Let $f : X\to Y$ be a syntomic morphism of locally Noetherian $S$-schemes (i.e. flat and lci) and assume $X$ and $Y$ are smooth over a locally Noetherian scheme $S$.

Q1: is $\Omega^1_{X/Y}$ a flat $\mathcal{O}_Y$-module?

The first answer here Flatness of sheaf of relative Kahler differentials contains an example of a syntomic morphism with flat $\Omega^1$ as in the question.

If the answer to the question is "no", as I expect,

**Q2:**is it "yes" under the additional assumption that $S$ is the spectrum of a local Artin ring and the dimension of the fibers of $\Omega^1_{X/Y}$ at the closed points of $Y$ is constant?

A counterexample to Q1?

A simple counterexample to Q1 has been described in the comments. However I'm now a bit confused by it, so maybe I'm missing something at the beginning.

The example is the squaring map on $\mathbf{A}^1_{\mathbf{Z}_{(2)}}$.

However, let's consider the squaring map $f$ on $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$ and the question is: is $\Omega^1_f$ flat over the base $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$?

$\Omega^1_f$ is finitely presented over the source $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$, which is in turn finite over the target $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$ via $f$, so $\Omega^1_f$, renamed as $\mathcal{F}$, is a finitely presented module over the base $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$. The locus of points of $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$ at which $\mathcal{F}$ is flat is open, so by properness of $\mathbf{P}^1$, in order to check flatness of $\mathcal{F}$, it is enough to check it on the special fiber.

Is $\Omega^1_{f_0}$, the module of differentials of the squaring map $f_0$ on $\mathbf{P}^1_{\mathbf{F}_2}$, flat as a module on the base copy of $\mathbf{P}^1_{\mathbf{F}_2}$? The answer is yes, as can be checked locally on basic affines (this is essentially done in the comments).

But then $\mathcal{F}$ is flat on the base $\mathbf{P}^1_{\mathbf{Z}_{(2)}}$, and so is its restriction to $\mathbf{A}^1_{\mathbf{Z}_{(2)}} = \mathbf{P}^1_{\mathbf{Z}_{(2)}}-\{\infty\}$. Since formation of $\Omega^1$ is local on the base, this is the $\Omega^1$ discussed in the comments.

The question is, where is the mistake located? Somewhere in the comments, or in this argument? I think I'd be already fully satisfied by an answer to this question. Thanks!

(I guess all this shows is that $f_*\Omega^1_f$ is flat over the base $\mathbf{P}^1$, which is consistent with the comments too. I'll leave this here anyway, in case further comments are warranted)

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    $\begingroup$ Did you check any examples of non-smooth morphisms? For example, the squaring map on $\mathbf{A}^1_S$ with $S=\operatorname{Spec}(k)$? $\endgroup$ Jun 15, 2021 at 20:10
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    $\begingroup$ @PiotrAchinger In that case for $A = k[x]$, $B = k[t,x]/(t^2-x)$, we have $\Omega^1_{B/A} = (k[t,x]dt)/(2tdt)$. If $k$ has char $2$ this is $k[t,x]dt$, so flat over $k[x]$. If $k$ has char not $2$, then this is $k[t,x]dt/(tdt)=k[x]dt$, flat over $k[x]$. Maybe I'm missing something? I agree we shouldn't expect the question to have a positive answer (at least the first), but I'd like to see why $\endgroup$
    – user290895
    Jun 15, 2021 at 20:23
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    $\begingroup$ Doesn't your example work as a counterexample when you take $k=\mathbb{Z}$? Or does that violate some condition I'm missing? $\endgroup$ Jun 15, 2021 at 22:10
  • $\begingroup$ @AchimKrause Yes, you're right it does. $\endgroup$
    – user290895
    Jun 15, 2021 at 22:32
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    $\begingroup$ Your guess about support might be incorrect: you can always take a direct sum with a free module to make the support "full". $\endgroup$
    – Z. M
    Jun 16, 2021 at 20:14

1 Answer 1


I think the calculation in the comment is incorrect. Let $S=\operatorname{Spec}(k)$ for a field $k$, let $m\geq 2$ be an integer invertible in $k$, and let $f\colon X\to Y$ be the $m$-th power map on $\mathbf{A}^1_k$, i.e. $\operatorname{Spec} k[x] \to \operatorname{Spec} k[y]$ with $f^*(y) = x^m$. Then $\Omega^1_{k[x]/k[y]} = k[x]dx/k[x]dy = k[x]dx/(k[x]mx^{m-1}dx)\simeq k[x]/(x^{m-1})$. As a $k[y]$-module, this is torsion: we have $y\cdot \Omega^1_{k[x]/k[y]}=0$. So it is not flat over $k[y]$. This gives a counterexample to both questions.


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