The Stacks project

Proof. By More on Morphisms, Lemma 37.22.7 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. $\square$

Proof. Recall that the set of points where $\pi $ is smooth is open. Thus the complement, i.e., the singular locus, is closed. By Chevalley's theorem (in the form of Morphisms, Lemma 29.22.2) the image of the singular locus is constructible. Hence if the generic point of $\mathbf{A}^{c(n + 1)}_ k$ is not in the image of the singular locus, then the lemma follows (by Topology, Lemma 5.15.15 for example). Thus we have to show there is no point $y \in Y$ where $\pi $ is not smooth mapping to the generic point of $\mathbf{A}^{c(n + 1)}_ k$. Consider the matrix of partial derivatives

The image of this matrix in $\kappa (y)$ must have rank $< c$ since otherwise $\pi $ would be smooth at $y$, see discussion in Smoothing Ring Maps, Section 16.2. Thus we can find $\lambda _1, \ldots , \lambda _ c \in \kappa (y)$ not all zero such that the vector $(\lambda _1, \ldots , \lambda _ c)$ is in the kernel of this matrix. After renumbering we may assume $\lambda _1 \not= 0$. Dividing by $\lambda _1$ we may assume our vector has the form $(1, \lambda _2, \ldots , \lambda _ c)$. Then we obtain

in $\kappa (y)$ for $i = 1, \ldots , n$. Moreover, since $y \in Y$ we also have

in $\kappa (y)$. This means that the subfield of $\kappa (y)$ generated by $a_{ij}$ is contained in the subfield of $\kappa (y)$ generated by the images of $x_1, \ldots , x_ n, \lambda _2, \ldots , \lambda _ c$, and $a_{ij}$ except for $a_{i1}$ and $a_{0j}$. We count and we see that the transcendence degree of this is at most $c(n + 1) - 1$. Hence $y$ cannot map to the generic point as desired. $\square$

Proof. Write $A = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ as in Algebra, Definition 10.135.1. We are going to choose $a_{ij} \in (t) \subset k[[t]]$ and set

After doing this we take $B = k[[t]][x_1, \ldots , x_ n]/(g_1, \ldots , g_ c)$. We claim that $k[[t]] \to B$ is flat at every prime ideal lying over $(t)$. Namely, the elements $f_1, \ldots , f_ c$ form a regular sequence in the local ring at any prime ideal $\mathfrak p$ of $k[x_1, \ldots , x_ n]$ containing $f_1, \ldots , f_ c$ (Algebra, Lemma 10.135.4). Thus $g_1, \ldots , g_ c$ is locally a lift of a regular sequence and we can apply Algebra, Lemma 10.99.3. Flatness at primes lying over $(0) \subset k[[t]]$ is automatic because $k((t)) = k[[t]]_{(0)}$ is a field. Thus $B$ is flat over $k[[t]]$.

All that remains is to show that for suitable choices of $a_{ij}$ the generic fibre $B_{(0)}$ is smooth over $k((t))$. For this we have to show that we can choose our $a_{ij}$ so that the induced morphism

maps into the nonempty Zariski open of Lemma 93.17.2. This is clear because there is no nonzero polynomial in the $a_{ij}$ which vanishes on $(t)^{\oplus c(n + 1)}$. (We leave this as an exercise to the reader.) $\square$

Proof. Since $A$ is Artinian (Algebra, Lemma 10.53.2), we can write $A$ as a product of local Artinian rings (Algebra, Lemma 10.53.6). Thus it suffices to prove the lemma if $A$ is local (this uses that being a local complete intersection is preserved under taking principal localizations, see Algebra, Lemma 10.135.2). In this case $A$ is a global complete intersection. Consider the algebra $B$ constructed in Lemma 93.17.3. Then $k[[t]] \to B$ is quasi-finite at the unique prime of $B$ lying over $(t)$ (Algebra, Definition 10.122.3). Observe that $k[[t]]$ is a henselian local ring (Algebra, Lemma 10.153.9). Thus $B = B' \times C$ where $B'$ is finite over $k[[t]]$ and $C$ has no prime lying over $(t)$, see Algebra, Lemma 10.153.3. Then $B'$ is the ring we are looking for (recall that étale is the same thing as smooth of relative dimension $0$). $\square$

Proof. By Algebra, Lemma 10.135.7 we can write $A$ as the localization at a prime $\mathfrak p$ of a global complete intersection $P$ over $k$. Observe that $\dim (P) = d$ by Algebra, Lemma 10.116.3. By Lemma 93.17.3 we can find a flat, finite type ring map $k[[t]] \to Q$ such that $P \cong Q/tQ$ and such that $k((t)) \to Q[1/t]$ is smooth. It follows from the construction of $Q$ in the lemma that $k[[t]] \to Q$ is a relative global complete intersection of relative dimension $d$; alternatively, Algebra, Lemma 10.136.15 tells us that $Q$ or a suitable principal localization of $Q$ is such a global complete intersection. Hence by Divisors, Lemma 31.10.3 the $d$th Fitting ideal $I \subset Q$ of $\Omega _{Q/k[[t]]}$ cuts out the singular locus of $\mathop{\mathrm{Spec}}(Q) \to \mathop{\mathrm{Spec}}(k[[t]])$. Thus $t^ n \in I$ for some $n$. Let $\mathfrak q \subset Q$ be the inverse image of $\mathfrak p$. Set $B = Q_\mathfrak q$. The lemma is proved. $\square$

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. $\square$

Proof. If $X$ is reduced, then we can choose an embedding $X \subset \overline{X}$ as in Varieties, Lemma 33.43.6. Writing $X = \overline{X} \setminus \{ x_1, \ldots , x_ n\} $ we see that $\mathcal{O}_{\overline{X}, x_ i}$ is a discrete valuation ring and hence in particular a local complete intersection (Algebra, Definition 10.135.5). Thus $\overline{X}$ is a local complete intersection over $k$ because this holds over the open $X$ and at the points $x_ i$ by Algebra, Lemma 10.135.7. Thus we may apply Lemma 93.17.6 to find a projective flat morphism $\overline{Y} \to \mathop{\mathrm{Spec}}(k[[t]])$ whose generic fibre is smooth and whose special fibre is $\overline{X}$. Then we remove $x_1, \ldots , x_ n$ from $\overline{Y}$ to obtain $Y$.

In the general case, write $X = X' \amalg X''$ where with $\dim (X') = 0$ and $X''$ equidimensional of dimension $1$. Then $X''$ is reduced and the first paragraph applies to it. On the other hand, $X'$ can be dealt with as in the proof of Lemma 93.17.6. Some details omitted. $\square$