Proof.
Set $X = \mathop{\mathrm{Spec}}(A)$, $X_0 = \mathop{\mathrm{Spec}}(A_0)$, $Y_0 = \mathop{\mathrm{Spec}}(B_0)$, $Z = \mathop{\mathrm{Spec}}(C)$, $Z_0 = \mathop{\mathrm{Spec}}(C_0)$. Furthermore, denote $U \subset X$, $U_0 \subset X_0$, $V_0 \subset Y_0$, $W \subset Z$, $W_0 \subset Z_0$ the complement of the vanishing set of $I$. Here is a picture to help visualize the situation:
\[ \xymatrix{ Z \ar[dd] & Z_0 \ar[l] \ar[d] \\ & Y_0 \ar[d] \\ X & X_0 \ar[l] } \quad \quad \quad \xymatrix{ W \ar[dd] & W_0 \ar[l] \ar[d] \\ & V_0 \ar[d] \\ U & U_0 \ar[l] } \]
The conditions in the lemma guarantee that
\[ \xymatrix{ W_0 \ar[r] \ar[d] & Z_0 \ar[d] \\ V_0 \ar[r] & Y_0 } \]
is an elementary distinguished square, see Derived Categories of Spaces, Definition 73.9.1. In addition we know that $W_0 \to U_0$ and $V_0 \to U_0$ are étale. The morphism $X_0 \subset X$ is a finite order thickening as $J$ is assumed nilpotent. By the topological invariance of the étale site we can find a unique étale morphism $V \to X$ of schemes with $V_0 = V \times _ X X_0$ and we can lift the given morphism $W_0 \to V_0$ to a unique morphism $W \to V$ over $X$. See Étale Morphisms, Theorem 41.15.2. Since $W_0 \to V_0$ is separated, the morphism $W \to V$ is separated too, see for example More on Morphisms, Lemma 37.10.3. By Pushouts of Spaces, Lemma 79.9.2 we can construct an elementary distinguished square
\[ \xymatrix{ W \ar[r] \ar[d] & Z \ar[d] \\ V \ar[r] & Y } \]
in the category of algebraic spaces over $X$. Since the base change of an elementary distinguished square is an elementary distinguished square (Derived Categories of Spaces, Lemma 73.9.2) we see that
\[ \xymatrix{ W_0 \ar[r] \ar[d] & Z_0 \ar[d] \\ V_0 \ar[r] & Y \times _ X X_0 } \]
is an elementary distinguished square. It follows that there is a unique isomorphism $Y \times _ X X_0 = Y_0$ compatible with the two squares involving these spaces because elementary distinguished squares are pushouts (Pushouts of Spaces, Lemma 79.9.1). It follows that $Y$ is affine by Limits of Spaces, Proposition 68.15.2. Write $Y = \mathop{\mathrm{Spec}}(B)$. It is clear that $B$ fits into the desired diagram and satisfies all the properties required of it.
$\square$
Second proof.
This proof uses a little bit of deformation theory to construct $B$. By induction on the smallest $n$ such that $J^ n = 0$ we reduce to the case $J^2 = 0$. Denote by a subscript zero the base change of objects to $A_0 = A/J$. Since $J^2 = 0$ we see that $JC$ is a $C_0$-module.
Consider the canonical map
\[ \gamma : J \otimes _{A_0} C_0 \longrightarrow JC \]
Since $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over the complement of $V(I)$ (and hence flat) we see that $\gamma $ is an isomorphism away from $V(IC_0)$, see More on Morphisms, Lemma 37.10.1. In particular, the kernel and cokernel of $\gamma $ are annihilated by a power of $I$ (use that $C_0$ is Noetherian and that the modules in question are finite). Observe that $J \otimes _{A_0} C_0 = (J \otimes _{A_0} B_0) \otimes _{B_0} C_0$. Hence by More on Algebra, Lemma 15.84.16 there exists a unique $B_0$-module homomorphism
\[ c : J \otimes _{A_0} B_0 \to N \]
with $c \otimes \text{id}_{C_0} = \gamma $ and $\mathop{\mathrm{Ker}}(\gamma ) = \mathop{\mathrm{Ker}}(c)$ and $\mathop{\mathrm{Coker}}(\gamma ) = \mathop{\mathrm{Coker}}(c)$. Moreover, $N$ is a finite $B_0$-module, see More on Algebra, Remark 15.84.19.
Choose a presentation $B_0 = A[x_1, \ldots , x_ r]/K$. To construct $B$ we try to find the dotted arrow $m$ fitting into the following pushout diagram
\[ \xymatrix{ 0 \ar[r] & N \ar@{..>}[r] & B \ar@{..>}[r] & B_0 \ar[r] & 0 \\ 0 \ar[r] & K/K^2 \ar[r] \ar@{..>}[u]_ m & A[x_1, \ldots , x_ r]/K^2 \ar[r] \ar@{..>}[u] & A[x_1, \ldots , x_ r]/K \ar@{=}[u] \ar[r] & 0 \\ & J \otimes _{A_0} B_0 \ar[u] \ar@/^2pc/[uu] |!{[lu];[u]}\hole } \]
where the curved arrow is the map $c$ constructed above and the map $J \otimes _{A_0} B_0 \to K/K^2$ is the obvious one.
As $B_0 \to C_0$ is étale we can write $C_0 = B_0[y_1, \ldots , y_ r]/(g_{0, 1}, \ldots , g_{0, r})$ such that the determinant of the partial derivatives of the $g_{0, j}$ is invertible in $C_0$, see Algebra, Lemma 10.142.2. We combine this with the chosen presentation of $B_0$ to get a presentation $C_0 = A[x_1, \ldots , x_ r, y_1, \ldots , y_ s]/L$. Choose a lift $\psi : A[x_ i, y_ j] \to C$ of the map to $C_0$. Then it is the case that $C$ fits into the diagram
\[ \xymatrix{ 0 \ar[r] & JC \ar[r] & C \ar[r] & C_0 \ar[r] & 0 \\ 0 \ar[r] & L/L^2 \ar[r] \ar[u]_\mu & A[x_ i, y_ j]/L^2 \ar[r] \ar[u] & A[x_ i, y_ j]/L \ar@{=}[u] \ar[r] & 0 \\ & J \otimes _{A_0} C_0 \ar[u] \ar@/^2pc/[uu] |!{[lu];[u]}\hole } \]
where the curved arrow is the map $\gamma $ constructed above and the map $J \otimes _{A_0} C_0 \to L/L^2$ is the obvious one. By our choice of presentations and the fact that $C_0$ is a complete intersection over $B_0$ we have
\[ L/L^2 = K/K^2 \otimes _{B_0} C_0 \oplus \bigoplus C_0 g_ j \]
where $g_ j \in L$ is any lift of $g_{0, j}$, see More on Algebra, Lemma 15.33.6.
Consider the three term complex
\[ K^\bullet : J \otimes _{A_0} B_0 \to K/K^2 \to \bigoplus B_0 \text{d}x_ i \]
where the second arrow is the differential in the naive cotangent complex of $B_0$ over $A$ for the given presentation and the last term is placed in degree $0$. Since $\mathop{\mathrm{Spec}}(B_0) \to \mathop{\mathrm{Spec}}(A_0)$ is étale away from $V(I)$ the cohomology modules of this complex are supported on $V(IB_0)$. Namely, for $a \in I$ after inverting $a$ we can apply More on Algebra, Lemma 15.33.6 for the ring maps $A_ a \to A_{0, a} \to B_{0, a}$ and use that $\mathop{N\! L}\nolimits _{A_{0, a}/A_ a} = J_ a$ and $\mathop{N\! L}\nolimits _{B_{0, a}/A_{0, a}} = 0$ (some details omitted). Hence these cohomology groups are annihilated by a power of $I$.
Similarly, consider the three term complex
\[ L^\bullet : J \otimes _{A_0} C_0 \to L/L^2 \to \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j \]
By our direct sum decomposition of $L/L^2$ above and the fact that the determinant of the partial derivatives of the $g_{0, j}$ is invertible in $C_0$ we see that the natural map $K^\bullet \to L^\bullet $ induces a quasi-isomorphism
\[ K^\bullet \otimes _{B_0} C_0 \longrightarrow L^\bullet \]
Applying Dualizing Complexes, Lemma 47.9.8 we find that
86.7.2.1
\begin{equation} \label{restricted-equation-go-down} \mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , E) = \mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , E \otimes _{B_0} C_0) \end{equation}
for any object $E \in D(B_0)$.
The maps $\text{id}_{J \otimes _{A_0} C_0}$ and $\mu $ define an element in
\[ \mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , (J \otimes _{A_0} C_0 \to JC)) \]
(the target two term complex is placed in degree $-2$ and $-1$) such that the composition with the map to $J \otimes _{A_0} C_0[2]$ is the element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , J \otimes _{A_0} C_0[2])$ corresponding to $\text{id}_{J \otimes _{A_0} C_0}$. Picture
\[ \xymatrix{ J \otimes _{A_0} C_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} C_0}} & L/L^2 \ar[r] \ar[d]^\mu & \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j \\ J \otimes _{A_0} C_0 \ar[r]^-\gamma & JC } \]
Applying (86.7.2.1) we obtain a unique element
\[ \xi \in \mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , (J \otimes _{A_0} B_0 \to N)) \]
Its composition with the map to $J \otimes _{A_0} B_0[2]$ is the element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(K^\bullet , J \otimes _{A_0} B_0[2])$ corresponding to $\text{id}_{J \otimes _{A_0} B_0}$. By Lemma 86.3.4 we can find a map of complexes $K^\bullet \to (J \otimes _{A_0} B_0 \to N)$ representing $\xi $ and equal to $\text{id}_{J \otimes _{A_0} B_0}$ in degree $-2$. Denote $m : K/K^2 \to N$ the degree $-1$ part of this map. Picture
\[ \xymatrix{ J \otimes _{A_0} B_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} B_0}} & K/K^2 \ar[r] \ar[d]^ m & \bigoplus B_0 \text{d}x_ i \\ J \otimes _{A_0} B_0 \ar[r]^-c & N } \]
Thus we can use $m$ to create an algebra $B$ by push out as explained above. However, we may still have to change $m$ a bit to make sure that $B$ maps to $C$ in the correct manner.
Denote $m \otimes \text{id}_{C_0} \oplus 0 : L/L^2 \to JC$ the map coming from the direct sum decomposition of $L/L^2$ (see above), using that $N \otimes _{B_0} C_0 = JC$, and using $0$ on the second factor. By our choice of $m$ above the maps of complexes $(\text{id}_{J \otimes _{A_0} C_0}, \mu , 0)$ and $(\text{id}_{J \otimes _{A_0} C_0}, m \otimes \text{id}_{C_0} \oplus 0, 0)$ define the same element of $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , (J \otimes _{A_0} C_0 \to JC))$. By Lemma 86.3.4 there exist maps $h : L^{-1} \to J \otimes _{A_0} C_0$ and $h' : L^0 \to JC$ which define a homotopy between $(\text{id}_{J \otimes _{A_0} C_0}, \mu , 0)$ and $(\text{id}_{J \otimes _{A_0} C_0}, m \otimes \text{id}_{C_0} \oplus 0, 0)$. Picture
\[ \xymatrix{ J \otimes _{A_0} C_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} C_0}} & K/K^2 \otimes _{B_0} C_0 \oplus \bigoplus C_0 g_ j \ar@{..>}[ld]_ h \ar[r] \ar@<-1ex>[d]_\mu \ar[d]^{m \otimes \text{id}_{C_0} \oplus 0} & \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j \ar@{..>}[ld]_{h'}\\ J \otimes _{A_0} C_0 \ar[r]^-\gamma & JC } \]
Since $h$ precomposed with $d_ L^{-2}$ is zero it defines an element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , J \otimes _{A_0} C_0[1])$ which comes from a unique element $\chi $ of $\mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , J \otimes _{A_0} B_0[1])$ by (86.7.2.1). Applying Lemma 86.3.4 again we represent $\chi $ by a map $g : K/K^2 \to J \otimes _{A_0} B_0$. Then the base change $g \otimes \text{id}_{C_0}$ and $h$ differ by a homotopy $h'' : L^0 \to J \otimes _{A_0} C$. Hence if we modify $m$ into $m + c \circ g$, then we find that $m \otimes \text{id}_{C_0} \oplus 0$ and $\mu $ just differ by a map $h' : L^0 \to JC$.
Changing our choice of the map $\psi : A[x_ i, y_ j] \to C$ by sending $x_ i$ to $\psi (x_ i) + h'(\text{d}x_ i)$ and sending $y_ j$ to $\psi (y_ j) + h'(\text{d}y_ j)$, we find a commutative diagram
\[ \xymatrix{ N \ar[r] & JC \\ K/K^2 \ar[r] \ar[u]_ m & L/L^2 \ar[u]_\mu \\ J \otimes _{A_0} B_0 \ar[u] \ar@/^2pc/[uu]^ c \ar[r] & J \otimes _{A_0} C_0 \ar[u] \ar@/_2pc/[uu]_\gamma } \]
At this point we can define $B$ as the pushout in the first commutative diagram of the proof. The commutativity of the diagram just displayed, shows that there is an $A$-algebra map $B \to C$ compatible with the given map $N = JB \to JC$. As $N \otimes _{B_0} C_0 = JC$ it follows from More on Morphisms, Lemma 37.10.1 that $B \to C$ is flat. From this it easily follows that it is étale. We omit the proof of the other properties as they are mostly self evident at this point.
$\square$
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