# If $f :R \to S$ is an isomorphism from ring $R$ to ring $S$, both with identity, does $f(1_R) = 1_S$?

My attempted proof:

Let $$a \in R$$, then $$f(a) = f(a \cdot 1_R)$$ = $$f(a) \cdot f(1_R)$$, and for $$f(a) \ne 0$$, it gets canceled, so $$1_S = f(1_R)$$

Is this correct? Thanks.

• What do you mean by 'it gets canceled'? Note that $S$ is a ring. Aug 26, 2021 at 12:59
• Setting $a=1_R$ only shows that $f(1_R)$ is an idempotent... Aug 26, 2021 at 13:23
• I think "it gets cancelled" was intended to mean "acts like the identity." So the OP succeeds in showing that $f(1)$ acts like the identity on $f(R)$, and as long as the OP realizes $f(R)=S$ that would be enough. Aug 26, 2021 at 15:13

## 3 Answers

You don't have cancelation in arbitrary rings. You have to show that $$f(1_R)s=s$$ for all $$s \in S$$. Hint: Use the fact that $$f$$ is surjective.

Let $$R,S$$ be rings and $$f$$ the isomorphism. To prove $$f(1_R)=1_S$$, we need to prove that for each $$s\in S$$, $$f(1_R)s=s$$. Therefore from uniqueness of $$1_S$$, we get $$1_S=f(1_R)$$.

Let $$s\in S$$. $$f$$ is an isomorphism therefore there exists $$r\in R$$ s.t $$f(r)=s$$. So: $$f(1_R)s=f(1_R)f(r)=f(1_Rr)=f(r)=s$$

as desired.

You have $$f(r)=f(1_R)f(r)$$. Since $$f$$ is surjective, we may choose $$r$$ so that $$f(r)=1_S$$. The result follows.