Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 2E / Solution

The fraction field of an integral domain $R$ is a field $F$ containing an identified copy of $R$ in which every element has the form $a/b$ with $a,b\in R$ and $b\ne 0$. Equality means $a/b=c/d$ exactly when $ad=bc$; the operations are the usual addition and multiplication of fractions. We use unital ring homomorphisms.

Since $\phi$ is injective, $\phi (b)\ne 0$ whenever $b\ne 0$. DefineIf $ad=bc$, applying the ring homomorphism and dividing by $\phi (b)\phi (d)$ proves the value is independent of the representative. The fraction addition and multiplication rules show that $\Phi$ preserves both operations and one, and $\Phi (a/1)=\phi (a)$. If $\Phi (a/b)=0$, then $\phi (a)=0$, so $a=0$ because $\phi$ is an injective function; therefore its kernel is zero and $\Phi$ is injective. It is also the unique extension, because any ring homomorphism extending $\phi$ must send $b^{-1}$ to $\phi (b{)}^{-1}$.

For a counterexample, take $R=\mathbb{Z}$, $F=\mathbb{Q}$, and reduction modulo two $\psi :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$. An extension would give $1=\Psi (2)\Psi (1/2)=0$, a contradiction. Nonzero denominators must become invertible in the target; an arbitrary ring homomorphism need not have that property.