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.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
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: $$f(1_R)s=f(1_R)f(r)=f(1_Rr)=f(r)=s$$