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I am working on the following exercise:

Let $S$ be a commutative ring and let $R \subset S$ be an integral ring extension and $I \vartriangleleft$ an ideal in $S$. Then $R/(I \cap R) \subset S/I$ is an integral extension.

I do recognize that $R/(I \cap R) \simeq \ker(f)$ for $f:R \rightarrow S/I$ by the First Isomorphism Theorem, so $R/(I \cap R)$ is a subring of $S/I$, but I do not see why there should be a monic polynomial $f \in R/(I \cap R)[X]$ for every $a \in S/I$ sucht that $f(a) = 0$. Could you give me a hint?

3nondatur
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    Consider an element $a+ I$ of $S/I$. The element $a \in S$ satisfies a monic polynomial with coefficients in $R$. Project this monic polynomial to $R/(I\cap R)$ to get an integral relation for $a+ I$ – Badam Baplan Oct 03 '20 at 21:48
  • I wouldn't say that $R/(I\cap R)$ is a subring of $S/I$. The first isomorphism theorem tells you that it is isomorphic to a subring of $S/I$. But whether the distinction matters depends on the context. – Servaes Oct 03 '20 at 22:46

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To show that $R/(I\cap R)\subset S/I$ is an integral extension, it suffices to show that every $\overline{s}\in S/I$ is integral over $R/(I\cap R)$. That is to say, that $\overline{s}$ satisfies a monic polynomial with coefficients in $R/(I\cap R)$.

For every $\overline{s}\in S/I$ there exists some $s\in S$ such that $\overline{s}=s+I$. We are given that $S$ is integral over $R$, so $f(s)=0$ for some monic $f\in R[X]$. Then also $f(\overline{s})=0$, hence also $\overline{f}(\overline{s})=0$, where $\overline{f}=f+I\cap R$.

To see that indeed $\overline{f}(\overline{s})=0$, note that $f(\overline{s})=0$ implies that $f$ is in the kernel of the quotient map $$R[X]\ \longrightarrow\ \big(S/(I+(s))\big)[X],$$ where $I+(s)$ denotes the ideal generated by $I$ and $s$. Of course this map factors as $$R[X]\ \longrightarrow\ \big(R/(I\cap R)\big)[X]\ \longrightarrow\ \big(S/(I+(s))\big)[X],$$ so also the image $\overline{f}$ of $f$ in $R/(I\cap R)$ satisfies $\overline{f}(\overline{s})=0$.

Servaes
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  • Sorry, but I do not quite get you answer. Could you please explain what $f+I \cap R$ is supposed to mean? What does $\overline{f}(x)$ look like? – 3nondatur Oct 03 '20 at 22:34
  • It is the image of $f$ in the quotient $(R/(I\cap R))[X]$. It is simply $f$ with all coefficients taken mod $I\cap R$. – Servaes Oct 03 '20 at 22:39
  • But then we have $\overline{f}(\overline{s}) = 0 + I \cap R$ and we want $0+I$, correct? – 3nondatur Oct 03 '20 at 22:42
  • No, that doesn't even make sense because $\overline{f}(\overline{s})$ is an element of $S/I$, not if $R/(I\cap R)$. I've added a (coordinate-free) explanation. – Servaes Oct 03 '20 at 22:43