Lemma 38.2.5. Let $S$ be a scheme and $s \in S$ a point. Denote $\mathcal{O}_{S, s}^ h$ (resp. $\mathcal{O}_{S, s}^{sh}$) the henselization (resp. strict henselization), see Algebra, Definition 10.155.3. Let $M^{sh}$ be a $\mathcal{O}_{S, s}^{sh}$-module. The following are equivalent

1. $M^{sh}$ is flat over $\mathcal{O}_{S, s}$,

2. $M^{sh}$ is flat over $\mathcal{O}_{S, s}^ h$, and

3. $M^{sh}$ is flat over $\mathcal{O}_{S, s}^{sh}$.

If $M^{sh} = M^ h \otimes _{\mathcal{O}_{S, s}^ h} \mathcal{O}_{S, s}^{sh}$ this is also equivalent to

1. $M^ h$ is flat over $\mathcal{O}_{S, s}$, and

2. $M^ h$ is flat over $\mathcal{O}_{S, s}^ h$.

If $M^ h = M \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S, s}^ h$ this is also equivalent to

1. $M$ is flat over $\mathcal{O}_{S, s}$.

Proof. By More on Algebra, Lemma 15.45.1 the local ring maps $\mathcal{O}_{S, s} \to \mathcal{O}_{S, s}^ h \to \mathcal{O}_{S, s}^{sh}$ are faithfully flat. Hence (3) $\Rightarrow$ (2) $\Rightarrow$ (1) and (5) $\Rightarrow$ (4) follow from Algebra, Lemma 10.39.4. By faithful flatness the equivalences (6) $\Leftrightarrow$ (5) and (5) $\Leftrightarrow$ (3) follow from Algebra, Lemma 10.39.8. Thus it suffices to show that (1) $\Rightarrow$ (2) $\Rightarrow$ (3) and (4) $\Rightarrow$ (5). To prove these we may assume $S$ is an affine scheme.

Assume (1). By Lemma 38.2.4 we see that $M^{sh}$ is flat over $\mathcal{O}_{T, t}$ for any étale neighbourhood $(T, t) \to (S, s)$. Since $\mathcal{O}_{S, s}^ h$ and $\mathcal{O}_{S, s}^{sh}$ are directed colimits of local rings of the form $\mathcal{O}_{T, t}$ (see Algebra, Lemmas 10.155.7 and 10.155.11) we conclude that $M^{sh}$ is flat over $\mathcal{O}_{S, s}^ h$ and $\mathcal{O}_{S, s}^{sh}$ by Algebra, Lemma 10.39.6. Thus (1) implies (2) and (3). Of course this implies also (2) $\Rightarrow$ (3) by replacing $\mathcal{O}_{S, s}$ by $\mathcal{O}_{S, s}^ h$. The same argument applies to prove (4) $\Rightarrow$ (5). $\square$

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