hcju/stacksqa huggingface dataset:
Let $I = \Ker(A' \to A)$. By induction on the integer $n > 1$ such that $I^n = 0$ we reduce to the case where $I$ is an ideal of square zero; details omitted. We translate the condition of liftability of $M$ into the problem of finding an object of $D^-(P')$ as in Lemma \ref{lemma-lift-equivalence-module-derived}. The obstruction to doing this is the element $$ \omega(M) \in \text{Ext}^2_P(M, M \otimes_P^\mathbf{L} IP) = \text{Ext}^2_P(M, M \otimes_P IP) $$ constructed in Deformation Theory, Lemma \ref{defos-lemma-canonical-class-algebra}. The equality in the displayed formula holds as $M \otimes_P^\mathbf{L} IP = M \otimes_P IP$ since $M$ and $P$ are $A$-flat\footnote{Choose a resolution $F_\bullet \to I$ by free $A$-modules. Since $A \to P$ is flat, $P \otimes_A F_\bullet$ is a free resolution of $IP$. Hence $M \otimes_P^\mathbf{L} IP$ is represented by $M \otimes_P P \otimes_A F_\bullet = M \otimes_A F_\bullet$. This only has cohomology in degree $0$ as $M$ is $A$-flat.}. The obstruction for lifting $M \otimes_P Q$ is similarly the element $$ \omega(M \otimes_P Q) \in \text{Ext}^2_Q(M \otimes_P Q, (M \otimes_P Q) \otimes_Q IQ) $$ which is the image of $\omega(M)$ by the functoriality of the construction $\omega(-)$ of Deformation Theory, Lemma \ref{defos-lemma-canonical-class-algebra}. By More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom} we have $$ \text{Ext}^2_Q(M \otimes_P Q, (M \otimes_P Q) \otimes_Q IQ) = \text{Ext}^2_P(M, M \otimes_P IP) \otimes_P Q $$ here we use that $P$ is Noetherian and $M$ finite. Our assumption on $P' \to Q'$ guarantees that for an $P$-module $E$ the map $E \to E \otimes_P Q$ is bijective on $J'$-power torsion, see More on Algebra, Lemma \ref{more-algebra-lemma-neighbourhood-equivalence}. Thus we conclude that it suffices to show $\omega(M)$ is $J'$-power torsion. In other words, it suffices to show that $\omega(M)$ dies in $$ \text{Ext}^2_P(M, M \otimes_P IP)_g = \text{Ext}^2_{P_g}(M_g, M_g \otimes_{P_g} IP_g) $$ for all $g \in J'$. Howeover, by the compatibility of formation of $\omega(M)$ with base change again, we conclude that this is true as $M_g$ is assumed to have a lift (of course you have to use the whole string of equivalences again).
Choose an $A$-algebra surjection $P = A[x_1, \ldots, x_n] \to B$. Let $\mathfrak p \subset P$ be the inverse image of $\mathfrak b$. Set $P' = A'[x_1, \ldots, x_n]$ and denote $\mathfrak p' \subset P'$ the inverse image of $\mathfrak p$. (Of course $\mathfrak p$ and $\mathfrak p'$ do not designate prime ideals here.) We will denote $P^\wedge$ and $(P')^\wedge$ the respective completions. \medskip\noindent Suppose $A' \to B'$ is a flat lift of $A \to B$, in other words, $A' \to B'$ is flat and there is an $A$-algebra isomorphism $B = B' \otimes_{A'} A$. Then we can choose an $A'$-algebra map $P' \to B'$ lifting the given surjection $P \to B$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we find that $B'$ is a quotient of $P'$. In particular, we find that we can endow $B'$ with an $A'$-flat $P'$-module structure lifting $B$ as an $A$-flat $P$-module. Conversely, if we can lift $B$ to a $P'$-module $M'$ flat over $A'$, then $M'$ is a cyclic module $M' \cong P'/J'$ (using Nakayama again) and setting $B' = P'/J'$ we find a flat lift of $B$ as an algebra. \medskip\noindent Set $C = B^\wedge$ and $\mathfrak c = \mathfrak bC$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Then $C'$ is complete with respect to the inverse image $\mathfrak c'$ of $\mathfrak c$ (Algebra, Lemma \ref{algebra-lemma-complete-modulo-nilpotent}). We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. These maps pass through completions to give surjections $P^\wedge \to C$ and $(P')^\wedge \to C'$ (for the second again using Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $(P')^\wedge$-module structure lifting $C$ as an $A$-flat $P^\wedge$-module. Conversely, if we can lift $C$ to a $(P')^\wedge$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong (P')^\wedge/\tilde J$ (using Nakayama again) and setting $C' = (P')^\wedge/\tilde J$ we find a flat lift of $C$ as an algebra. \medskip\noindent Observe that $P' \to (P')^\wedge$ is a flat ring map which induces an isomorphism $P'/\mathfrak p' = (P')^\wedge/\mathfrak p'(P')^\wedge$. We conclude that our lemma is a consequence of Lemma \ref{lemma-lift-equivalence-module} provided we can show that $B_g$ lifts to an $A'$-flat $P'_g$-module for $g \in \mathfrak p'$. However, the ring map $A \to B_g$ is syntomic and hence lifts to an $A'$-flat algebra $B'$ by Smoothing Ring Maps, Proposition \ref{smoothing-proposition-lift-smooth}. Since $A' \to P'_g$ is smooth, we can lift $P_g \to B_g$ to a surjective map $P'_g \to B'$ as before and we get what we want.
We know that $\Deformationcategory_P$ and $\Deformationcategory_{P^\wedge}$ are deformation categories by Lemma \ref{lemma-rings-RS}. Thus it suffices to check our functor identifies tangent spaces and a correspondence between liftability, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-easy-check-smooth}. The property on liftability is proven in Lemma \ref{lemma-lift-equivalence} and the isomorphism on tangent spaces is the special case of Lemma \ref{lemma-first-order-completion} where $N = B$.
This proof is the same as the proof of Lemma \ref{lemma-lift-equivalence} but easier. We suggest the reader to skip the proof. Choose an $A$-algebra surjection $P = A[x_1, \ldots, x_n] \to B$. Let $S_P \subset P$ be the inverse image of $S$. Set $P' = A'[x_1, \ldots, x_n]$ and denote $S_{P'} \subset P'$ the inverse image of $S_P$. \medskip\noindent Suppose $A' \to B'$ is a flat lift of $A \to B$, in other words, $A' \to B'$ is flat and there is an $A$-algebra isomorphism $B = B' \otimes_{A'} A$. Then we can choose an $A'$-algebra map $P' \to B'$ lifting the given surjection $P \to B$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we find that $B'$ is a quotient of $P'$. In particular, we find that we can endow $B'$ with an $A'$-flat $P'$-module structure lifting $B$ as an $A$-flat $P$-module. Conversely, if we can lift $B$ to a $P'$-module $M'$ flat over $A'$, then $M'$ is a cyclic module $M' \cong P'/J'$ (using Nakayama again) and setting $B' = P'/J'$ we find a flat lift of $B$ as an algebra. \medskip\noindent Set $C = S^{-1}B$. Suppose that $A' \to C'$ is a flat lift of $A \to C$. Elements of $C'$ which map to invertible elements of $C$ are invertible. We choose an $A'$-algebra map $P' \to C'$ lifting the $A$-algebra map $P \to C$. By the remark above these maps pass through localizations to give surjections $S_P^{-1}P \to C$ and $S_{P'}^{-1}P' \to C'$ (for the second use Nakayama's lemma). In particular, we find that we can endow $C'$ with an $A'$-flat $S_{P'}^{-1}P'$-module structure lifting $C$ as an $A$-flat $S_P^{-1}P$-module. Conversely, if we can lift $C$ to a $S_{P'}^{-1}P'$-module $N'$ flat over $A'$, then $N'$ is a cyclic module $N' \cong S_{P'}^{-1}P'/\tilde J$ (using Nakayama again) and setting $C' = S_{P'}^{-1}P'/\tilde J$ we find a flat lift of $C$ as an algebra. \medskip\noindent The syntomic locus of a morphism of schemes is open by definition. Let $J_B \subset B$ be an ideal cutting out the set of points in $\Spec(B)$ where $\Spec(B) \to \Spec(A)$ is not syntomic. Denote $J_P \subset P$ and $J_{P'} \subset P'$ the corresponding ideals. Observe that $P' \to S_{P'}^{-1}P'$ is a flat ring map which induces an isomorphism $P'/J_{P'} = S_{P'}^{-1}P'/J_{P'}S_{P'}^{-1}P'$ by our assumption on $S$ in the lemma, namely, the assumption in the lemma is exactly that $B/J_B = S^{-1}(B/J_B)$. We conclude that our lemma is a consequence of Lemma \ref{lemma-lift-equivalence-module} provided we can show that $B_g$ lifts to an $A'$-flat $P'_g$-module for $g \in J_B$. However, the ring map $A \to B_g$ is syntomic and hence lifts to an $A'$-flat algebra $B'$ by Smoothing Ring Maps, Proposition \ref{smoothing-proposition-lift-smooth}. Since $A' \to P'_g$ is smooth, we can lift $P_g \to B_g$ to a surjective map $P'_g \to B'$ as before and we get what we want.
This proof is the same as the proof of Lemma \ref{lemma-first-order-completion} but easier. We suggest the reader to skip the proof. The map is given by localization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_k(B, N)$ we send it to the localization $S_C^{-1}C$ of $C$ with respect to the inverse image $S_C \subset C$ of $S$. Compare with the proof of Lemma \ref{lemma-localization}. \medskip\noindent The smooth locus of a morphism of schemes is open by definition. Let $J \subset B$ be an ideal cutting out the set of points in $\Spec(B)$ where $\Spec(B) \to \Spec(A)$ is not smooth. Since $k \to B$ is of finite presentation the complex $\NL_{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^i$ is a finite $B$-module, see Algebra, Section \ref{algebra-section-netherlander} and in particular Algebra, Lemma \ref{algebra-lemma-NL-homotopy}. As $B$ is Noetherian, this means that $\NL_{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_g$ is smooth and hence $(\NL_{B/k})_g = \NL_{B_g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^i_B(\NL_{B/k}, N)_g = 0$ for $i \geq 1$ and any $B$-module $N$. Finally, we have $$ \text{Ext}^1_{S^{-1}B}(\NL_{S^{-1}B/k}, S^{-1}N) = \text{Ext}^1_B(\NL_{B/k}, N) \otimes_B S^{-1}B = \text{Ext}^1_B(\NL_{B/k}, N) $$ The first equality by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom} and Algebra, Lemma \ref{algebra-lemma-localize-NL}. The second because $\text{Ext}^1_B(\NL_{B/k}, N)$ is $J$-power torsion and elements of $S$ act invertibly on $J$-power torsion modules. This concludes the proof by the description of $\text{Exal}_A(B, N)$ as $\text{Ext}^1_B(\NL_{B/A}, N)$ given just above Lemma \ref{lemma-first-order-completion}.
We know that $\Deformationcategory_P$ and $\Deformationcategory_{S^{-1}P}$ are deformation categories by Lemma \ref{lemma-rings-RS}. Thus it suffices to check our functor identifies tangent spaces and a correspondence between liftability, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-easy-check-smooth}. The property on liftability is proven in Lemma \ref{lemma-lift-equivalence-localization} and the isomorphism on tangent spaces is the special case of Lemma \ref{lemma-first-order-localization} where $N = B$.
[First proof] This proof is a cheat. Namely, if $B$ has a flat lift $B'$, then taking the henselization $(B')^h$ we obtain a flat lift of $B^h$ (compare with the proof of Lemma \ref{lemma-henselization}). Conversely, suppose that $C'$ is an $A'$-flat lift of $(B')^h$. Then let $\mathfrak c' \subset C'$ be the inverse image of the ideal $\mathfrak b^h$. Then the completion $(C')^\wedge$ of $C'$ with respect to $\mathfrak c'$ is a lift of $B^\wedge$ (details omitted). Hence we see that $B$ has a flat lift by Lemma \ref{lemma-lift-equivalence}.
This proof is the same as the proof of Lemma \ref{lemma-first-order-completion} but easier. We suggest the reader to skip the proof. The map is given by henselization: given $0 \to N \to C \to B \to 0$ in $\text{Exal}_k(B, N)$ we send it to the henselization $C^h$ of $C$ with respect to the inverse image $J_C \subset C$ of $J$. Compare with the proof of Lemma \ref{lemma-henselization}. \medskip\noindent Since $k \to B$ is of finite presentation the complex $\NL_{B/k}$ can be represented by a complex $N^{-1} \to N^0$ where $N^i$ is a finite $B$-module, see Algebra, Section \ref{algebra-section-netherlander} and in particular Algebra, Lemma \ref{algebra-lemma-NL-homotopy}. As $B$ is Noetherian, this means that $\NL_{B/k}$ is pseudo-coherent. For $g \in J$ the $k$-algebra $B_g$ is smooth and hence $(\NL_{B/k})_g = \NL_{B_g/k}$ is quasi-isomorphic to a finite projective $B$-module sitting in degree $0$. Thus $\text{Ext}^i_B(\NL_{B/k}, N)_g = 0$ for $i \geq 1$ and any $B$-module $N$. Finally, we have \begin{align*} \text{Ext}^1_{B^h}(\NL_{B^h/k}, N^h) & = \text{Ext}^1_{B^h}(\NL_{B/k} \otimes_B B^h, N \otimes_B B^h) \\ & = \text{Ext}^1_B(\NL_{B/k}, N) \otimes_B B^h \\ & = \text{Ext}^1_B(\NL_{B/k}, N) \end{align*} The first equality by More on Algebra, Lemma \ref{more-algebra-lemma-henselization-NL} (or rather its analogue for henselizations of pairs). The second by More on Algebra, Lemma \ref{more-algebra-lemma-base-change-RHom}. The third because $\text{Ext}^1_B(\NL_{B/k}, N)$ is $J$-power torsion, the map $B \to B^h$ is flat and induces an isomorphism $B/J \to B^h/JB^h$ (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-flat}), and More on Algebra, Lemma \ref{more-algebra-lemma-neighbourhood-equivalence}. This concludes the proof by the description of $\text{Exal}_A(B, N)$ as $\text{Ext}^1_B(\NL_{B/A}, N)$ given just above Lemma \ref{lemma-first-order-completion}.
We know that $\Deformationcategory_P$ and $\Deformationcategory_{P^h}$ are deformation categories by Lemma \ref{lemma-rings-RS}. Thus it suffices to check our functor identifies tangent spaces and a correspondence between liftability, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-easy-check-smooth}. The property on liftability is proven in Lemma \ref{lemma-lift-equivalence-henselization} and the isomorphism on tangent spaces is the special case of Lemma \ref{lemma-first-order-henselization} where $N = B$.
The tangent space is finite dimensional by Lemma \ref{lemma-finite-type-rings-TI}. The functors between the categories are constructed in Lemmas \ref{lemma-localization}, \ref{lemma-henselization}, and \ref{lemma-completion} (we omit some verifications of the form: the completion of the henselization is the completion). \medskip\noindent Set $J = \mathfrak m_1 \cap \ldots \cap \mathfrak m_n$ and apply Lemma \ref{lemma-smooth-completion} to get that $\Deformationcategory_P \to \Deformationcategory_{P^\wedge}$ is smooth and induces an isomorphism on tangent spaces where $P^\wedge$ is the $J$-adic completion of $P$. However, since $P^\wedge = \prod P_{\mathfrak m_i}^\wedge$ we see that the map $\Deformationcategory_P \to \prod \Deformationcategory_{P_{\mathfrak m_i}^\wedge}$ is smooth and induces an isomorphism on tangent spaces. \medskip\noindent Let $(P^h, J^h)$ be the henselization of the pair $(P, J)$. Then $P^h = \prod P_{\mathfrak m_i}^h$ (look at idempotents and use More on Algebra, Lemma \ref{more-algebra-lemma-characterize-henselian-pair}). Hence we can apply Lemma \ref{lemma-smooth-henselization} to conclude as in the case of completion. \medskip\noindent To get the final case it suffices to show that $\Deformationcategory_{P_{\mathfrak m_i}} \to \Deformationcategory_{P_{\mathfrak m_i}^\wedge}$ is smooth and induce isomorphisms on tangent spaces for each $i$ separately. To do this, we may replace $P$ by a principal localization whose only singular point is a maximal ideal $\mathfrak m$ (corresponding to $\mathfrak m_i$ in the original $P$). Then we can apply Lemma \ref{lemma-smooth-localization} with multiplicative subset $S = P \setminus \mathfrak m$ to conclude. Minor details omitted.
Let $(A, Q) \to (k, P)$ be an object of $\Deformationcategory_P$. Then we see that $A \to Q$ is a syntomic ring map by Algebra, Definition \ref{algebra-definition-lci}. Hence for any surjection $A' \to A$ in $\mathcal{C}_\Lambda$ we see that there is a morphism $(A', Q') \to (A, Q)$ lifting $A' \to A$ by Smoothing Ring Maps, Proposition \ref{smoothing-proposition-lift-smooth}. This proves the lemma.
The equality signs hold by Lemma \ref{lemma-affine}. Let us think of $\mathcal{C}_\Lambda$ as a deformation category over $\mathcal{C}_\Lambda$ as in Formal Deformation Theory, Section \ref{formal-defos-section-smooth}. Then $$ \Deformationcategory_{P_1} \times \Deformationcategory_{P_2} = \Deformationcategory_{P_1} \times_{\mathcal{C}_\Lambda} \Deformationcategory_{P_2}, $$ see Formal Deformation Theory, Remarks \ref{formal-defos-remarks-cofibered-groupoids} (\ref{formal-defos-item-product}). Using Lemma \ref{lemma-glueing} the first statement is that the functor $$ \Deformationcategory_{P_1} \times_{\Deformationcategory_{P_{12}}} \Deformationcategory_{P_2} \longrightarrow \Deformationcategory_{P_1} \times_{\mathcal{C}_\Lambda} \Deformationcategory_{P_2} $$ is smooth. This follows from Formal Deformation Theory, Lemma \ref{formal-defos-lemma-map-fibre-products-smooth} as long as we can show that $T\Deformationcategory_{P_{12}} = (0)$. This vanishing follows from Lemma \ref{lemma-smooth} as $P_{12}$ is smooth over $k$. For the second statement it suffices to show that $\Deformationcategory_{P_1} \to \mathcal{C}_\Lambda$ is smooth, see Formal Deformation Theory, Lemma \ref{formal-defos-lemma-smooth-properties}. In other words, we have to show $\Deformationcategory_{P_1}$ is unobstructed, which is Lemma \ref{lemma-lci-unobstructed}.
Choose an affine open $U_2 \subset X$ containing $p_1, \ldots, p_n$ and the generic point of every irreducible component of $X$. This is possible by Varieties, Lemma \ref{varieties-lemma-dim-1-quasi-projective} and Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}. Then $X \setminus U_2$ is finite and we can choose an affine open $U_1 \subset X \setminus \{p_1, \ldots, p_n\}$ such that $X = U_1 \cup U_2$. Set $U_{12} = U_1 \cap U_2$. Then $U_1$ and $U_{12}$ are smooth affine schemes over $k$. We conclude that $$ \Deformationcategory_X \longrightarrow \Deformationcategory_{U_2} $$ is smooth by Lemma \ref{lemma-glueing-smooth}. Applying Lemmas \ref{lemma-affine} and \ref{lemma-isolated} we win.
Let $p_1, \ldots, p_n \in X$ be the points where $X \to \Spec(k)$ isn't smooth. Choose an affine open $U_2 \subset X$ containing $p_1, \ldots, p_n$ and the generic point of every irreducible component of $X$. This is possible by Varieties, Lemma \ref{varieties-lemma-dim-1-quasi-projective} and Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}. Then $X \setminus U_2$ is finite and we can choose an affine open $U_1 \subset X \setminus \{p_1, \ldots, p_n\}$ such that $X = U_1 \cup U_2$. Set $U_{12} = U_1 \cap U_2$. Then $U_1$ and $U_{12}$ are smooth affine schemes over $k$. We conclude that $$ \Deformationcategory_X \longrightarrow \Deformationcategory_{U_2} $$ is smooth by Lemma \ref{lemma-glueing-smooth}. Applying Lemmas \ref{lemma-affine} and \ref{lemma-lci-unobstructed} we win.
By More on Morphisms, Lemma \ref{more-morphisms-lemma-flat-finite-presentation-CM-open} we see that $Y \to S$ is a Cohen-Macaulay morphism. By Morphisms, Lemma \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension} we see that $Y \to S$ has relative dimension $d$. By Divisors, Lemma \ref{divisors-lemma-d-fitting-ideal-omega-smooth} 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 \ref{divisors-lemma-base-change-and-fitting-ideal-omega} 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.
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 \ref{morphisms-lemma-chevalley}) 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 \ref{topology-lemma-generic-point-in-constructible} 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 $$ (\frac{\partial g_j}{\partial x_i}) = (\frac{\partial f_j}{\partial x_i} + a_{ij}) $$ 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 \ref{smoothing-section-singular-ideal}. 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 $$ a_{i1} = - \frac{\partial f_j}{\partial x_1} - \sum\nolimits_{j = 2, \ldots, c} \lambda_j(\frac{\partial f_j}{\partial x_i} + a_{ij}) $$ in $\kappa(y)$ for $i = 1, \ldots, n$. Moreover, since $y \in Y$ we also have $$ a_{0j} = -f_j - a_{1j}x_1 - \ldots - a_{nj}x_n $$ 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.
Write $A = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ as in Algebra, Definition \ref{algebra-definition-lci-field}. We are going to choose $a_{ij} \in (t) \subset k[[t]]$ and set $$ g_j = f_j + a_{0j} + a_{1j}x_1 + \ldots + a_{nj}x_n \in k[[t]][x_1, \ldots, x_n] $$ 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 \ref{algebra-lemma-lci}). Thus $g_1, \ldots, g_c$ is locally a lift of a regular sequence and we can apply Algebra, Lemma \ref{algebra-lemma-grothendieck-regular-sequence}. 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 $k[[t]]$. \medskip\noindent 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 $$ (a_{ij}) : \Spec(k[[t]]) \longrightarrow \mathbf{A}^{c(n + 1)}_k $$ maps into the nonempty Zariski open of Lemma \ref{lemma-jouanolou-type-thing}. 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.)
Since $A$ is Artinian (Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra}), we can write $A$ as a product of local Artinian rings (Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}). 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 \ref{algebra-lemma-localize-lci}). In this case $A$ is a global complete intersection. Consider the algebra $B$ constructed in Lemma \ref{lemma-smoothing-affine-lci}. Then $k[[t]] \to B$ is quasi-finite at the unique prime of $B$ lying over $(t)$ (Algebra, Definition \ref{algebra-definition-quasi-finite}). Observe that $k[[t]]$ is a henselian local ring (Algebra, Lemma \ref{algebra-lemma-complete-henselian}). Thus $B = B' \times C$ where $B'$ is finite over $k[[t]]$ and $C$ has no prime lying over $(t)$, see Algebra, Lemma \ref{algebra-lemma-characterize-henselian}. Then $B'$ is the ring we are looking for (recall that \'etale is the same thing as smooth of relative dimension $0$).