The Stacks project

Proof. Observe that $X$ is Cohen-Macaulay, see Algebra, Lemma 10.135.3. Thus $X = X' \amalg X''$ with $\dim (X') = 0$ and $X''$ equidimensional of dimension $1$, see Morphisms, Lemma 29.29.4. Since $X'$ is finite over $k$ (Varieties, Lemma 33.20.2) we can find $Y' \to \mathop{\mathrm{Spec}}(k[[t]])$ with special fibre $X'$ and generic fibre smooth by Lemma 93.17.4. Thus it suffices to prove the lemma for $X''$. After replacing $X$ by $X''$ we have $X$ is Cohen-Macaulay and equidimensional of dimension $1$.

We are going to use deformation theory for the situation $\Lambda = k \to k$. Let $p_1, \ldots , p_ r \in X$ be the closed singular points of $X$, i.e., the points where $X \to \mathop{\mathrm{Spec}}(k)$ isn't smooth. For each $i$ we pick an integer $n_ i$ and a flat, essentially of finite type ring map

with $B_ i/tB_ i \cong \mathcal{O}_{X, p_ i}$ such that $t^{n_ i}$ is in the $1$st Fitting ideal of $\Omega _{B_ i/k[[t]]}$. This is possible by Lemma 93.17.5. Observe that the system $(B_ i/t^ nB_ i)$ defines a formal object of $\mathcal{D}\! \mathit{ef}_{\mathcal{O}_{X, p_ i}}$ over $k[[t]]$. By Lemma 93.16.3 the map

is a smooth map between deformation categories. Hence by Formal Deformation Theory, Lemma 90.8.8 there exists a formal object $(X_ n)$ in $\mathcal{D}\! \mathit{ef}_ X$ mapping to the formal object $\prod _ i (B_ i/t^ n)$ by the arrow above. By More on Morphisms of Spaces, Lemma 76.43.5 there exists a projective scheme $Y$ over $k[[t]]$ and compatible isomorphisms $Y \times _{\mathop{\mathrm{Spec}}(k[[t]])} \mathop{\mathrm{Spec}}(k[t]/(t^ n)) \cong X_ n$. By More on Morphisms, Lemma 37.12.4 we see that $Y \to \mathop{\mathrm{Spec}}(k[[t]])$ is flat. Since $X$ is Cohen-Macaulay and equidimensional of dimension $1$ we may apply Lemma 93.17.1 to check $Y$ has smooth generic fibre¹. Choose $n$ strictly larger than the maximum of the integers $n_ i$ found above. It we can show $t^{n - 1}$ is in the first Fitting ideal of $\Omega _{X_ n/S_ n}$ with $S_ n = \mathop{\mathrm{Spec}}(k[t]/(t^ n))$, then the proof is done. To do this it suffices to prove this is true in each of the local rings of $X_ n$ at closed points $p$. However, if $p$ corresponds to a smooth point for $X \to \mathop{\mathrm{Spec}}(k)$, then $\Omega _{X_ n/S_ n, p}$ is free of rank $1$ and the first Fitting ideal is equal to the local ring. If $p = p_ i$ for some $i$, then

Since taking Fitting ideals commutes with base change (with already used this but in this algebraic setting it follows from More on Algebra, Lemma 15.8.4), and since $n - 1 \geq n_ i$ we see that $t^{n - 1}$ is in the Fitting ideal of this module over $B_ i/t^ nB_ i$ as desired. $\square$