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Zero Frobenius tangent map does not imply formal etaleness

Statement refuted

“A morphism of schemes over a field is formally etale whenever the induced map on absolute differentials $d F$ vanishes.”

Counterexample

Let $k = F_{p}$ and let $F ⁣ : A_{k}^{1} \to A_{k}^{1}$ be the absolute Frobenius, the endomorphism of $Spec k [t]$ induced by the $k$ -algebra map $k [u] \to k [t]$ , $u \mapsto t^{p}$ ; since $a^{p} = a$ for $a \in F_{p}$ , the map is a morphism of $k$ -schemes. Then $d F ⁣ : F^{*} Ω_{A_{k}^{1} / k} ⟶ Ω_{A_{k}^{1} / k}$ sends the generator $1 \otimes d u$ to $d (t^{p}) = p t^{p - 1} d t = 0$ and is therefore the zero map, yet the relative module of the morphism is $Ω_{A_{k}^{1} / A_{k}^{1}, F} = Ω_{k [t] / k [u]} ≅ k [t] d t \neq 0,$ so $F$ is not formally unramified and hence not formally etale. The vanishing of $d F$ concerns the two absolute modules over $k$ ; formal etaleness concerns the relative module of the morphism, and the two must not be confused.

Facts & Assumptions

Given: The field $k = F_{p}$ , the affine line $X = Y = A_{k}^{1} = Spec k [t]$ with coordinate ring $k [t]$ , and the Frobenius morphism $F ⁣ : X \to Y$ induced by the $k$ -algebra map $k [u] \to k [t]$ , $u \mapsto t^{p}$ .

[F1]

Differential of an S-morphism: for a morphism $f ⁣ : X \to Y$ of $S$ -schemes there is a unique $O_{X}$ -linear map $d f ⁣ : f^{*} Ω_{Y / S} \to Ω_{X / S}$ with $1 \otimes d_{Y / S} (g) \mapsto d_{X / S} (g \circ f)$ ; its fibre at $x$ has source the cotangent space at $f (x)$ extended to $κ (x)$ , and its dual has the corresponding extended cotangent dual as target.

[F2]

Polynomial differentials are free: $Ω_{k [t] / k}$ is free with basis $d t$ and $d g = g^{'} (t) d t$ for every $g \in k [t]$ ; in particular $d (t^{p}) = p t^{p - 1} d t$ , which is $0$ over $k$ of characteristic $p$ .

[F3]

Jacobian presentation of Ω: for a commutative ring $A$ , $P = A [x]$ and $B = P / (f)$ one has $Ω_{B / A} ≅ B / (f^{'})$ , the cokernel of multiplication by the derivative, and no flatness or surjectivity of the presentation map is assumed.

[F4]

Formal unramifiedness iff Omega vanishes: a morphism of schemes is formally unramified if and only if its sheaf of relative differentials vanishes; no finiteness hypothesis is imposed.

[F5]

Formally etale morphism: a morphism is formally etale exactly when it is formally smooth and formally unramified.

Verification

1.1

The morphism $F$ is a morphism of $k$ -schemes: the ring map $k [u] \to k [t]$ is the identity on $k = F_{p}$ , where every element satisfies $a^{p} = a$ , so it is $k$ -linear and corresponds to a morphism of $k$ -schemes $Spec k [t] \to Spec k [u]$ .

given

1.2

The induced map on absolute differentials: by [F1] applied to $F$ over the base $S = Spec k$ , the map $d F ⁣ : F^{*} Ω_{Y / k} \to Ω_{X / k}$ sends $1 \otimes d u$ to $d_{X / k} (u \circ F) = d (t^{p})$ , and by [F2] one has $d (t^{p}) = p t^{p - 1} d t = 0$ because $p = 0$ in $k$ . Since $Ω_{Y / k}$ is free with basis $d u$ by [F2], the pullback is generated as an $O_{X}$ -module by $1 \otimes d u$ , so $d F = 0$ ; at every point $x$ , its dual fibre map $T_{X / k, x} \to {Hom}_{κ (x)} (Ω_{Y / k, F (x)} \otimes_{O_{Y, F (x)}} κ (x), κ (x))$ is zero.

F1F2given

1.3

The relative module of the morphism: the ring $k [t]$ is the quotient of the polynomial algebra $k [u] [v]$ in the variable $v$ by the single element $v^{p} - u$ , under the identification $v = t$ , since $t^{p} = u$ in the $k [u]$ -algebra structure. Applying [F3] with $A = k [u]$ , $P = k [u] [v]$ and $f = v^{p} - u$ gives $Ω_{k [t] / k [u]} ≅ k [t] / (f^{'})$ where $f^{'} = p v^{p - 1} = 0$ in characteristic $p$ ; hence $Ω_{k [t] / k [u]} ≅ k [t] \cdot d t$ is a free $k [t]$ -module of rank one and, in particular, nonzero.

F3given

2.1

Consequence for the lifting properties: by [F4], the nonzero relative module of step 1.3 means that $F$ is not formally unramified; by [F5] a morphism that fails to be formally unramified is not formally etale. So although $d F = 0$ by step 1.2, the Frobenius is not formally etale; the failure is detected by $Ω_{A_{k}^{1} / A_{k}^{1}, F} \neq 0$ and not by the zero map on absolute differentials.

F4F5step 1.3

3.1

Consequently the displayed statement is false: the vanishing of the map induced on absolute differentials, and hence of the dual fibre map at every point, is not a criterion for formal etaleness, because it tests a different module from the relative one; the Frobenius of step 1.2 is the witness, with $d F = 0$ and $Ω_{X / Y} ≅ k [t] d t \neq 0$ .

step 1.2step 1.3step 2.1∎

Depends on

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Sources

Stacks Morphisms 29.35.13 warning and 29.33 (standard reference, not scraped)