Suppose $A\subseteq B$ are commutative rings, $B$ integral over $A$. Let $\mathfrak{b}$ be an ideal of $B$, and set $\mathfrak{a}=A\cap\mathfrak{b}$.
Apparently, $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$.
If $x\in B$, then it satisfies some $ x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0. $ Reducing modulo $\mathfrak{b}$ shows $ (x+\mathfrak{b})^n+(a_{n-1}+\mathfrak{b})(x+\mathfrak{b})^{n-1}+\cdots+(a_1+\mathfrak{b})(x+\mathfrak{b})+(a_0+\mathfrak{b})=0+\mathfrak{b}. $ This would show that $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$ if the coefficients are in $A/\mathfrak{a}$. Why is $a+\mathfrak{b}=a+\mathfrak{a}$ for $a\in A$ here? Isn't $a+\mathfrak{b}$ the set of all $y\in B$ such that $y-a\in\mathfrak{b}$? Maybe I've interpreted the image of $a$ incorrectly.
If $y\in a+\mathfrak{b}$, then $y-a\in\mathfrak{b}$, so if $y\in A$ also, then $y-a\in A$, so $y-a\in\mathfrak{a}$. But is there some reason why $y\in A$, if at all, to see explicitly that these cosets are equal? Thanks.