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.
Best Answer
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?