Consider the universal property for the fraction field of an integer domain:

Let $R$ be a integral domain, $F(R)$, its fraction field, $K$ some field and $f:R\rightarrow K$ a injective ring homomorphism, i.e. $R$ is embedded via $f$ in $K$. Then there exists a unique field homomorphism $g:F(R)\rightarrow K$ such that $g\circ \varepsilon =f$, where $ \varepsilon $ is the map that embeds $R$ in $F(R)$.

To make things clear, this is what I understand to be a field resp. ring homomorphism:

$\bullet$ Let $R,S$ be fields. Then a field homomorphism $f$ is a map $f:\rightarrow S$ such that $f(x+y)=f(x)+f(y)$, $f(xy)=f(x)f(y)$ and $f(1)=1$.

$\bullet$ Let $R,S$ be (not necessarily unitary) rings. Then a ring homomorphism $f$ is a map $f:\rightarrow S$ such that $f(x+y)=f(x)+f(y)$, $f(xy)=f(x)f(y)$.

Notice that a ring homomorphism doesn't have to fulfill $f(1)=1$, even if the rings are both unitary. In particular, a ring homomorphism doesn't have to be injective. (These definitions are in conformity with the Mathworld definition)

My question is, what happens if we require in the above proposition that $g$ is only a ring homomorphism ? Does then $f$ have to be still injective, such that this proposition has to hold ? I wasn't able to construct a (nontrivial) counterexample where $f$ is not injective and $g$ is a ring homomorphism and the above holds.

  • 3
    $\begingroup$ Well, precisely the link you wrote defines homom. of rings as mapping unit to unit in case we have unitary rings, so why you say otherwise? $\endgroup$
    – DonAntonio
    Dec 7, 2012 at 19:06
  • $\begingroup$ The universal property doesn't just talk about rings ( = always unital), but rather it takes place in the category of rings, where the morphisms are of course the homomorphisms of rings, not just the homomorphisms of the underlying rngs ( = no unit required). Why do people always keep applying this forgetful functor (probably because they think that forgetful functors are identities). $\endgroup$ Dec 7, 2012 at 20:18
  • $\begingroup$ @DonAntonio I'm afraid this is a terminology issue: It is true that a homomorphism between unitary rings has to map the unit to the unit. But it's also true, that we can forget about the unitary structure between the rings and view them only as rings. Then a homomorphism. doesn't have to map the unit to the unit (although units are present in the rings). [...] $\endgroup$
    – temo
    Dec 8, 2012 at 13:14
  • $\begingroup$ [...] Consider for example the mapping $f:\mathbb{Z}\rightarrow \mathbb{Z}, \ x\mapsto 0$. The $\mathbb{Z}$'s are unitary rings, but we don't take the units into consideration, so $f$ is only a ring homomorphism (and not a unitary ring homomorphism). $\endgroup$
    – temo
    Dec 8, 2012 at 13:17
  • $\begingroup$ Ohm, that's fine @temo. It's just that you said that in the link above they so and so, and they didn't. $\endgroup$
    – DonAntonio
    Dec 8, 2012 at 13:19

1 Answer 1


Let $f: R \rightarrow K$ be a rng homomorphism with $R$ an integral domain and $K$ a field. Note that $f(1)f(1)=f(1)$ so $f(1)$ is a solution to the equation $x^2-x$ in particular $f(1)=1$ or $f(1)=0$.


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