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 $S$.



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> **Q1:** is $\Omega^1\_{X/Y}$ a flat $\mathcal{O}\_Y$-module?
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The first answer here [Flatness of sheaf of relative Kahler differentials](https://mathoverflow.net/questions/75410/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,



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> \*\*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?
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**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)



Flatness of $\Omega^1_{X/S}$
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.


