Ring homomorphism may not preserve $1$.

Question

Suppose $R$ is a ring with unity and $f:R\to R'$ is a ring homomorphism. Then, $f(R)$ must have an identity $1_{f(R)}.$ But $1_{f(R)}$ may not be the identity of $R'$. Even $R'$ may not contain any identity. Suppose $R'$ contains $1_{R'};$ even still, the ring homomorphism may not necessarily map $1_R$ to $1_{R'}.$

Why does this occur? I am unable to find exactly why it is so. Can this be explained by semigroup homomorphisms or monoid homomorphisms, i.e., maps from semigroup $S_1$ to $S_2$ such that $f(a \cdot b)=f(a) \cdot f(b)$?

Answer 1

Today's mathematicians always suppose a ring homomorphism to preserve identity, i.e. $\phi:R\to S$ supposed to have $f(1_R)=1_S$ (of course we suppose $R$ and $S$ unital).

But there is some cases which the property $f(1_R)=1_S$ can be concluded.

For example,

suppose $\phi:R\to S$ is onto and $R$ is unital (has identity element), then $\phi(1_R)$ is the identity of $S$.

For any $s\in S$, there exists an $r\in R$ with $\phi(r)=s$ and thus $$\phi(1_R)s=\phi(1_r)\phi(r)=\phi(1_Rr)=\phi(r)=s$$ and similarly, $s\phi(1_R)=s$, which shows $\phi(1_R)$ is the identity of $S$.

For another one,

Suppose $S$ is an integral domain and $\phi:R\to S$ is a ring homomorphism. Then $\phi(1_R)=1_S$ or $\phi(1_R)=0$.

We have $$\phi(1_R)=\phi(1_R1_R)=\phi(1_R)\phi(1_R)\implies \phi(1_R)(1_S-\phi(1_R))=0$$ thus either $\phi(1_R)=0$ or $1_S=\phi(1_R)$.