The Stacks project

Proof. The remarks following Definition 87.5.1 show that (2) implies (1). Conversely, assume $\varphi$ is taut. If $I \subset A$ is a weak ideal of definition, then the closure of $\varphi (I)B$ is open by definition of tautness and consists of topologically nilpotent elements by Lemma 87.4.10. Hence the closure of $\varphi (I)B$ is a weak ideal of definition. Furthermore, by definition of tautness these ideals form a fundamental system of open ideals and we see that (2) is true. $\square$

Proof. Let $I_\lambda$ be a fundamental system of open ideals of $A$. Recall that $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/I_\lambda$ with the limit topology, which means that the kernels $J_\lambda = \mathop{\mathrm{Ker}}(A^\wedge \to A/I_\lambda )$ form a fundamental system of open ideals of $A^\wedge$. Since $J_\lambda$ is the closure of $I_\lambda A^\wedge$ (compare with Lemma 87.4.11) we conclude. $\square$

Proof. Omitted. Hint: if $I \subset A$ is an ideal and $J$ is the closure of $IB$, then the closure of $JC$ is equal to the closure of $IC$. $\square$

Proof. Let $J \subset B$ be an open ideal with inverse image $I \subset A$. Then the closure of $JC$ contains the closure of $IC$. Hence this closure is open as $A \to C$ is taut. Let $I_\lambda$ be a fundamental system of open ideals of $A$. Let $K_\lambda$ be the closure of $I_\lambda C$. Since $A \to C$ is taut, these form a fundamental system of open ideals of $C$. Denote $J_\lambda \subset B$ the inverse image of $K_\lambda$. Then the closure of $J_\lambda C$ is $K_\lambda$. Hence we see that the closures of the ideals $JC$, where $J$ runs over the open ideals of $B$ form a fundamental system of open ideals of $C$. $\square$

Proof. Let $K \subset C$ be an open ideal. Choose any open ideal $I \subset A$ whose image in $C$ is contained in $J$. By assumption the closure $J$ of $IB$ is open. Since $A \to B$ is taut we see that $B \widehat{\otimes }_ A C$ is the limit of the rings $B/J \otimes _{A/I} C/K$ over all choices of $K$ and $I$, i.e, the ideals $J(B \widehat{\otimes }_ A C) + K(B \widehat{\otimes }_ A C)$ form a fundamental system of open ideals. Now, since $B \to B \widehat{\otimes }_ A C$ is continuous we see that $J$ maps into the closure of $K(B \widehat{\otimes }_ A C)$ (as $I$ maps into $K$). Hence this closure is equal to $J(B \widehat{\otimes }_ A C) + K(B \widehat{\otimes }_ A C)$ and the proof is complete. $\square$

Proof. Immediate from the definitions. $\square$

Proof. Let $I \subset A$ be a weak ideal of definition. Then the closure $J$ of $IB$ is open and consists of topologically nilpotent elements by Lemma 87.4.10. Hence $J$ is a weak ideal of definition of $B$. $\square$

Proof. Let $I \subset A$ be an ideal of definition. Let $I_\lambda \subset A$ be a fundamental system of open ideals. Then the closure $J$ of $IB$ is open and the closures $J_\lambda$ of $I_\lambda B$ are open and form a fundamental system of open ideals of $B$. For every $\lambda$ there is an $n$ such that $I^ n \subset I_\lambda$. Observe that $J^ n$ is contained in the closure of $I^ nB$. Thus $J^ n \subset J_\lambda$ and we conclude $J$ is an ideal of definition. $\square$

Proof. By the open mapping lemma (More on Algebra, Lemma 15.36.5) combined with tautness of $\varphi$, we see the map $\varphi$ is open. Since the image of $\varphi$ is dense, we see that $\varphi$ is surjective. The kernel $K$ of $\varphi$ is closed as $\varphi$ is continuous. It follows that $B = A/K$ is complete, see for example Lemma 87.4.4. $\square$