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.
Best Answer
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 $f(1)=0$.