The Stacks project

Lemma 87.21.10. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Let $B$ be an $I$-adically complete $A$-algebra. Assume that

  1. the $I$-torsion in $A$ is $0$,

  2. $A/I^ n \to B/I^ nB$ is flat and of finite type for all $n$.

Then $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective if and only if $A/I \to B/IB$ is faithfully flat.

Proof. Faithful flatness implies rig-surjectivity by Lemma 87.21.9. To prove the converse we will use without further mention that the vanishing of $I$-torsion is equivalent to the vanishing of $I$-power torsion (More on Algebra, Lemma 15.88.3). We will also use without further mention that morphisms between formal spectra are given by continuous maps between the corresponding topological rings, see Formal Spaces, Lemma 86.9.10.

Assume $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective. Choose a maximal ideal $I \subset \mathfrak m \subset A$. The open $U = \mathop{\mathrm{Spec}}(A_\mathfrak m) \setminus V(I_\mathfrak m)$ of $\mathop{\mathrm{Spec}}(A_\mathfrak m)$ is nonempty as the $I_\mathfrak m$-torsion of $A_\mathfrak m$ is zero (use Algebra, Lemma 10.62.4). Thus we can find a prime $\mathfrak q \subset A_\mathfrak m$ which defines a point of $U$ (i.e., $IA_\mathfrak m \not\subset \mathfrak q$) and which corresponds to a closed point of $\mathop{\mathrm{Spec}}(A_\mathfrak m) \setminus \{ \mathfrak m\} $, see Properties, Lemma 28.6.4. Then $A_\mathfrak m/\mathfrak q$ is a dimension $1$ local domain. Thus we can find an injective local homomorphism of local rings $A_\mathfrak m/\mathfrak q \subset R$ where $R$ is a discrete valuation ring (Algebra, Lemma 10.119.13). By construction $IR \subset \mathfrak m_ R$ and we see that $A \to R$ extends to a continuous map $A \to R^\wedge $. Since the completion of a discrete valuation ring is a discrete valuation ring, we see that the assumption gives a commutative diagram of rings

\[ \xymatrix{ R' & B \ar[l] \\ R^\wedge \ar[u] & A \ar[l] \ar[u] } \]

Thus we find a prime ideal of $B$ lying over $\mathfrak m$. It follows that $\mathop{\mathrm{Spec}}(B/IB) \to \mathop{\mathrm{Spec}}(A/I)$ is surjective, whence $A/I \to B/IB$ is faithfully flat (Algebra, Lemma 10.39.16). $\square$


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