Chinese Remainder Theorem for integral ring homomorphisms

Question

Let $A\to B$ be an integral homomorphism of commutative rings, $\mathfrak p\subseteq A$ be a prime ideal and $\mathscr Q$ be a finite set of prime ideals of $B$ lying above $\mathfrak p$. I ask if the canonical ring homomorphism $$B\to\prod_{\mathfrak q\in\mathscr Q}B/\mathfrak q$$ is surjective.

My attempt: By localizing at $\mathfrak p$ we get the ring homomorphism $$B_{\mathfrak p}\to\prod_{\mathfrak q\in\mathscr Q}B_{\mathfrak p}/\mathfrak qB_{\mathfrak p}$$ which is surjective because $\mathfrak qB_{\mathfrak p}$ are distinct maximal ideals of $B_{\mathfrak p}$ since they lies above the maximal ideal $\mathfrak pA_{\mathfrak p}$ of $A_{\mathfrak p}$.

For every prime $\mathfrak p'\neq\mathfrak p$ of $A$, the ring homomorphism $$B_{\mathfrak p'}\to\prod_{\mathfrak q\in\mathscr Q}B_{\mathfrak p'}/\mathfrak qB_{\mathfrak p'}$$ is surjective as well, because $\mathfrak qB_{\mathfrak p'}=B_{\mathfrak p'}$, hence the right-handed side is zero.

Since being surjective for an $A$-module homomorphism is a local property, the assertion is proved.

Question. Is the statement true and this approach correct?

Answer 1

The mistake is in the paragraph:

For every prime $\mathfrak p'\neq\mathfrak p$ of $A$, the ring homomorphism $$B_{\mathfrak p'}\to\prod_{\mathfrak q\in\mathscr Q}B_{\mathfrak p'}/\mathfrak qB_{\mathfrak p'}$$ is surjective as well, because $\mathfrak qB_{\mathfrak p'}=B_{\mathfrak p'}$, hence the right-handed side is zero.

beacuse, in general, $\mathfrak p'\neq\mathfrak p$ doesn't implies $\mathfrak qB_{\mathfrak p'}=B_{\mathfrak p'}$, for example, when $\mathfrak p'\supset\mathfrak p$.