Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.

Question

Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.

My attempt: I think if S is going to be unitary, then it's unit element should be the image of the unit element of R. but no idea for starting...

Answer 1

If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:

its unit element should be the image of the unit element of R

This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.

Answer 2

Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) \neq 0$. \ Let $a \in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.