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?



Squares in a triquadratic field
The field $L=\mathbb Q(\sqrt{-3},\sqrt{5},\sqrt{-7})$ has a lot of intermediate fields which we can exploit. Pick one of its index two subextensions, say $K=\mathbb Q(\sqrt{-3},\sqrt{5})$. The element $\alpha=1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$ of $L$ has as its only conjugate over $K$ the element $-1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$, and so the norm of $\alpha$ with respect to the extension $L/K$ is $$N\_{L/K}(\alpha)=-1170^2\cdot(-3)\cdot 5\cdot(-7)+19110^2=221457600 = 2^6\cdot 3^2\cdot 5^2\cdot 7\cdot13^3.$$
This norm is not a square in $K$ - for otherwise $K$ would ramify at $7$ and $13$. Therefore $\alpha$ is not a square in $L$.


This method should generalize quite nicely if you consider elements of the form $a+b\sqrt{n}$ inside a multiquadratic extension $L/\mathbb Q$, as picking a subfield $K$ of index $2$ not containing $n$, the norm will be given by $a^2-bn^2$, and if the factorization of this integer contains some primes in odd exponents which do not ramify in $K$, you know it will not be a square in $K$. It probably won't always give you the right answer but it should work in many cases of interest.


