I was given to this hint to try.
Q(D) ∼= Q(i) = {r + si | r, s ∈ Q} by showing that the
map f : Q(i) → Q(D) given by
f(a/b+c/di) = [ad + bci, bd]
We want to show 1-1, onto, f(a+b)=f(a)+f(b) an f(ab)=f(a)f(b)
1-1:Assume a/b+c/di=e/f+g/hi
f(a/b+c/di)=f(e/f+g/hi)
[ad+bci,bd]=[eg+fgi,fh]
onto:I always get hung up onto.
f(a+b)=f((a/b+c/di)+(e/f+g/hi))=f(af+be/bf+chi+gdi/-dh)=[(af+be)(-dh)+bf(chi+gdi)(-dh)]
f(a*b):
Consider f((a/b+c/di)(e/f+g/hi))=f(ae/bf+ga/hi+ce/di+cg/-hd)
This is where I've gotten so far
Honestly, what you have written is really hard to follow.
We are trying to determine the quotient field of (what you have written as ).
First, notice that , so that . Since is the quotient field of , this means .
Obviously . This and the previous containment force (if are both inside of the field , then so is the smallest field containing and ).
But then is a field satisfying . Since the quotient field of an integral domain is the smallest field containing that integral domain, this forces equality: .