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I was given the following question (in my undergraduate Abstract Algebra module):

Let $f:R\to S$ be a ring homomorphism. Prove that: the image of $f$ is a subring of $S$ if $R$ is a ring with unity and $f$ is surjective.

The following is my attempt:
The image of $f=\{s\in S\mid s=f(r)$ for some $r\in R\}$. Let $x,y\in R$ and $f(x), f(y)\in f(R)$
$$f(x)-f(y)=f(x-y)$$ $x-y\in R$ (since $R$ is a group). Thus the image of $f$ is closed under subtraction.

$$f(x)*f(y)=f(xy) $$ $xy\in R$ (since $R$ is a group). Thus the image of $f$ is closed under multiplication.

Therefore the image of $f$ is a subring of $S$.

Is this even correct? I never used the fact that $R$ is a ring with unity and that $f$ is surjective so it is confusing me? Thank you.

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    $\begingroup$ Wait, you are assuming that $f$ is surjective and talking about the image of $f$? Wouldn't that be just $S$ in this case? $\endgroup$ – Dirk Apr 11 '17 at 9:17
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    $\begingroup$ I guess so. Does this change my answer? (I must add that I am terrible at this subject and nothing seems obvious to me. But i am still trying to understand as best as I can) $\endgroup$ – Mieke Möller Apr 11 '17 at 9:37
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Your proof is correct (unless I am missing something). There is no need to assume $R$ has a unit, or that $f$ is surjective. In fact, assuming $f$ is surjective is a bit weird because in that case its image is all of $S$ and then it's obviously a subring.

Are you sure you copied the question correctly?

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  • $\begingroup$ Thanks a lot! Yes, it is copied correctly. But we were already told to ignore one of the other questions as it was typed without enough info, so I am assuming that here the second part of the question must go with the next question, as that would make more sense. $\endgroup$ – Mieke Möller Apr 11 '17 at 9:50

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