TITLE: How to show that if two integral domains are isomorphic, then their corresponding field of quotients are isomorphic?
QUESTION [6 upvotes]: If two integral domains $D$ and $D'$ are isomorphic show that their corresponding field of quotients (fractions) $Q(D)$ and $Q(D')$ are isomorphic.

REPLY [5 votes]: Let $\psi\colon D\to E$ be an isomorphism of integral domains. Let $\iota\colon D\to Q(D)$ be the embedding of $D$ into its field of fractions (via $\iota(a) = \frac{a}{1}$, for instance), and $j\colon E\to Q(E)$ the embedding fo $E$ into its field of fractions (via $j(b)=\frac{b}{1}$).
Then $j\circ\psi\colon D\to Q(E)$ is an injective ring homomorphism (composition of injective ring homomorphisms) that maps $D$ into a field. Therefore, $j\circ\psi$ induces a ring homomorphism $\Psi\colon Q(D)\to Q(E)$ such that $\Psi\circ\iota = j\circ\psi$.
Symmetrically, $\iota\circ\psi^{-1}\colon E\to Q(D)$ is an injective ring homomorphism from $E$ into a field, hence it induces a unique ring homomorphism $\Phi\colon Q(E)\to Q(D)$ such that $\Phi\circ j = \iota\circ\psi^{-1}$. 
The map $D\to Q(D)$ given by $\iota$ should induce the identity map on $Q(D)$; that is, the only map $f\colon Q(D)\to Q(D)$ such that $f\circ\iota = \iota$ is $f=\mathrm{id}_{Q(D)}$, by the uniqueness clause of the universal property of $Q(D)$. Since
Now, 
$$\Phi\circ\Psi\circ\iota = \Phi\circ j \circ\psi = \iota\circ\psi^{-1}\circ\psi = \iota = \mathrm{id}_{Q(D)}\circ\iota,$$
it follows that $\Phi\circ\Psi = \mathrm{id}_{Q(D)}$. A symmetric argument using the universal property of $Q(E)$ and $\Psi\circ\Phi\circ j$ shows that $\Psi\circ\Phi=\mathrm{id}_{Q(E)}$. 
Therefore, $\Psi$ and $\Phi$ are inverses of each other, hence they are both isomorphisms, showing that $Q(D)\cong Q(E)$. 
Note the use of the universal properties, as opposed to an "element-by-element" check; although I specified the definitions of $\iota$ and $j$, those do not need to be specified: all that matters is that they are part of the "universal property" definition.