Lemma 86.14.6. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak p \subset A$ be rig-closed. Let $\mathfrak a \subset A$ and $\mathfrak b \subset B$ be the ideals of topologically nilpotent elements. If $\varphi$ is flat, then the following are equivalent

1. the maximal ideal of $A/\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(B/\mathfrak b) \to \mathop{\mathrm{Spec}}(A/\mathfrak a)$,

2. there exists a rig-closed prime ideal $\mathfrak q \subset B$ such that $\mathfrak p = \varphi ^{-1}(\mathfrak q)$.

and if so then $\varphi$, $\mathfrak p$, and $\mathfrak q$ satisfy the conclusions of Lemma 86.14.4.

Proof. The implication (2) $\Rightarrow$ (1) is immediate. Assume (1). To prove the existence of $\mathfrak q$ we may replace $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ (some details omitted). Thus we may assume $(A, \mathfrak m, \kappa )$ is a local complete $1$-dimensional Noetherian ring, $\mathfrak m = \mathfrak a$, and $\mathfrak p = (0)$. Condition (1) just says that $B_0 = B \otimes _ A \kappa = B/\mathfrak m B = B/\mathfrak a B$ is nonzero. Note that $B_0$ is of finite type over $\kappa$. Hence we can use induction on $\dim (B_0)$. If $\dim (B_0) = 0$, then any minimal prime $\mathfrak q \subset B$ will do (flatness of $A \to B$ insures that $\mathfrak q$ will lie over $\mathfrak p = (0)$). If $\dim (B_0) > 0$ then we can find an element $b \in B$ which maps to an element $b_0 \in B_0$ which is a nonzerodivisor and a nonunit, see Algebra, Lemma 10.130.4. By Algebra, Lemma 10.99.2 the ring $B' = B/bB$ is flat over $A$. Since $B'_0 = B' \otimes _ A \kappa = B_0/(b_0)$ is not zero, we may apply the induction hypothesis to $B'$ and conclude. The final statement of the lemma is clear from Lemma 86.14.4. $\square$

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