Universal Property of Quotient Field

Question

I have following question about Universal Property for Quotient Field that leads me to following contradiction:

The statement is: Let $R$ be an intergal domain and $\phi: R \to F$ is a ring morphism. Futhermore, let $i: R \to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $\bar{\phi}: Q(R) \to F$ with $\phi = \bar{\phi} \circ i$.

Then I consider following example: Let $R = \mathbb{Z}, F = \mathbb{Z}/(p)$ for a prime $p$. Let $\phi: \mathbb{Z} \to \mathbb{Z}/(p)$ be the canonic projection. Obviously, $\phi \neq 0$. But on the other hand the extension $\bar{\phi}: \mathbb{Q}= Q(\mathbb{Z}) \to \mathbb{Z}/(p)$ is obviously zero map since in $\mathbb{Q}= Q(\mathbb{Z})$ we have $1 = \frac{p}{p}$ and therefore $\bar{\phi(1)}=0$. Where is the error in my reasonings?

Answer 1

The ring homomorphisms $\phi:R\to F$ is required to be injective (and $F$ is required to be a field). See here for example. More generally, a ring homomorphism $\phi:R\to F$ extends to a ring homomorphism $Q(R)\to F$ if and only if $\phi$ maps every non-zero element of $R$ into a unit in $F$.

Answer 2

Your statement of the universal property is incorrect. It only applies when the homomorphism $\phi:R\to F$ is injective.

Answer 3

The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $\,\Bbb Z\,$ into $\,\Bbb Z/p\,$ extends to the subring of $\,\Bbb Q\,$ of fractions writable with denominator coprime to $p$, via $\,a/b \mapsto ab^{-1}\pmod p.\,$ This implies that it is valid to use such fractions when working in $\,\Bbb Z/p.\,$ You can find many examples of such modular fraction arithmetic in my posts here.