The Stacks project

Proof. The horizontal implications in the diagram are trivial. Let $S \subset A$ be the set of nonzerodivisors so that $Q(A) = S^{-1}A$ and $Q(A) \otimes _ A B = S^{-1}B$. Recall that $S = A \setminus \bigcup _{\mathfrak p \in \text{Ass}(A)} \mathfrak p$ by Algebra, Lemma 10.63.9. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$.

Assume (2). If $\mathfrak q \in \text{Ass}(B)$ then $\mathfrak q$ consists of zerodivisors, hence (1) implies the same is true for $\mathfrak p$. Hence $\mathfrak p$ corresponds to a prime of $S^{-1}A$. Hence $A \to B_\mathfrak q$ is flat by our assumption (2). If $\mathfrak q$ lies over an associated prime $\mathfrak p$ of $A$, then certainly $\mathfrak p \in \mathop{\mathrm{Spec}}(S^{-1}A)$ and the same argument works.

Assume (3). Let $f \in A$ be a nonzerodivisor. If $f$ were a zerodivisor on $B$, then $f$ is contained in an associated prime $\mathfrak q$ of $B$. Since $A \to B_\mathfrak q$ is flat by assumption, we conclude that $\mathfrak p$ is an associated prime of $A$ by Algebra, Lemma 10.65.3. This would imply that $f$ is a zerodivisor on $A$, a contradiction.

Assume (4) and going up for $A \to B$. We already know (1) holds. If $\mathfrak q$ corresponds to a prime of $S^{-1}B$ then $\mathfrak p$ is contained in an associated prime $\mathfrak p'$ of $A$. By going up there exists a prime $\mathfrak q'$ containing $\mathfrak q$ and lying over $\mathfrak p$. Then $A \to B_{\mathfrak q'}$ is flat by (4). Hence $A \to B_{\mathfrak q}$ is flat as a localization. Thus $A \to S^{-1}B$ is flat and so is $S^{-1}A \to S^{-1}B$, see Algebra, Lemma 10.39.18. $\square$