How to take the conjugate of a number with more than 2 square roots I was doing some abstract algebra and I came across the problem of figuring out if $\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} : a,b,c,d \in \mathbb{Q}\}$ is a ring and further if it is a field. Part of this is proving that every element is invertible for a field which I believe to be the case in this example. 
My question is, is there a general way to take a conjugate of a number of this form and more generally numbers of the form,
$$x=a_{0}+ \sum_{n \neq k^2} a_{i}\sqrt{n}$$
Where each $a_i \in \mathbb{Q}$? 
In other words a number $y$ such that $\frac{1}{x} \times \frac{y}{y}$ involves no square roots in the denominator.

It's not really difficult to find the “rationalization” the hard way:
$$
\frac{1}{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}}
=
\frac{a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}}{(a+b\sqrt{2})^2-(c\sqrt{3}+d\sqrt{6})^2}
$$
The denominator becomes
$$
a^2+2ab\sqrt{2}+2b^2-3c^2-6cd\sqrt{2}-6d^2
$$
and now a further step will give you the final result. Not a pretty formula.
If you instead consider the map $x\mapsto(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})x$, its matrix is
$$
\begin{bmatrix}
a & b & 3c & 6d \\
b & a & 3d & 3c \\
c & 2d & a & 2b \\
d & c & b & a
\end{bmatrix}
$$
The determinant is exactly the denominator you're looking for as you can easily check.
