Definition 37.64.1. A morphism of schemes $X \to Y$ is weakly étale or absolutely flat if both $X \to Y$ and the diagonal morphism $X \to X \times _ Y X$ are flat.
37.64 Weakly étale morphisms
A ring homomorphism $A \to B$ is weakly étale if both $A \to B$ and $B \otimes _ A B \to B$ are flat, see More on Algebra, Definition 15.104.1. The analogous notion for morphisms of schemes is the following.
An étale morphism is weakly étale and conversely it turns out that a weakly étale morphism is indeed somewhat like an étale morphism. For example, if $X \to Y$ is weakly étale, then $L_{X/Y} = 0$, as follows from Cotangent, Lemma 92.8.4. We will prove a very precise result relating weakly étale morphisms to étale morphisms later (see Pro-étale Cohomology, Section 61.9). In this section we stick with the basics.
Lemma 37.64.2. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent
$X \to Y$ is weakly étale, and
for every $x \in X$ the ring map $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ is weakly étale.
Proof. Observe that under both assumptions (1) and (2) the morphism $f$ is flat. Thus we may assume $f$ is flat. Let $x \in X$ with image $y = f(x)$ in $Y$. There are canonical maps of rings
where the first map is a localization (hence flat) and the second map is a surjection. Condition (1) means that the second arrow is flat for all $x$. Condition (2) is that the composition is flat for all $x$. Thus the equivalence by Algebra, Lemma 10.39.18 part (2). $\square$
Lemma 37.64.3. Let $X \to Y$ be a morphism of schemes such that $X \to X \times _ Y X$ is flat. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. If $\mathcal{F}$ is flat over $Y$, then $\mathcal{F}$ is flat over $X$.
Proof. Let $x \in X$ with image $y = f(x)$ in $Y$. Since $X \to X \times _ Y X$ is flat, we see that $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \to \mathcal{O}_{X, x}$ is flat. Hence the result follows from More on Algebra, Lemma 15.104.2 and the definitions. $\square$
Lemma 37.64.4. Let $f : X \to S$ be a morphism of schemes. The following are equivalent
The morphism $f$ is weakly étale.
For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is weakly étale.
There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is weakly étale.
There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is of weakly étale, for all $j\in J, i\in I_ j$.
Moreover, if $f$ is weakly étale then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is weakly-étale.
Proof. Suppose given open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$. Then $U \times _ V U \subset X \times _ Y X$ is open (Schemes, Lemma 26.17.3) and the diagonal $\Delta _{U/V}$ of $f|_ U : U \to V$ is the restriction $\Delta _{X/Y}|_ U : U \to U \times _ V U$. Since flatness is a local property of morphisms of schemes (Morphisms, Lemma 29.25.3) the final statement of the lemma is follows as well as the equivalence of (1) and (3). If $X$ and $Y$ are affine, then $X \to Y$ is weakly étale if and only if $\mathcal{O}_ Y(Y) \to \mathcal{O}_ X(X)$ is weakly étale (use again Morphisms, Lemma 29.25.3). Thus (1) and (3) are also equivalent to (2) and (4). $\square$
Lemma 37.64.5. Let $X \to Y \to Z$ be morphisms of schemes.
If $X \to X \times _ Y X$ and $Y \to Y \times _ Z Y$ are flat, then $X \to X \times _ Z X$ is flat.
If $X \to Y$ and $Y \to Z$ are weakly étale, then $X \to Z$ is weakly étale.
Proof. Part (1) follows from the factorization
of the diagonal of $X$ over $Z$, the fact that
the fact that a base change of a flat morphism is flat, and the fact that the composition of flat morphisms is flat (Morphisms, Lemmas 29.25.8 and 29.25.6). Part (2) follows from part (1) and the fact (just used) that the composition of flat morphisms is flat. $\square$
Lemma 37.64.6. Let $X \to Y$ and $Y' \to Y$ be morphisms of schemes and let $X' = Y' \times _ Y X$ be the base change of $X$.
If $X \to X \times _ Y X$ is flat, then $X' \to X' \times _{Y'} X'$ is flat.
If $X \to Y$ is weakly étale, then $X' \to Y'$ is weakly étale.
Proof. Assume $X \to X \times _ Y X$ is flat. The morphism $X' \to X' \times _{Y'} X'$ is the base change of $X \to X \times _ Y X$ by $Y' \to Y$. Hence it is flat by Morphisms, Lemmas 29.25.8. This proves (1). Part (2) follows from (1) and the fact (just used) that the base change of a flat morphism is flat. $\square$
Lemma 37.64.7. Let $X \to Y \to Z$ be morphisms of schemes. Assume that $X \to Y$ is flat and surjective and that $X \to X \times _ Z X$ is flat. Then $Y \to Y \times _ Z Y$ is flat.
Proof. Consider the commutative diagram
The top horizontal arrow is flat and the vertical arrows are flat. Hence $X$ is flat over $Y \times _ Z Y$. By Morphisms, Lemma 29.25.13 we see that $Y$ is flat over $Y \times _ Z Y$. $\square$
Lemma 37.64.8. Let $f : X \to Y$ be a weakly étale morphism of schemes. Then $f$ is formally unramified, i.e., $\Omega _{X/Y} = 0$.
Proof. Recall that $f$ is formally unramified if and only if $\Omega _{X/Y} = 0$ by Lemma 37.6.7. Via Lemma 37.64.4 and Morphisms, Lemma 29.32.5 this follows from the case of rings which is More on Algebra, Lemma 15.104.12. $\square$
Lemma 37.64.9. Let $f : X \to Y$ be a morphism of schemes. Then $X \to Y$ is weakly étale in each of the following cases
$X \to Y$ is a flat monomorphism,
$X \to Y$ is an open immersion,
$X \to Y$ is flat and unramified,
$X \to Y$ is étale.
Proof. If (1) holds, then $\Delta _{X/Y}$ is an isomorphism (Schemes, Lemma 26.23.2), hence certainly $f$ is weakly étale. Case (2) is a special case of (1). The diagonal of an unramified morphism is an open immersion (Morphisms, Lemma 29.35.13), hence flat. Thus a flat unramified morphism is weakly étale. An étale morphism is flat and unramified (Morphisms, Lemma 29.36.5), hence (4) follows from (3). $\square$
Lemma 37.64.10. Let $f : X \to Y$ be a morphism of schemes. If $Y$ is reduced and $f$ weakly étale, then $X$ is reduced.
Proof. Via Lemma 37.64.4 this follows from the case of rings which is More on Algebra, Lemma 15.104.8. $\square$
The following lemma uses a nontrivial result about weakly étale ring maps.
Lemma 37.64.11. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent
$f$ is weakly étale, and
for $x \in X$ the local ring map $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ induces an isomorphism on strict henselizations.
Proof. Let $x \in X$ be a point with image $y = f(x)$ in $Y$. Choose a separable algebraic closure $\kappa ^{sep}$ of $\kappa (x)$. Let $\mathcal{O}_{X, x}^{sh}$ be the strict henselization corresponding to $\kappa ^{sep}$ and $\mathcal{O}_{Y, y}^{sh}$ the strict henselization relative to the separable algebraic closure of $\kappa (y)$ in $\kappa ^{sep}$. Consider the commutative diagram
local homomorphisms of local rings, see Algebra, Lemma 10.155.10. Since the strict henselization is a filtered colimit of étale ring maps, More on Algebra, Lemma 15.104.14 shows the horizontal maps are weakly étale. Moreover, the horizontal maps are faithfully flat by More on Algebra, Lemma 15.45.1.
Assume $f$ weakly étale. By Lemma 37.64.2 the left vertical arrow is weakly étale. By More on Algebra, Lemmas 15.104.9 and 15.104.11 the right vertical arrow is weakly étale. By More on Algebra, Theorem 15.104.24 we conclude the right vertical map is an isomorphism.
Assume $\mathcal{O}_{Y, y}^{sh} \to \mathcal{O}_{X, x}^{sh}$ is an isomorphism. Then $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}^{sh}$ is weakly étale. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^{sh}$ is faithfully flat we conclude that $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is weakly étale by More on Algebra, Lemma 15.104.10. Thus (2) implies (1) by Lemma 37.64.2. $\square$
Lemma 37.64.12. Let $f : X \to Y$ be a morphism of schemes. If $Y$ is a normal scheme and $f$ weakly étale, then $X$ is a normal scheme.
Proof. By More on Algebra, Lemma 15.45.6 a scheme $S$ is normal if and only if for all $s \in S$ the strict henselization of $\mathcal{O}_{S, s}$ is a normal domain. Hence the lemma follows from Lemma 37.64.11. $\square$
Lemma 37.64.13. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. If $X$, $Y$ are weakly étale over $S$, then $f$ is weakly étale.
Proof. We will use Morphisms, Lemmas 29.25.8 and 29.25.6 without further mention. Write $X \to Y$ as the composition $X \to X \times _ S Y \to Y$. The second morphism is flat as the base change of the flat morphism $X \to S$. The first is the base change of the flat morphism $Y \to Y \times _ S Y$ by the morphism $X \times _ S Y \to Y \times _ S Y$, hence flat. Thus $X \to Y$ is flat. The morphism $X \times _ Y X \to X \times _ S X$ is an immersion. Thus Lemma 37.64.3 implies, that since $X$ is flat over $X \times _ S X$ it follows that $X$ is flat over $X \times _ Y X$. $\square$
The following is a scheme theoretic generalization of the observation that a field extension that is simultaneously separable and purely inseparable must be an isomorphism.
Lemma 37.64.14. Let $f : X \to Y$ be a morphism of schemes. If $f$ is weakly étale and a universal homeomorphism, it is an isomorphism.
Proof. Since $f$ is a universal homeomorphism, the diagonal $\Delta : X \to X \times _ Y X$ is a surjective closed immersion by Morphisms, Lemmas 29.45.4 and 29.10.2. Since $\Delta $ is also flat, we see that $\Delta $ must be an isomorphism by Morphisms, Lemma 29.26.1. In other words, $f$ is a monomorphism (Schemes, Lemma 26.23.2). Since $f$ is a universal homeomorphism it is certainly quasi-compact. Hence by Descent, Lemma 35.25.1 we find that $f$ is an isomorphism. $\square$
The following is a weakly étale generalization of Étale Morphisms, Lemma 41.14.3.
Lemma 37.64.15. Let $U \to X$ be a weakly étale morphism of schemes where $X$ is a scheme in characteristic $p$. Then the relative Frobenius $F_{U/X} : U \to U \times _{X, F_ X} X$ is an isomorphism.
Proof. The morphism $F_{U/X}$ is a universal homeomorphism by Varieties, Lemma 33.36.6. The morphism $F_{U/X}$ is weakly étale as a morphism between schemes weakly étale over $X$ by Lemma 37.64.13. Hence $F_{U/X}$ is an isomorphism by Lemma 37.64.14. $\square$
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