Showing that if $ker(\theta) \subseteq A$ (an ideal in R) then $R/A \simeq S/\theta(A)$ $\theta: R \to S$ 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. 
 A: If you prove that $\text{ker }\alpha=A$ and that $\alpha$ is surjective, by the First Isomorphism Theorem, you are done.
Surjectivity of $\theta$ implies that $\alpha$ is surjective.
Now, let us show that $\text{ker }\alpha=A$.
If $r\in A$, $\theta(r)\in \theta(A)$, so $\alpha(r)=0$.
On the contrary, if $r\in R$ with $\alpha(r)=\theta(r)+\theta(A)=0$, then $\theta(r)\in \theta(A)$, so $\theta(r)=\theta(y)$ for some $y\in A$. Then, $r-y\in \text{ker }(\theta)\in A$, and $y\in A$, so $r\in A$.
A: How I thought about it:
Applying the first isomorphism theorem: $S \cong R/\ker \theta$, and this isomorphism restricts $\theta (A) \cong A/(\ker \theta)$, so $S/\theta (A) \cong  (R/\ker\theta)/(A/\ker \theta )=(R/A)$ by the third isomorphism theorem.
Following your suggestion, however:
Note that $r \in \ker \alpha \iff r \in A$, so the first isomorphism theorem implies the result.
A: Your textbook's hint applies as follows.
The kernel of the surjective ring homomorphism $\alpha:R\twoheadrightarrow S\twoheadrightarrow S/\theta(A)$ is $\theta^{-1}(\theta(A))=A$ because $A\supseteq\operatorname{Ker}(\theta)$.
Thus the first isomorphism theorem applies giving rise to a ring isomorphism $R/A\xrightarrow\sim S/\theta(A)$.
