Recently, I came upon a solution that solves this problem:
We have $A$ is a subring of $B$, where $B$ is a commutative ring and is integral over $A$. Now, $Q$ is an ideal of $B$. We may take $P = Q\cap A$. We want to prove that $B/Q$ is integral over $A/P$.
The solution can be found here on this platform: Show that $B/Q$ is integral over $A/P$.
However, I will write it out:
Since $b\in B$ is integral over $A$, then:
$b^n + a_{n-1}b^{n-1} + ... + a_0 = 0.$
i.e. $(b^n + Q) + (a_{n-1}b^{n-1} + Q) + ... + (a_0 + Q) = Q$
i.e. $(b + Q)^n + (a_{n-1} + Q)(b + Q)^{n-1} + ... + (a_0 + Q) = Q$
i.e. $(b + Q)^n + (a_{n-1} + P)(b + Q)^{n-1} + ... + (a_0 + Q) = Q$
Therefore, $b + Q$ is integral over $A/P$.
QUESTIONS:
(1) how did we come about the second step? I applied $\phi : A\to A/P$ on $f(s)$, but still didn't get the result.
(2) how did we get the last step? That is, how did $a_{n-1} + Q$ become $a_{n-1} + P$?
(3) For $b + Q$ to be integral over $A/P$, shouldn't the RHS of the equation be zero?
Thanks in advance.
Let $A\subset B$ be the two rings, let $Q\subset B$ be an ideal and set $P:=A\cap Q$; call the quotient maps $\pi_B:B\to B/Q$ and $\pi_A:A\to A/P$.
An element of $B/Q$ is equal to $\pi_B(b)$ for a $b\in B$, as $\pi_B$ is surjective, and $b$ satisfies: $$0=b^n+a_1b^{n-1}+\dots+a_{n-1}b+a_0$$ hence $\pi_B(b)$ satisfies:$$0=\pi_B(b^n+a_1b^{n-1}+\dots+a_{n-1}b+a_0)=\pi_B(b)^n+\pi_B(a_1)\pi_B(b)^{n-1}+\dots+\pi_B(a_{n-1})\pi_B(b)+\pi_B(a_0).$$ Moreover any element $\pi_B(a)\in B/Q$, for $a\in A$, is equal to the element $\pi_A(a)\in A/P$ viewed as an element of $B/Q$, thanks to the following commutative diagram of maps, where the horizontal maps are injective. $\require{AMScd}$ $$\begin{CD} A @>>> B\\ @VV\pi_AV @VV\pi_BV\\ A/P@>>> B/Q \end{CD}$$ So finally $\pi_B(b)$ satisfies the equation $$0=\pi_B(b)^n+\pi_A(a_1)\pi_B(b)^{n-1}+\dots+\pi_A(a_{n-1})\pi_B(b)+\pi_A(a_0),$$ meaning that $\pi_B(b)$ is integral over $A/P$.
In your solution, $b+Q$ and $a+P$ are another way to denote the images of $b\in B$ and $a\in A$ in the quotients $B/Q$ and $A/P$, i.e. to denote $\pi_B(b)$ and $\pi_A(a)$. In particular $Q\in B/Q$ stands for $0+Q=\pi_B(0)$, which is in fact the zero element of $B/Q$.