QUESTION:
Let $S$ be a commutative ring and $R$ a subring of $S$, so that $1\in R$. If $I$ is an ideal of $S$, prove that $S/I$ is integral over $R/(R\cap I)$.
ATTEMPT:
Let $x + I \in S/I$. This means $x\in S$. Since $S$ is integral over $R$, there exists monic polynomial $f(t) = t^n + a_{n-1}t^{n-1} + ... + a_0$ with coefficients in $R$, and $f(x) = 0$. Consider the homomorphism $\phi : R\to R/(R\cap I)$ given by $\phi (r) = r + (R\cap I)$. This is a well-defined surjective homomorphism. Now, consider $\phi (f(t)) = \tilde{f}(t)$ in $R/(R\cap I)[t]$, where $\tilde{f} (t) = t^n + \tilde{a_{n-1}}t^{n-1} + ... + \tilde{a_0} \in R/(R\cap I)[t]$ and each $\tilde{a_i}$ is in the residue class of $a_i$ modulo $R\cap I$. Now, $\tilde{f} (x + I) = \phi (f(x)) = 0$. This shows $x + I$ is integral over $R/(R\cap I)$. Hence, $S/I$ is an integral extension of $R/(R\cap I)$.
Note: I am not confident about this proof: I feel it is not airtight, and that many details are missing, especially where I drew the conclusion, "This shows $x + I$ is integral over $R/(R\cap I)$". I would appreciate it if you could help me pick holes in my logic/presentation and suggest ways to improve this.
Thanks in advance.
First, as is commeted by KReiser, you need to add the hypothesis that $S$ is integral over $R$, but you did not edit the question.
When we say ring $A$ is integral over ring $B$, $B$ usually needs to be a subset of $A$ in the set-theoretical sense (that is, members of $B$ are members of $A$). But if you are careful enough, you may note that elements of $R/(R\cap I)$ are cosets of $R\cap I$, while members of $S/I $ are cosets of $I$. Therefore $R/(R\cap I)\not\subseteq S/I$ in the theoretical sense. When your book states $S/I$ being integral over $R/(R\cap I)$, it actually means that $S/I$ is integral over $(R+I)/I$, because we have the well-know isomorphism: $$\varphi: R/(R\cap I)\to (R+I)/I,\ \ a+(R\cap I)\mapsto a+I. $$ (I have to say "$S/I$ integral over $R/(R\cap I)$" is confusing to new learners.)
And $S/I$ is integral over $(R+I)/I$ is quite easy to prove, since if $x\in S$, $a_0,\cdots,a_{n-1}\in R$, and $$x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0\ \cdots(\star), $$ then consider the cosets of $I$ of both sides of eq $(\star)$, we have \begin{equation*} (x+I)^n+(a_{n-1}+I)(x+I)^{n-1}+\cdots+(a_1+I)(x+I)+(a_0+I)=0\ (=0+I). \end{equation*} Therefore any $x+I\in S/I$ is integral over $(R+I)/I$.