I would like to know how to calculate fractions in the field $\mathbb Q(\sqrt 2,\sqrt 3)$, for example:
$$\frac{A+B \sqrt 2+C \sqrt 3 + D \sqrt 6}{E+F \sqrt2+G \sqrt3 + H \sqrt6}$$
Couldn't this be done with techniques from linear algebra?
Clearly this can be done with the help of an algebra package like Mathematica, but I would like to get an idea how it is done "manually"? And on which theories it is based.
In general if you extend field $F$ as $F[\sqrt p]$, then you can construct inverse by doing the following: $$ \frac1{a+b\sqrt p} = \frac{a-b\sqrt{p}}{a^2-b^2p} = (a^2-b^2p)^{-1}(a-b\sqrt p). $$
You can consider $Q[\sqrt 2, \sqrt 3]=Q[\sqrt 2][\sqrt 3]$: $$ \frac1{E+F \sqrt2+G \sqrt3 + H \sqrt6} = \frac1{(E+F\sqrt2)+(G+H\sqrt2)\sqrt3} = \frac{(E+F\sqrt2)-(G+H\sqrt2)\sqrt3}{(E+F\sqrt2)^2-3(G+H\sqrt2)^2} $$
Now you have to invert the denominator from only $Q[\sqrt 2]$.
You can use base $\frac 12\sqrt 6+\frac 12\sqrt 2$, and a vertical row of $\sqrt 2$. That's what I normally use here.
If you want to do divisions, you do this by taking the three conjugates, and multiplying them together, and then multiply that by the fraction.
For $a+b\sqrt2+c\sqrt3+d\sqrt6$, the three conjugates keep a and one of the other letters positive, and change the other two. The product of the four is an integer, so you multiply the product of the three by the numerator, and divide by the product of the fourth.
Note that the integer set here allows halves on $\sqrt2+\sqrt6$ as well as integers.
Unless you can pick the factors, this method works for all instances of $\mathbb Z[\sqrt x, \sqrt y]$.
The denomator is E+Fr2+Gr3+Hr6 In terms of the matrix, we use
E 2F 3G 6H This matrix converts the vector
F E 3H 3G (E F G H) into four colum-vectors
G 2H E 2F v, v.r2, v.r3, v.r6
H G F E Any vector can be thus made to a matrix.
All of the numbers in the field can be so transformed.
But we're intrested in the product of vectors
E F -G -H
E -F G -H
E -F -G H
If we multiply this product by the denominator and numerator,
the denominator becomes a member of Z, and this can be divided
through the numerator.
Using (1, 2, 1, 0)/(2, 1, 1, 1) as the example, we get
2 2 -3 -6 | 2 | 5 = 4-2-3+6
1 2 -3 -3 | -1 | 0 = 2 -2 -3 +3
-1 -2 2 2 | 1 | 0 = -2+2+2-2
-1 -1 1 2 | -1 | -2 = -2+1+1-2
2 -2 -3 6 | 5 | -2=10-12
-1 2 3 -3 | 0 | -5+6=1
-1 2 2 -2 | 0 | -5+4=-1
1 -1 -1 2 | -2 | 5-4 = 1
This gives (-2,1,-1,1) for the number to multiply through
-2 2 -3 6 | 5
1 -2 3 -3 | 0
-1 2 -2 2 | 0
1 -1 1 -2 | -2
Hint: to rationalize the denominator of $\,a/b\,$ we seek a $\rm\color{#90f}{nonzero}$ rational multiple $\,bc\,$ of the denominator, then $\,a/b = ac/(bc).$ To obtain this multiple we use a norm (multiple) map to eliminate radicals one-at-time till we reach a rational. Namely, apply the Theorem below with $S = R_n := \Bbb Q[\sqrt r_1,\cdots \sqrt r_n],\ R_0 := \Bbb Q,\,$ $\, r_k\in R_{k-1}\backslash R_{k-1}^2;\: $ height $h(s)= $ least $k$ with $\,s\in R_k,\,$ i.e. index of the largest indexed radical occurring in $\,s,\,$ so height $0$ are in $\Bbb Q,\,$ height $1$ are in $R_1\backslash \Bbb Q$, height $2$ are in $R_2\backslash R_1,\,$ etc. To get a simpler (lower height) multiple take the $\color{#c00}{{\rm norm} = s\bar s}$ as below
$$h(s) = k\Rightarrow s \in R_k,\, s\not\in R_{k-1},\,\ {\rm so}\ \ s = a+b\sqrt r_k,\, a,b\in R_{k-1},\, b\neq 0,\ \,\rm so\qquad $$
$$ \color{#c00}{s\bar s} = (a+b\sqrt r_k)(a-b\sqrt r_k) = a^2- b^2 r_k \in R_{k-1}\Rightarrow h(s\bar s) \le k-1 < k\qquad$$
Theorem (Simpler Multiples) $ $ Suppose $S$ is a monoid, i.e. a set of numbers containing $\color{darkorange}1$ and $\rm\color{#0a0}{closed\ under\ associative\ multiplication}$, with height map $\,h:S\to \Bbb N.\,$ If all $s\in S$ of height $> 0$ have simpler (lower height) multiples in $S$ then all elements have a multiple with height $= 0$.
Proof $ $ By complete induction on height. Base case: $ $ if $\,h(s) = 0$ then $s = s\cdot \color{darkorange}1$ is a zero-height multiple of $\,s.\,$ Else $\,h(s) > 1.\,$ By hypothesis there is $\,t\in S$ such that $\,h(st) < h(s) \,$ so by induction there is $u \in S$ such that $h(stu) = 0,\,$ i.e. $\,s(tu)\,$ is zero-height multiple of $\,s\,$ (note $s,t,u\in S\Rightarrow st,tu,stu\in S$ by $\rm\color{#0a0}{closure\ and\ associativity}$).
Beware $ $ We need a $\rm\color{#90f}{nonzero}$ rational multiple of the denominator when using the above to rationalize denominators. To ensure that above we need to know that the norm map is nonzero for nonzero elements, and for that it suffices that $\, r_k\not\in R_{k-1}^2,\,$ i.e $\,x^2-r_k\,$ is irreducible over $R_{k-1}$. Indeed, if $\,0 = s\bar s = a^2-b^2r_k\,$ then $\,r_k = (a/b)^2\in R_{k-1}^2\,$ contra hypothesis.
