# (When) is isomorphism on differentials enough to guarantee that a map is étale?

I'm sorry if this is too easy for MO.

Let $S$ be a locally noetherian scheme, flat over $\mathrm{Spec}\,\mathbb{Z}$, $X$ and $Y$ be flat $S$-schemes locally of finite presentation, and let $f:X\to Y$ be a map over $S$. Suppose that the differential $df: f^* \Omega^1_{Y/S}\to \Omega^1_{X/S}$ is an isomorphism.

Can we conclude that $f$ is étale?

The answer is yes if $X$ is smooth over $S$ (because then $H_1(\mathbb{L}_{X/S})=0$, hence the kernel of $df$ equals $H_1(\mathbb{L}_{X/Y})$), or if $f$ is flat (because $df$ surjective means $f$ unramified, and étale means flat and unramified).

The assumption that $X$ and $Y$ are flat over $S$ are necessary, otherwise we can take $Y=S$, $f:X\to Y$ a closed immersion, in which case $\Omega^1_{Y/S} = 0$ and $\Omega^1_{X/S} = 0$ (the original version of the question missed this point).

The assumption that the $S$ is flat over $Spec(\mathbb{Z})$ is necessary: take $S=\mathrm{Spec}\,k$ with $k$ of characteristic $p>0$, $Y = \mathrm{Spec}\,k[x]$, $X = \mathrm{Spec}\,k[x]/(x^p)$.

Alternatively, we might ask the same question without the flatness over $\mathrm{Spec}\,\mathbb{Z}$ assumption, but assuming $X$ reduced.

EDIT. Here is an idea how one might proceed if $Y$ is smooth over $S$. Since $\Omega_{X/Y} = 0$, $f$ is unramified, hence étale locally on $X$ and $Y$, $f$ is a closed immersion. Moreover, $Y$ is locally étale over $\mathbb{A}^n_S$. It seems therefore that we should be able to reduce to the case $S=\mathrm{Spec}\,R$ (with $R$ a ring in which no integer is a zero divisor), $Y = \mathbb{A}^n_R$, $f:X\to Y$ a closed immersion, say $X= \mathrm{Spec}\,R[x_1, \ldots, x_n]/I$ for some ideal $I\subseteq R[x_1, \ldots, x_n]$. Let $P = R[x_1, \ldots, x_n]$. We have the conormal sequence $$I/I^2 \to \Omega_{P/R}\otimes_P P/I \to \Omega_{(P/I)/R} \to 0,$$ hence $df$ being an isomorphism means that the map $d:I/I^2 \to \Omega_{P/R}\otimes_P P/I$ is zero. This translates to the condition that $\frac{d}{dx_i}I\subseteq I$ for $i=1, \ldots, n$. Since integers are nonzero divisors in $R$, differentiating does not kill non-constant monomials in $P$, and considering degrees we find out that $I$ is generated by elements of $R$, so $X$ maps to a closed subscheme of $S$, which should contradict the fact that $X$ is flat over $S$ in case $S$ is connected (which we can assume).