The Stacks project

Proof. The statement makes sense because $X_{/T}$ is adic* by Lemma 87.20.8 (and hence we're allowed to use the terminology adic for morphisms, see Definition 87.23.2). Let $p$ be given. By Lemma 87.33.3 we get a unique morphism $g : \mathop{\mathrm{Spec}}(R) \to X$ corresponding to the composition $\text{Spf}(R) \to X_{/T} \to X$. Let $Z \subset X$ be the reduced induced closed subspace structure on $T$. The incusion morphism $Z \to X$ corresponds to a morphism $Z \to X_{/T}$. Since $p$ is adic it is representable by algebraic spaces and we find

is an algebraic space endowed with a closed immersion to $\text{Spf}(R)$. (Equality holds because $X_{/T} \to X$ is a monomorphism.) Thus this fibre product is equal to $\mathop{\mathrm{Spec}}(R/J)$ for some ideal $J \subset R$ wich contains $\mathfrak m_ R^{n_0}$ for some $n_0 \geq 1$. This implies that $\mathop{\mathrm{Spec}}(R) \times _ X Z$ is a closed subscheme of $\mathop{\mathrm{Spec}}(R)$, say $\mathop{\mathrm{Spec}}(R) \times _ X Z = \mathop{\mathrm{Spec}}(R/I)$, whose intersection with $\mathop{\mathrm{Spec}}(R/\mathfrak m_ R^ n)$ for $n \geq n_0$ is equal to $\mathop{\mathrm{Spec}}(R/J)$. In algebraic terms this says $I + \mathfrak m_ R^ n = J + \mathfrak m_ R^ n = J$ for all $n \geq n_0$. By Krull's intersection theorem this implies $I = J$ and we conclude. $\square$