Example 87.27.6. Let $S$ be a scheme. Let $A$ be a weakly admissible topological ring over $S$. Let $K \subset A$ be a closed ideal. Setting

$B = (A/K)^\wedge = \mathop{\mathrm{lim}}\nolimits _{I \subset A\ w.i.d.} A/(I + K)$

the morphism $\text{Spf}(B) \to \text{Spf}(A)$ is representable, see Example 87.19.11. If $T \to \text{Spf}(A)$ is a morphism where $T$ is a quasi-compact scheme, then this factors through $\mathop{\mathrm{Spec}}(A/I)$ for some weak ideal of definition $I \subset A$ (Lemma 87.9.4). Then $T \times _{\text{Spf}(A)} \text{Spf}(B)$ is equal to $T \times _{\mathop{\mathrm{Spec}}(A/I)} \mathop{\mathrm{Spec}}(A/(K + I))$ and we see that $\text{Spf}(B) \to \text{Spf}(A)$ is a closed immersion. The kernel of $A \to B$ is $K$ as $K$ is closed, but beware that in general the ring map $A \to B = (A/K)^\wedge$ need not be surjective.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).