The Stacks project

Proof. Choose a fundamental system $B \supset J_1 \supset J_2 \supset \ldots$ of open ideals. By tautness of $\varphi$, for every $n$ we can find an open ideal $I_ n$ such that $J_ n \supset I_ nB$. We claim that $I_ n$ is a fundamental system of open ideals of $A$. Namely, suppose that $I \subset A$ is open. As $\varphi$ is taut, the closure of $IB$ is open and hence contains $J_ n$ for some $n$ large enough. Hence $I_ nB \subset IB$. Let $J$ be the closure of $IB$ in $B$. Since $A/I \to B/J$ is faithfully flat, it is injective. Hence, since $I_ n \to A/I \to B/J$ is zero as $I_ nB \subset IB \subset J$, we conclude that $I_ n \to A/I$ is zero. Hence $I_ n \subset I$ and we win. $\square$