# Prove that the Gaussian rationals is the field of fractions of the Gaussian integers

I'm looking to prove that $\Bbb Q[i] = \{ p + qi : p, q \in \Bbb Q \}$ is the field of fractions of $\Bbb Z[i] = \{p + qi : p, q \in Z\}$.

I am familiar with definition of a field of fractions. For example, I understand that if one has an integral domain $D$, it can be embedded in a field of fractions $F_D$, and every element of $F_D$ can be written as the quotient of two elements in $D$. However I have been confused by two questions:

(1) If one has a field $F$, and one can show that any element in $F$ can be expressed as the quotient of two elements in an integral domain $D$, does that imply that $F$ is the field of fractions of $D$?

(i.e. Would it suffice in my case to show that every element in $\Bbb Q[i]$ can be written as a the quotient of 2 elements of $\Bbb Z[i]$?)

(2) How can I show that element in $\Bbb Q[i]$ can be written as a the quotient of 2 elements of $\Bbb Z[i]$? When I write an element of $\Bbb Q[i]$ like

$$q = \frac ab + \frac cdi$$

I get mixed up trying to come up with 2 elements of $\Bbb Z[i]$ that could equal that. I set $z = u+vi$ and $w = x+yi$, equate $q = z/w$ and I end up with these equations like this:

$$a = ux + vy; b = x^2 + y^2; c = vx - uv; d = x^2 + y^2$$

This doesn't strike me as the way to go. Possibly there is a way to take advantage of fact that the elements are equivalence classes, but I'm not seeing it. Thank you very much!

$(1)\$ Yes, since any field containing $D$ must contain all fractions over $D$ (see Remark below).
$(2)\ \ \dfrac{a}b + \dfrac{c}d\ i\, =\, \dfrac{ad+bc\,i}{bd}$
Presumably you have already shown that $\,\Bbb Q[i]\,$ is not only a ring, but also a field (e.g. by rationalizing denominators). That, combined with the above step of clearing denominators, does the trick.
Remark  One can deduce $(1)$ from the universal mapping property of the fraction field: that any field containing an image of the ring $D$ contains a canonical image of the fraction field of $D$. More precisely, if $\,i\, :\, D\to {\cal F}(D)\,$ is the inclusion monomorphism of $\,D$ in its fraction field then any ring monomorphism $\,h : D \to F\,$ to a field factors as $\,h = g \circ i\,$ for a unique field morphism $\,g,\,$ viz. $\, g(a/b) = h(a)\,(h(b))^{-1}.\,$ Therefore, the hypothesis that every element of $F$ is a fraction over $D$ implies that the map $\,g\,$ is surjective, which yields a field isomorphism $\, F\cong {\cal F}(D).$
• Glad it was helpful. Don't sweat over $(2),$ we all overlook things like that at one time or another in our mathematical career. – Bill Dubuque Feb 6 '14 at 1:36