Q about 2nd isomorphism Th for rings: Is this true $(S+I)/I\cong S/(S\cap I)\cong S/I$, for subring $S$ and ideal $I$?

Question

Let $R$ be a ring, $S\subseteq R$ a subring and $I\subseteq R$ an ideal.

Then Theorem 8 in chapter 7 from Abstract Algebra by Dummit and Foote states that $$ (S+I)/I\cong S/(S\cap I). $$

Let $(A+B)=\left\{a+b\,|\, a\in A, b\in B\right\}$.

I wonder if this additional statement is also true:

$$(S+I)/I\cong S/(S\cap I){\color{red}{\cong S/I}}\,\, ?$$

This seems obvious to me, but I would like to make sure that I'm not missing anything here. Thanks.

Answer 1

$I$ is not necessarily contained in $S$, so the expression $S/I$ does not necessarily make sense. When it does, it is equal to $S/(S \cap I)$ so nothing new is gained by your version of the theorem.