1

Given any positive prime $p$, there is some cyclotomic extension $\Bbb Q[\omega_n]$ of $\Bbb Q$ containing $\sqrt p$ as a consequence of Kronecker–Weber theorem. But more specifically, given any positive integer $n$ is there any nice way to tell if $\sqrt p\in\Bbb Q[\omega_n]$? I am looking for a way to compute the "yes or no" just like Legendre symbol, or any related results.

This question arises naturally when one attempts to compute the degree of the extension $\Bbb Q\left[\omega_n,\sqrt{p}\right]$ over $\Bbb Q$, in which case $\Bbb Q\left[\omega_n,\sqrt p\right]$ has degree $2$ over $\Bbb Q[\omega_n]$ if and only if $\sqrt p\in\Bbb Q[\omega_n]$. This computation is crucial in determining the degree of the splitting field of $x^n-p$ over $\Bbb Q$.

William Sun
  • 2,463
  • 9
  • 17

1 Answers1

1

The primes ramified in $\mathbb Q(\sqrt{p})$ are $p$ and, if $p\not\equiv1(4)$, also $2$. On the other hand, the primes ramified in $\mathbb Q(\omega_n$) are the odd primes dividing $n$, and also $2$ if $4|n$. So necessary conditions for $\mathbb Q(\sqrt{p})\subset\mathbb Q(\omega_n)$ are (primes ramified in the subfield should stay ramified in the containing field):

for $p\equiv1(4)$: $p|n$

for $p\equiv-1(4)$: $4p|n$

and by the comment of Jyrki Lahtonen (that $\sqrt{p}\in Q(\omega_p)$ in the 1st case and $\sqrt{p}\in Q(\omega_{4p})$ in the second case) they are also sufficient.

What remains is $p=2$, where the necessary & sufficient condition is $8|n$. Here the ramification argument only gives $4|n$, but in that case $1+i\in\mathbb{Q}(\omega_n)$, and if $\sqrt2$ is in that field too, then so is $\frac{1+i}{\sqrt2}=\omega_8$, so indeed $8|n$.

user8268
  • 21,348