Lemma 85.15.14. With notation and assumptions as in Lemma 85.15.13 let $f : X \to Y$ correspond to $B \to A^\wedge$. The following are equivalent

1. $f : X \to Y$ is surjective,

2. $B \to A$ is faithfully flat,

3. for every weak ideal of definition $J \subset B$ the ring map $B/J \to A/JA$ is faithfully flat, and

4. for some weak ideal of definition $J \subset B$ the ring map $B/J \to A/JA$ is faithfully flat.

Proof. Let $J \subset B$ be a weak ideal of definition. As every element of $J$ is topologically nilpotent, we see that every element of $1 + J$ is a unit. It follows that $J$ is contained in the Jacobson radical of $B$ (Algebra, Lemma 10.19.1). Hence a flat ring map $B \to A$ is faithfully flat if and only if $B/J \to A/JA$ is faithfully flat (Algebra, Lemma 10.39.16). In this way we see that (2) – (4) are equivalent. If (1) holds, then for every weak ideal of definition $J \subset B$ the morphism $\mathop{\mathrm{Spec}}(A/JA) = \mathop{\mathrm{Spec}}(B/J) \times _ Y X \to \mathop{\mathrm{Spec}}(B/J)$ is surjective which implies (3). Conversely, assume (3). A morphism $T \to Y$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(B/J)$ for some ideal of definition $J$ of $B$ (Lemma 85.5.4). Hence $X \times _ Y T = \mathop{\mathrm{Spec}}(A/JA) \times _{\mathop{\mathrm{Spec}}(B/J)} T \to T$ is surjective as a base change of the surjective morphism $\mathop{\mathrm{Spec}}(A/JA) \to \mathop{\mathrm{Spec}}(B/J)$. Thus (1) holds. $\square$

Comment #1953 by Brian Conrad on

In the statement of the lemma, in (3) and (4) the map should have source $B/J$ rather than $B$.

Comment #1954 by Brian Conrad on

In the statement of the lemma, for parts (3) and (4) the map should have source $B/J$ rather than $B$. In the first sentence of the proof, take $J$ to be a weak ideal of definition, not necessarily an ideal of definition (i.e., all we need is that each element of $J$ is topologically nilpotent, not that $J^n \rightarrow 0$, and for the proof that former generality in $J$ is essential).

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