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

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)

• Did you check any examples of non-smooth morphisms? For example, the squaring map on $\mathbf{A}^1_S$ with $S=\operatorname{Spec}(k)$? Jun 15, 2021 at 20:10
• @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
– user290895
Jun 15, 2021 at 20:23
• Doesn't your example work as a counterexample when you take $k=\mathbb{Z}$? Or does that violate some condition I'm missing? Jun 15, 2021 at 22:10
• @AchimKrause Yes, you're right it does.
– user290895
Jun 15, 2021 at 22:32
• Your guess about support might be incorrect: you can always take a direct sum with a free module to make the support "full".
– Z. M
Jun 16, 2021 at 20:14

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.