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Can you tell me if my reasoning is correct?

I want to prove if $S \subset R$ are rings and $R$ is integral over $S$ and $I$ is an ideal of $R$ then $R/I$ is integral over $S/ (S\cap I)$.

Let $R$ be integral over $S$. $(S \cap I) \subset I$ is an ideal of $S$ and hence of $R$.

Since $R$ is integral over $S$ we have that $R/(S \cap I)$ is integral over $S/(S \cap I)$ and since $(S \cap I) \subset I$ we have $R/I \subset R/(S\cap I)$ and hence $R/I$ is integral over $S/(S\cap I)$.

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    Why is $S \cap I$ an ideal of $R$?2012-07-06
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    @DylanMoreland It is not. : / It is only an ideal of $S$.2012-07-06
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    If you have a monic polynomial with coefficients in $S$, then reducing the coefficients modulo $S\cap I$ still gives you something monic.2012-07-06
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    @DylanMoreland But the multiplication isn't defined. $\bar{x} = x + I \in R/I$ and $\bar{s_i} = s_i + S \cap I \in S/(S \cap I)$. Then $\bar{x} \cdot \bar{s_i}$ is not defined.2012-07-06
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    It's defined in $R/I$, which contains an embedded copy of $S/(S \cap I)$.2012-07-06
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    @DylanMoreland Right. Of course. Thank you very much!2012-07-06
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    @DylanMoreland I tried to write this in an answer.2012-07-06

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Let $r+I\in R/I$. Find a monic polynomial $f(x)\in S[x]$ that $r$ satisfies.

It's natural then to look at the image of $f(x)$ in $S/(S\cap I)[x]$ to see if it works for $r+I$!

Here is the viewpoint from homomorphisms' perspective:

If $R$ is integral over $S$ and $\phi:R\rightarrow T$ is a ring homomorphism, then $\mathrm{Im}(R)$ is integral over $\mathrm{Im}(S)$.

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Let me post my new attempt (with help of Dylan's comment):

Let $S \subset R$, $I$ an ideal of $R$, $S$ a subring of $R$. Let $\bar{x} \in R/I$. Then $x$ in $R$ and since $R$ is integral over $S$ there are $s_i$ such that $0 = x^n + s_{n-1} x^{n-1} + \dots + s_1 x + s_0$.

The same equation holds modulo $I$. We can view $\bar{s_i} = s_i + (S/S \cap I)$ as an element of $R/I$ since there is an embedding $S/(S \cap I) \hookrightarrow R/I$ defined by $s + (S\cap I) \mapsto s + I$. This is injective since if $s + (S\cap I)$ maps to zero, $s \in I$ and hence $s \in I \cap S$, i.e. the only element mapped to $0$ is $0$.