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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.

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    $\begingroup$ What do you mean by 'it gets canceled'? Note that $S$ is a ring. $\endgroup$ Aug 26, 2021 at 12:59
  • $\begingroup$ Setting $a=1_R$ only shows that $f(1_R)$ is an idempotent... $\endgroup$ Aug 26, 2021 at 13:23
  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Aug 26, 2021 at 15:13

3 Answers 3

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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.

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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.

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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.

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