$\varphi : R → S$ is a epimorphism from $R$ to ring $S$, let $I$ be an ideal of $R$. Prove $\varphi (I) = S$ if and only if $R = I +Ker(\varphi)$

Question

Let $\varphi : R → S$ be an epimorphism from ring $R$ to ring $S$, and let $I$ be an ideal of $R$. Prove that $\varphi (I) = S$ if and only if $R = I +Ker(\varphi)$

I am quite confused on what exactly this question is asking me to do. Can someone please elaborate on this question or possibly re-word it to make it more understandable??

Answer 1

Note that $\ker(\phi) \subset R$ is an ideal, and $I +\ker(\phi)\subset$ is also an ideal in $R$.

Since $\phi(\ker(\phi)) = 0\in S$, we see that $\phi(I +\ker(\phi)) = \phi(I)$.

The question wants you to see that $\phi(I) = S,$ so the image of $I$ is the whode codomain iff $I +\ker(\phi) =R$.

Answer 2

It is true because $\varphi^{-1}\bigl(\varphi(I)\bigr)=I+\ker\varphi$ and $\varphi^ {-1}(S)=R$.

By the way, don't use the term epimorphism for rings when you mean surjective: it is a terminology borrowed from category theory and it happens that for rings it is not equivalent to surjective. For instance, the inclusion of $\mathbf Z$ in $\mathbf Q$ is an injective ring epimorphism.