# Ring homomorphism: Prove the image is a subring

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.

• Wait, you are assuming that $f$ is surjective and talking about the image of $f$? Wouldn't that be just $S$ in this case? – Dirk Apr 11 '17 at 9:17
• 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) – Mieke Möller Apr 11 '17 at 9:37

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.