Show that $B/Q$ is integral over $A/P$

Question

If $A$ is a subring of $B$ and $B$ is integral over $A$, let $Q$ be a prime ideal of $B$ and $P=Q\cap A$. Show that $B/Q$ is integral over $A/P$.

If $b\in B$ is integral over $A$ then for some polynomial;

$x^n+a_{n-1}x^{n-1}+\dots+a_0=0$ for $a_i\in A$

why does the same hold if we take $b+Q$ and $a_i+P$ instead of $b$ and $a_i$ ? If I substitute it in the equation I don't see it.

Answer 1

No need for $P,Q$ to be prime. For $b\in B $$$b^n+a_{n-1}b^{n-1}+\dots+a_0=0$$ so $$(b^n+Q)+(a_{n-1}b^{n-1}+Q)+\dots+(a_0+Q)=Q$$ so $$(b+Q)^n+(a_{n-1}+Q)(b+Q)^{n-1}+\dots+(a_0+Q)=Q$$ so $$(b+Q)^n+(a_{n-1}+P)(b+Q)^{n-1}+\dots+(a_0+P)=Q$$ Therefore $b+Q$ is integral over $A/P$

Answer 2

The difficulty may be that $A/P$ is set-theoretically not a subring of $A/Q$. You rather have an inclusion ring homomorphism $A/P → A/Q$ coming from the composition $A → B → B/Q$, factoring through its kernel which is $A ∩ Q = P$.

From your point of view, you should say that the equation $$b^n + a_{n-1}b^{n-1} + … + a_0 = 0$$ holds in $B$, and therefore also holds in $B/Q$ by replacing $b$ and $a_k$ by $b + Q$ and $a_k + Q$ respectively (not $a_k + P$).

But maybe it helps here to rather think of the surjective ring homomorphism $B → B/Q$ than to think of an explicit equation and replacing its terms with new ones:

If $f ∈ A[X]$ such that $f(b) = 0$, then $[f]_Q ([b]_Q) = 0$ and $[f]_P([b]_Q) = [f]_Q([b]_Q) = 0$ (where $[·]_•$ denotes the residue class of $·$ modulo $•$), so $[b]_Q$ is integral over $A/P$.