How to take the conjugate of a number with more than 2 square roots

Question

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.

Answer 1

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.

Answer 2

$\alpha=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ has three conjugates. They are among the numbers $$ a\pm b\sqrt{2}\pm c\sqrt{3}\pm d\sqrt{6} $$ You have to choose the right signs.

The simple route for finding inverses (and more) is to note that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a finite-dimensional vector space over $\mathbb{Q}$ and to consider the map $x \mapsto \alpha x$. Prove that the map is an injective linear transformation (when $\alpha\ne0$) and conclude that it is surjective.

If you want an explicit expression for the inverse of $\alpha$, write the matrix of the map with respect to the basis $(1,\sqrt{2},\sqrt{3},\sqrt{6})$ and find its inverse. Or just solve a linear system.

Answer 3

The background story here is that your field is a splitting field of the polynomial $(x^2-2)(x^2-3)$, and as such the Galois group $\operatorname{Gal}(\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q)$ acts on it. It turns out that this group has four elements:

• $I$: Identity,
• $\Phi_2$: Automorphism that maps $\sqrt 2$ to $-\sqrt 2$ but leaves $\sqrt 3$ in place,
• $\Phi_3$: Automorphism that maps $\sqrt 3$ to $-\sqrt 3$ but leaves $\sqrt 2$ in place,
• $\Phi_6$: Automorphism that maps $\sqrt 2$ to $-\sqrt 2$ and $\sqrt 3$ to $-\sqrt 3$.

Now, for an element $x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$, its conjugates are defined as the maps of $x$ using all those automorphisms. In our case, what you will get is:

• $I(x)=x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$
• $\Phi_2(x)=a-b\sqrt 2+c\sqrt 3-d\sqrt 6$
• $\Phi_3(x)=a+b\sqrt 2-c\sqrt 3-d\sqrt 6$
• $\Phi_6(x)=a-b\sqrt 2-c\sqrt 3+d\sqrt 6$

The product of all those: $N(x)=I(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)$ must be mapped into itself by all those automorphisms, because those automorphisms make up a group. For example,

$$\Phi_2(N(x))=\Phi_2(I(x))\Phi_2(\Phi_2(x))\Phi_2(\Phi_3(x))\Phi_2(\Phi_6(x))=\Phi_2(x)I(x)\Phi_6(x)\Phi_3(x)=N(x)$$

And similar for $\Phi_3$ and $\Phi_6$.

Thus, as the Galois theory teaches us, $N(x)$ belongs to the field fixed by all the automorphisms, which coincides with $\mathbb Q$. In other words, $N(x)$ is always rational.

Moral: if $x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$, with $a,b,c,d$ rational: to find $\frac{1}{x}$, multiply both the numerator and denominator by $\Phi_2(x)\Phi_3(x)\Phi_6(x)=(a-b\sqrt 2+c\sqrt 3-d\sqrt 6)(a+b\sqrt 2-c\sqrt 3-d\sqrt 6)(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)$.

Edited to add: Wolfram Alpha has calculated for me that you will end up with the following in the denominator:

$$a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4$$

(yep, I know, it's horrid, but it is rational!)