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> Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian are then $S$ and $R/I$ also Noetherian? 
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I have done the following: 

$R$ is Noetherian iff each increasing sequence of ideal $I\_1\subseteq I\_2 \subseteq I\_3 \subseteq \dots \subseteq I\_k\subseteq \dots $ stops, i.e., $\exists k$ such that $I\_k=I\_{k+1}$, right? 

Then since $S$ is a subring of $R$, not all $I\_i$ are contained in $S$. 

Therefore, the above condition isn't necessarily satisfied. 

So, $S$ is not necessarily Noetherian. 

Is this correct? 

What can we say in that case of $R/I$? Does the increasing sequence stop?