# Ring homomorphism may not preserve $1$.

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 $$f(a \cdot b)=f(a) \cdot f(b)$$?

• If you don' condition anything, then the map $\;f:R\to S\,,\;\;f(r)=0\;$ could be regarded as a ring homomorphism...and it could be if it fits some fixed goal. May 20, 2020 at 6:43
• – lhf
May 20, 2020 at 11:18

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 $$1_S=\phi(1_R)$$.