Show that $S/(I\cap S)$ is isomorphic to $(I+S)/I$

Question

Let $I$ be an ideal of $R$ and $S$ be a subring of $R$. We know that

$I+S$ is a subring of $R$ containing $I$, and

$I\cap S$ is an ideal of $S$.

Without direct use of the first isomorphism theorem, I would like to show that $S/(I\cap S)$ is isomorphic to $(I+S)/I$.

So, first we need to define the homomorphism and then show that it is bijective. Would we define it as $\theta :S/(I\cap S) \to (I+S)/I$ and go from there? How would we show that it is onto and one-to-one?

Answer 1

Reinvent the wheel:

Define a map: \begin{align} S/I\cap S&\longrightarrow R/I\\ s+I\cap S&\longmapsto s+I \end{align}

• This is well defined: if $s+I\cap S=t+I\cap S$, then a fortiori, $s+I=t+I$.
• It is trivial to check this is a morphism, and the image is $(S+I)/I$ by construction.
• It is injective: if $s+I=t+I$, $s-t\in I$. As it is also in $S$, it is indeed in $I\cap S$, so $s+I\cap S=t+I\cap S$.