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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...

rschwieb
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    You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$? – Arthur Jan 03 '19 at 13:48
  • f(1) isnt zero divisor @arthur –  Jan 03 '19 at 14:17
  • Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor? – Arthur Jan 03 '19 at 14:20
  • @arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1 –  Jan 03 '19 at 14:23
  • You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)\cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$? – Arthur Jan 03 '19 at 14:26
  • When you write a post, you should put the question in the body of the post. It's OK to put it both in the body and the title. But it is not good to leave it out of the body. – rschwieb Jan 03 '19 at 14:29
  • it is equal to f(1), so for every s in S, s.f(1)=s.f(1).f(1) since S has not zero divisor, we can ommit f(1) from right. so s=s. f(1), am i right? @arthur –  Jan 03 '19 at 14:31
  • Yup. That's it. If you are not assuming the rings are commutative you also have to do $f(1)s = f(1)f(1)s$, but that's about it. – Arthur Jan 03 '19 at 14:34

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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.

rschwieb
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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 $f(1) = 1_S$.

Daniel W.
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