Lemma 91.17.1. Let $k$ be a field. Set $S = \mathop{\mathrm{Spec}}(k[[t]])$ and $S_ n = \mathop{\mathrm{Spec}}(k[t]/(t^ n))$. Let $Y \to S$ be a proper, flat morphism of schemes whose special fibre $X$ is Cohen-Macaulay and equidimensional of dimension $d$. Denote $X_ n = Y \times _ S S_ n$. If for some $n \geq 1$ the $d$the Fitting ideal of $\Omega _{X_ n/S_ n}$ contains $t^{n - 1}$, then the generic fibre of $Y \to S$ is smooth.

**Proof.**
By More on Morphisms, Lemma 37.20.7 we see that $Y \to S$ is a Cohen-Macaulay morphism. By Morphisms, Lemma 29.29.4 we see that $Y \to S$ has relative dimension $d$. By Divisors, Lemma 31.10.3 the $d$th Fitting ideal $\mathcal{I} \subset \mathcal{O}_ Y$ of $\Omega _{Y/S}$ cuts out the singular locus of the morphism $Y \to S$. In other words, $V(\mathcal{I}) \subset Y$ is the closed subset of points where $Y \to S$ is not smooth. By Divisors, Lemma 31.10.1 formation of this Fitting ideal commutes with base change. By assumption we see that $t^{n - 1}$ is a section of $\mathcal{I} + t^ n\mathcal{O}_ Y$. Thus for every $x \in X = V(t) \subset Y$ we conclude that $t^{n - 1} \in \mathcal{I}_ x$ where $\mathcal{I}_ x$ is the stalk at $x$. This implies that $V(\mathcal{I}) \subset V(t)$ in an open neighbourhood of $X$ in $Y$. Since $Y \to S$ is proper, this implies $V(\mathcal{I}) \subset V(t)$ as desired.
$\square$

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## Comments (1)

Comment #6020 by Will Chen on