$\theta: R \to S$ is a surjective ring homomorphism. A is an ideal of R and $ker(\theta) \subseteq A$.
I need to show that $R/A \simeq S/\theta(A)$. The textbook suggests that I use the first isomorphism theorem where $\alpha: R \to S/\theta(A)$ is defined by $\alpha(r) = \theta(r) + \theta(A)$ for all $r \in R$.
I know that:
- $\theta(A)$ is an ideal of S (by a previous problem)
- $R/ker(\theta) \simeq \theta(R)$ (by the isomorphism theorem)
My idea is that this follows if $ker(\theta)$ is a maximal ideal? In which case, $ker(\theta) = A$, but I'm not sure if this is correct nor am I completely sure how to proceed if that is, in fact, true.
Upon further thought, this seems like the wrong way to solve it. If $ker(\theta)$ is a maximal ideal then $R/ker(\theta)$ is a field, which seems to put me on the wrong track.