# Ring homomorphism with $\phi(1_R) \neq1_S$

Let $R$ and $S$ be rings with unity $1_R$ and $1_S$ respectively. Let $\phi\colon R\to S$ be a ring homomorphism. Give an example of a non-zero $\phi$ such that $\phi(1_R)\neq 1_S$

In trying to find a non-zero $\phi$ I've done the following observation:
Since for $\forall r\in R$ $\phi(r) = \phi(r\times1_R) = \phi(r)\times\phi(1_R)$ we must have that $\phi(1_R)$ is an identity of $\phi(R)$ but not an identity of $S$. We must therefor construct a $\phi$ that is not onto and which have this property. I can't come up with any explicit example though, please help me.

-
 Some conditions when homomorphism must map unity to unity are given here: Homomorphisms from a unital ring to a ring with no zero divisors preserve unity?. – Martin Sleziak Aug 7 '12 at 10:38

Let $R$ be any ring and $S=R\times R$. Then the inclusion map $r\mapsto (r,0)$ gives you such a homomorphism. (Note that some authors require that $\phi(1_r)=1_S$ for $\phi$ to be a homomorphism).