmod 4$ and $4n \mid m$ when $n \equiv 3 \bmod 4$, then $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$.

When $n = -1$ this is immediate: $4 \mid m$ and $\mathbf Q(\sqrt{n}) = \mathbf Q(i)\subset \mathbf Q(\omega_m)$. Now let $n$ be odd and not be $\pm 1$, so $n$ has prime factors.

Write $n = \pm p_1\cdots p_r$ where the $p_i$'s are distinct odd primes. In order to show $\sqrt{n} \in \mathbf Q(\omega_m)$, we're going to show $\sqrt{\pm p_i} \in \mathbf Q(\omega_m)$ for a careful choice of signs on each prime and then multiply together those square roots to show $\sqrt{n} \in \mathbf Q(\omega_m)$.

To handle signs nicely, it is convenient to consider, when $n$ is odd, the odd number $n^* = (-1)^{(n-1)/2}n$. So $n^* = \pm n$ with $n^* \equiv 1 \bmod 4$. Check $(-n)^* = -(n^*)$. Thus $(\varepsilon n)^* = \varepsilon (n^*)$ when $\varepsilon = \pm 1$. Also check $(n_1n_2)^* = n_1^*n_2^*$ when $n_1$ and $n_2$ are odd.

Let the prime factorization of $n$ be $\varepsilon p_1\cdots p_r$ where $\varepsilon = \pm 1$ and the $p_i$'s are distinct odd primes. Then $\varepsilon n = p_1\cdots p_r$, so $(\varepsilon n)^* = p_1^*\cdots p_r^*$. Since $(\varepsilon n)^* = \varepsilon n^*$,
$$ n^* = \varepsilon p_1^*\cdots p_r^*. $$

Case 1: $n \equiv 1 \bmod 4$, so $n \mid m$. Then $n^* = n$, so $n = p_1^*\cdots p_r^*$. Thus $\sqrt{p_1^*}\cdots\sqrt{p_r^*}$ is a square root of $n$. Using Gauss sums, $\sqrt{p^*} \in \mathbf Q(\omega_p)$ when $p$ is an odd prime. (Explicitly, $(\sum_{k=1}^{p-1} (\frac{k}{p})\omega^k)^2 = p^*$, so the Gauss sum $\sum_{k=1}^{p-1} (\frac{k}{p})\omega^k$ is a square root of $p^*$ and its very formula shows it is in $\mathbf Q(\omega_p)$.) When $n \equiv 1 \bmod 4$, each $\sqrt{p_i^*}$ is in $\mathbf Q(\omega_{p_i})$, which is in $\mathbf Q(\omega_m)$, so their product $\sqrt{n}$ is in $\mathbf Q(\omega_m)$.

Case 2: $n \equiv 3 \bmod 4$, so $4n \mid m$. Then $i \in \mathbf Q(\omega_m)$. We have $n^* = -n$, so $-n = p_1^*\cdots p_r^*$ and $\sqrt{p_1^*}\cdots\sqrt{p_r^*}$ is a square root of $-n$. Using Gauss sums, $\sqrt{p_i^*} \in \mathbf Q(\omega_{p_i}) \subset \mathbf Q(\omega_m)$. so $\sqrt{-n} \in \mathbf Q(\omega_m)$. Since $i \in \mathbf Q(\omega_m)$, we also get $\sqrt{n} \in \mathbf Q(\omega_m)$.

Remark. When $n$ is squarefree, the least $m$ such that $\mathbf Q(\sqrt{n}) \subset \mathbf Q(\omega_m)$ is $|{\rm disc}(\mathbf Q(\sqrt{n})|$, which is $|n|$ when $n \equiv 1 \bmod 4$ and is $4|n|$ when $n \not\equiv 1 \bmod 4$.