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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?

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    If you try to use the definition of "integral ring extension" for $R/R\cap I\subseteq S/I$, there is one obvious idea... Have you tried it?2011-10-18

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There a ring homomorphism $\phi:R\to R/R\cap I$ such that $\phi(r)=r+R\cap I$ for all $r\in R$. Using it, we can construct another homomorphism, $\Phi:R[X]\to(R/R\cap I)[X]$ which on a polynomial $p$ with coefficients in $R$ applies the function $\phi$ to each coefficient.

Now, if $\sigma\in S/I$ is any element, there exists an element $s\in S$ such that $\sigma=s+I$. Since $I$ is integral over $R$, there exists a polynomial $f\in R[X]$ such that $f(s)=0$. Now you should check that if $q=\Phi(p)\in(R/R\cap I)[X]$, then $q(\sigma)=0$.