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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.