TITLE: Squares in a triquadratic field
QUESTION [12 upvotes]: I would like to know (as part of an attempt to streamline some calculations in the cohomology of a Morava stabiliser group) whether $1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$ is a square in $\mathbb{Q}(\sqrt{-3},\sqrt{5},\sqrt{-7})$.  What is an efficient method for this kind of question?

REPLY [10 votes]: Set $K = \mathbf Q(\sqrt{-3},\sqrt{5},\sqrt{-7})$ and $\alpha = 1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$.
I think the "best" way to show $\alpha$ is not a square in $K$ is not to show something else is not a square in a (smaller) number field, but to show something else is not a square in a finite field, and more precisely in $\mathbf Z/p\mathbf Z$ for some prime $p$.  We'll wind up doing this with $p = 79$.
To prove $\alpha$ is not a square in $K$, or equivalently in $\mathcal O_K$, it suffices to find a (nonzero) prime ideal $\mathfrak p$ in $\mathcal O_K$ such that $\alpha \bmod \mathfrak p$ is not a square. There are many such $\mathfrak p$ if $\alpha$ is not a square: the Chebotarev density theorem implies that a nonsquare in $\mathcal O_K$ is not a square mod $\mathfrak p$ for a set of $\mathfrak p$ with density $1/2$ (natural density or Dirichlet density).  That means if such $\mathfrak p$ exist, it should not take very long to find one, just like for a nonsquare integer $b$ it should not take long to find a prime $p$ such that $b \bmod p$ is not a square (the set of such $p$ has density $1/2$).
How can we go about searching for a prime ideal $\mathfrak p$ where $\alpha \bmod \mathfrak p$ is not a square? In practice it's a good idea to look for $\mathfrak p$ lying over a prime number $p$ that splits completely in $\mathcal O_K$ since then $\mathcal O_K/\mathfrak p \cong \mathbf Z/p\mathbf Z$ and it's computationally easy to work in a field of prime order. The set of such $p$ has (natural) density $1/[K:\mathbf Q] = 1/8$ and $p$ splits completely in $K$ if and only if it splits completely in the quadratic fields $\mathbf Q(\sqrt{-3})$, $\mathbf Q(\sqrt{5})$, and $\mathbf Q(\sqrt{-7})$.  These splitting conditions are the congruences
$$
p \equiv 1 \bmod 3, \ \ 
p \equiv 1, 4 \bmod 5, \ \ 
p \equiv 1, 2, 4 \bmod 7.
$$
By the Chinese remainder theorem, these conditions correspond to six  congruence conditions mod $105$, which turn out to be
$$
p \equiv 1, 4, 16, 46, 64, 79 \bmod 105.
$$
The first prime fitting each of these six conditions is, respectively,
$$
211, \ \ 109, \ \ 331, \ \ 151, \ \ 379, \ \ 79.
$$
The smallest of these primes is $79$, so $79$ is the smallest prime splitting completely in $K$. Let's work with $p = 79$. Since it splits completely in $K$, there are $8$ primes $\mathfrak p_1, \ldots, \mathfrak p_8$ lying over $79$ in $\mathcal O_K$. For each $\mathfrak p_i$, $\mathcal O_K/\mathfrak p_i \cong \mathbf Z/79\mathbf Z$. We will reduce $\alpha$ modulo some of these $\mathfrak p_i$ until we find a nonsquare in $\mathbf Z/79\mathbf Z$. Since $1170 \equiv 64 \bmod 79$ and
$19110 \equiv 71 \bmod 79$, for each $\mathfrak p_i$ lying over $79$ we have
$$
\alpha \equiv 64\sqrt{-3}\sqrt{5}\sqrt{-7}-71 \bmod \mathfrak p_i.
$$
How can we describe the primes $\mathfrak p_i$?  The congruences
$a^2 \equiv -3 \bmod 79$, $b^2 \equiv 5 \bmod 79$, and $c^2 \equiv -7 \bmod 79$ each have two solutions: $a \equiv 32, 47 \bmod 79$,
$b \equiv 20, 59 \bmod 79$, and $c \equiv 25, 54 \bmod 79$.
Picking one solution for each of these three congruences mod $79$ distinguishes the $8$ different prime ideals $\mathfrak p_i$ from each other.
Example 1. There is a prime $\mathfrak p_1$ over $79$ where
$$
\sqrt{-3} \equiv 32 \bmod \mathfrak p_1, \ \ 
\sqrt{5} \equiv 20 \bmod \mathfrak p_1,  \ \ 
\sqrt{-7} \equiv 25 \bmod \mathfrak p_1.
$$
That makes
$$
\alpha \equiv 64(32)(20)(25)-71 \equiv 10 \bmod 79, 
$$
but $(\frac{10}{79}) = 1$ by quadratic reciprocity. Explicitly, $10 \equiv 22^2 \bmod 79$, so $\alpha \equiv 22^2 \bmod \mathfrak p_1$. That didn't work out as we had hoped.
Example 2. There is a prime $\mathfrak p_2$ over $79$ where
$$
\sqrt{-3} \equiv 47 \bmod \mathfrak p_2, \ \ 
\sqrt{5} \equiv 20 \bmod \mathfrak p_2,  \ \ 
\sqrt{-7} \equiv 25 \bmod \mathfrak p_2.
$$
That makes
$$
\alpha \equiv 64(47)(20)(25)-71 \equiv 6 \bmod 79, 
$$
and $(\frac{6}{79}) = -1$ by quadratic reciprocity. Thus $\alpha \bmod \mathfrak p_2$ is not a square, so $\alpha$ is not a square in $\mathcal O_K$ and we are done.