# Question about the First Isomorphism Theorem for rings

$$\DeclareMathOperator{\im}{im}$$The First Isomorphism Theorem for rings is usually stated as follows:

Let $$f\colon R\to S$$ be a surjective homomorphism of rings with kernel $$K$$. Then the quotient ring $$R/K$$ is isomorphic to $$S$$.

In some other textbook, the theorem would be stated slightly differently as:

Let $$f\colon R\to S$$ be a surjective homomorphism of rings with kernel $$K$$. Then the quotient ring $$R/K$$ is isomorphic to $$\im(f)$$ where $$\im(f)$$ denotes the direct image of $$f$$.

In the second form, if $$R/K \cong\im(f)$$ and $$\im(f)$$ is a subring of $$S$$, then how would I show that $$R/K\cong S$$. I can prove the theorem in both formulations, but from the second formulation of the theorem, I don't know how to how to show $$R/K \cong S$$ from $$R/K \cong\im(f)$$. I know that $$f$$ is a surjective homomorphic mapping from $$R$$ to $$S$$ and that if $$R/K\cong\im(f)$$, can Is that enough to show that $$\im(f)=S.$$

Thank you in advance.

• The second version is true without the assumption on surjectivity. The equivalence just using the fact that if $f$ is surjective, then by definition $\mathrm{Im}(f) = S$. Apr 19 '21 at 10:08
• @Mathmo123 because of $f$ is a surjective mapping and in the proof of the theorem, by constructing a function $\phi(r+ker f)=f(r)$, then $\phi$ is surjective from $R/I$ to $S$ because $f$ is surjective from $R$ to $S$. Would that be a correct reasoning. Apr 19 '21 at 10:12

If $$f$$ is surjective then $$\operatorname{im}f=S$$ by definition. Hence the second formulation implies the first one when we consider a surjective map. Conversly, any map $$f\colon R\to S$$ "can be made surjective" by restricting the codomain $$S$$ to the subring $$\operatorname{im}f$$ onto which $$f$$ is trivially surjective. In this case the first formulation implies the second one.
Your argument in the comments is correct. Alternatively, if we write $$\pi\colon R\to R/I$$ for the canonical projection (defined by $$r\mapsto r+I$$) we have that $$\phi\circ\pi=f$$. By elementary set theory the surjectivity of $$f$$ then implies the surjectivity of $$\phi$$ (this is a nice standard fact to keep in mind).