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If two integral domains $D$ and $D'$ are isomorphic show that their corresponding field of quotients (fractions) $Q(D)$ and $Q(D')$ are isomorphic.

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closed as off-topic by user26857, Adam Hughes, Daniel W. Farlow, Chris Godsil, USER91500 Mar 29 at 4:07

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Here's one approach. Let $\text{Frac}(R)$ denote the fraction field of an integral domain, and for an injective arrow $f:R\to S$ let $\text{Frac}(f)$ be the arrow $\text{Frac}(R)\to\text{Frac}(S)$ defined by $\text{Frac}(f)\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$. Show that this has the property that $\text{Frac}(g\circ F)=\text{Frac}(g)\circ\text{Frac}(f)$ and that $\text{Frac}(\text{id}_R)=\text{id}_{\text{Frac}(R)}$. Conclude that if $f:R\to S$ is an isomorphism then so is $\text{Frac}(f)$ for $\text{Frac}(f^{-1})$ is an inverse for $\text{Frac}(f)$

This is a general techinque--show that your construction is functorial.

If you actually care to learn it, what you have shown is that $\text{Frac}$ is a functor from the category $\mathbf{Dom}$ of integral domains an injective maps to the category $\mathbf{Field}$ of fields and field(ring) maps.

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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.

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Couold you please explqin something to me? How does$j\circ\psi$ induce a ring homomorphism $\Psi\colon Q(D)\to Q(E)$? Specifically what does $\Psi$ to do elements not in the image of $\iota$ ? – Bysshed Sep 5 '15 at 22:37

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