Suppose $S$ is an integral extension of $R$ and $I$ an ideal in $S$. Why is $S/I$ an integral ring extension of $R/(R \cap I)$?
To this question, Dummit and Foote says: Reducing the monic polynomial over $R$ satisfied by $s \in S$ modulo $I$ gives the monic polynomial satisfied by $\bar{s} \in S/I$ over $R/(R \cap I)$.
Can someone explain this remark in more detail?