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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.

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    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?2012-12-07
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    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).2012-12-07
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    @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). [...]2012-12-08
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    [...] 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).2012-12-08
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    Ohm, that's fine @temo. It's just that *you said* that in the link above they so and so, and they didn't.2012-12-08
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    @DonAntonio Well...I still can't see the difference between the link and my definition. The only place I see where I may have not been precise enough - and I think this caused the confusion - is that when I say "ring homomorphism" I mean that we look at the rings as rings and not unitary rings. So if I want $f(1)=1)$ I should say "unitary ring homomorphism" (which I directly called "field homomorphism"), whereas in the link a "ring homomorphism" is either a ring homomorphism as I see it or a unitary ring homomorphism. So indeed I may have had to say [...]2012-12-08
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    [...] "Notice that a ring homomorphism **(where we look at the rings only as rings and not unitary rings)** *doesn't* have to fulfill $f(1)=1$. Thus my definition were in concordance with the link, since I'm only looking at rings whose unitary structure I ignore and for those my definition coincides with the one in the link - but it probably would have been clearer if I would have pointed it out more clearly that when I say "ring" I ignore the fact that there may be a unit in it. Was this the source of the confusion ?2012-12-08

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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 $f(1)=0$.