The Stacks project

Proof. We will use without further mention that morphisms between formal spectra are given by continuous maps between the corresponding topological rings, see Formal Spaces, Lemma 87.9.10. Let $\varphi : A \to R$ be a continuous map into a complete discrete valuation ring $A$. This implies that $\varphi (I) \subset \mathfrak m_ R$. On the other hand, since we only need to produce the lift $\varphi ' : B' \to R'$ in the case that $\varphi $ corresponds to an adic morphism, we may assume that $\varphi (I) \not= 0$. Thus we may consider the base change $C = B \widehat{\otimes }_ A R$, see Remark 88.2.3 for example. Then $C$ is an $\mathfrak m_ R$-adically complete $R$-algebra such that $C/\mathfrak m_ R^ n C$ is of finite type and flat over $R/\mathfrak m_ R^ n$ and such that $C/\mathfrak m_ R C$ is nonzero. Pick any maximal ideal $\mathfrak m \subset C$ lying over $\mathfrak m_ R$. By flatness (which implies going down) we see that $\mathop{\mathrm{Spec}}(C_\mathfrak m) \setminus V(\mathfrak m_ R C_\mathfrak m)$ is a nonempty open. Hence We can pick a prime $\mathfrak q \subset \mathfrak m$ such that $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(C_\mathfrak m) \setminus \{ \mathfrak m\} $ and such that $\mathfrak q \not\in V(IC_\mathfrak m)$, see Properties, Lemma 28.6.4. Then $C/\mathfrak q$ is a dimension $1$-local domain and we can find $C/\mathfrak q \subset R'$ with $R'$ a discrete valuation ring (Algebra, Lemma 10.119.13). By construction $\mathfrak m_ R R' \subset \mathfrak m_{R'}$ and we see that $C \to R'$ extends to a continuous map $C \to (R')^\wedge $ (in fact we can pick $R'$ such that $R' = (R')^\wedge $ in our current situation but we do not need this). Since the completion of a discrete valuation ring is a discrete valuation ring, we see that the assumption gives a commutative diagram of rings

which gives the desired lift. $\square$