hcju/stacksqa huggingface dataset:
Choose a surjection $D''' = D''[x_1, \ldots, x_n] \to D'$ with finitely generated kernel $J$. By Algebra, Lemma \ref{algebra-lemma-finite-finitely-presented-extension} it suffices to show that $N'$ is finitely presented as a $D'''$-module. Moreover, $D''' \otimes_{B'} B \to D' \otimes_{B'} B = D$ and $D''' \otimes_{B'} A' \to D' \otimes_{B'} A' = C'$ are surjections whose kernels are generated by the image of $J$, hence $N$ is a finitely presented $D''' \otimes_{B'} B$-module and $M'$ is a finitely presented $D''' \otimes_{B'} A'$-module by Algebra, Lemma \ref{algebra-lemma-finite-finitely-presented-extension} again. Thus we may replace $D'$ by $D'''$ and $D$ by $D''' \otimes_{B'} B$, etc. Since $D'''$ is flat over $B'$, it follows that we may assume that $B' \to D'$ is flat. \medskip\noindent Assume $B' \to D'$ is flat. By Lemma \ref{lemma-relative-finite-module-over-fibre-product} the module $N'$ is finite over $D'$. Choose a surjection $(D')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $D^{\oplus n} \to N$, $(C')^{\oplus n} \to M'$, and $C^{\oplus n} \to M$ with kernels $K_D$, $K_{C'}$, and $K_C$. There is a canonical map $$ K' \longrightarrow K_D \times_{K_C} K_{C'} $$ On the other hand, since $N' = N \times_M M'$ and $D' = D \times_C C'$ (by Lemma \ref{lemma-flat-module-over-fibre-product}; applied to the flat $B'$-module $D'$) there is also a canonical map $K_D \times_{K_C} K_{C'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma \ref{algebra-lemma-extension} the modules $K_D$ and $K_{C'}$ are finite. We conclude from Lemma \ref{lemma-relative-finite-module-over-fibre-product} that $K'$ is a finite $D'$-module provided that $K_D \to K_C$ and $K_{C'} \to K_C$ induce isomorphisms $K_D \otimes_B A = K_C = K_{C'} \otimes_{A'} A$. This is true because the flatness assumptions implies the sequences $$ 0 \to K_D \to D^{\oplus n} \to N \to 0 \quad\text{and}\quad 0 \to K_{C'} \to (C')^{\oplus n} \to M' \to 0 $$ stay exact upon tensoring, see Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}.
Let $K \subset R^{\oplus n}$ be the kernel of the surjection $R^{\oplus n} \to M$. Pick $z_1, \ldots, z_{n - k} \in K$ and write $z_j = (z_{1j}, \ldots, z_{nj})$. Another description of the ideal $\text{Fit}_k(M)$ is that it is the ideal generated by the $(n - k) \times (n - k)$ minors of all the matrices $(z_{ij})$ we obtain in this way. \medskip\noindent Suppose we change the surjection into the surjection $R^{\oplus n + n'} \to M$ with kernel $K'$ where we use the original map on the first $n$ standard basis elements of $R^{\oplus n + n'}$ and $0$ on the last $n'$ basis vectors. Then the corresponding ideals are the same. Namely, if $z_1, \ldots, z_{n - k} \in K$ as above, let $z'_j = (z_{1j}, \ldots, z_{nj}, 0, \ldots, 0) \in K'$ for $j = 1, \ldots, n - k$ and $z'_{n + j'} = (0, \ldots, 0, 1, 0, \ldots, 0) \in K'$. Then we see that the ideal of $(n - k) \times (n - k)$ minors of $(z_{ij})$ agrees with the ideal of $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$. This gives one of the inclusions. Conversely, given  $z'_1, \ldots, z'_{n + n' - k}$ in $K'$ we can project these to $R^{\oplus n}$ to get $z_1, \ldots, z_{n + n' - k}$ in $K$. By Lemma \ref{lemma-ideals-generated-by-minors} we see that the ideal generated by the $(n + n' - k) \times (n + n' - k)$ minors of $(z'_{ij})$ is contained in the ideal generated by the $(n - k) \times (n - k)$ minors of $(z_{ij})$. This gives the other inclusion. \medskip\noindent Let $R^{\oplus m} \to M$ be another surjection with kernel $L$. By Schanuel's lemma (Algebra, Lemma \ref{algebra-lemma-Schanuel}) and the results of the previous paragraph, we may assume $m = n$ and that there is an isomorphism $R^{\oplus n} \to R^{\oplus m}$ commuting with the surjections to $M$. Let $C = (c_{li})$ be the (invertible) matrix of this map (it is a square matrix as $n = m$). Then given $z'_1, \ldots, z'_{n - k} \in L$ as above we can find $z_1, \ldots, z_{n - k} \in K$ with $z_1' = Cz_1, \ldots, z'_{n - k} = Cz_{n - k}$. By Lemma \ref{lemma-ideals-generated-by-minors} we get one of the inclusions. By symmetry we get the other.
Part (1) follows from the fact that $I_0(A) = R$ in Lemma \ref{lemma-ideals-generated-by-minors}. \medskip\noindent Part (2) follows form the corresponding statement in Lemma \ref{lemma-ideals-generated-by-minors}. \medskip\noindent Part (3) follows from the fact that $\otimes_R R'$ is right exact, so the base change of a presentation of $M$ is a presentation of $M \otimes_R R'$. \medskip\noindent Proof of (4). Let $R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0$ be a presentation. Then $\text{Fit}_k(M)$ is the ideal generated by the $n - k \times n - k$ minors of the matrix $A$. \medskip\noindent Part (5) is immediate from the definition. \medskip\noindent Proof of (6). Choose a presentation of $M$ with matrix $A$ as in Lemma \ref{lemma-fitting-ideal}. Let $J' \subset J$ be a subset of cardinality $n$. It suffices to show that $f = \det(a_{ij})_{i = 1, \ldots, n, j \in J'}$ annihilates $M$. This is clear because the cokernel of $$ R^{\oplus n} \xrightarrow{A' = (a_{ij})_{i = 1, \ldots, n, j \in J'}} R^{\oplus n} \to M \to 0 $$ is killed by $f$ as there is a matrix $B$ with $A' B = f1_{n \times n}$. \medskip\noindent Proof of (7). Choose a presentation of $M$ with matrix $A$ as in Lemma \ref{lemma-fitting-ideal}. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we have $$ M_\mathfrak p \not = 0 \Leftrightarrow M \otimes_R \kappa(\mathfrak p) \not = 0 \Leftrightarrow \text{rank}(\text{image }A\text{ in }\kappa(\mathfrak p)) < n $$ Clearly $\text{Fit}_0(M)$ exactly cuts out the set of primes with this property.
By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we see that $M_f$ can be generated by $k$ elements over $R_f$ for some $f \in R$, $f \not \in \mathfrak p$ if $M \otimes_R \kappa(\mathfrak p)$ can be generated by $k$ elements. Hence (2), (3), and (4) are equivalent. Using Lemma \ref{lemma-fitting-ideal-basics} part (3) this reduces the problem to the case where $R$ is a field and $\mathfrak p = (0)$. In this case the result follows from Example \ref{example-fitting-free}.
It is immediate that (2) is equivalent to (3) because the Fitting ideals form an increasing sequence of ideals. Since the formation of $\text{Fit}_k(M)$ commutes with base change (Lemma \ref{lemma-fitting-ideal-basics}) we see that (1) implies (2) by Example \ref{example-fitting-free} and glueing results (Algebra, Section \ref{algebra-section-more-glueing}). Conversely, assume (2). By Lemma \ref{lemma-fitting-ideal-generate-locally} we may assume that $M$ is generated by $r$ elements. Thus a presentation $\bigoplus_{j \in J} R \to R^{\oplus r} \to M \to 0$. But now the assumption that $\text{Fit}_{r - 1}(M) = 0$ implies that all entries of the matrix of the map $\bigoplus_{j \in J} R \to R^{\oplus r}$ are zero. Thus $M$ is free.
Choose generators $x_1, \ldots, x_n \in M$ corresponding to the surjection $R^{\oplus n} \to M$. Since $R$ is local if a set of elements $E \subset (f)$ generates $(f)$, then some $e \in E$ generates $(f)$, see Algebra, Lemma \ref{algebra-lemma-NAK}. Hence we may pick $z_1, \ldots, z_{n - k}$ in the kernel of $R^{\oplus n} \to M$ such that some $(n - k) \times (n - k)$ minor of the $n \times (n - k)$ matrix $A = (z_{ij})$ generates $(f)$. After renumbering the $x_i$ we may assume the first minor $\det(z_{ij})_{1 \leq i, j \leq n - k}$ generates $(f)$, i.e., $\det(z_{ij})_{1 \leq i, j \leq n - k} = uf$ for some unit $u \in R$. Every other minor is a multiple of $f$. By Algebra, Lemma \ref{algebra-lemma-matrix-right-inverse} there exists a $n - k \times n - k$ matrix $B$ such that $$ AB = f \left( \begin{matrix} u 1_{n - k \times n - k} \\ C \end{matrix} \right) $$ for some matrix $C$ with coefficients in $R$. This implies that for every $i \leq n - k$ the element $y_i = ux_i + \sum_j c_{ji}x_j$ is annihilated by $f$. Since $M/\sum Ry_i$ is generated by the images of $x_{n - k + 1}, \ldots, x_n$ we win.
Choose a presentation $$ R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0 $$ We claim the image of $A$ is finite locally free of rank $n - k$. If the claim holds, then the lemma is true by definition of projective dimension. To prove the claim we may replace $R$ by the localization at a prime, see Algebra, Lemma \ref{algebra-lemma-finite-projective}. This reduces us the the case discussed in the next paragraph. \medskip\noindent Assume $R$ is local. Set $M' = \{x \in M \mid fx = 0\}$. By Lemma \ref{lemma-principal-fitting-ideal} we can choose $x_1, \ldots, x_k \in M$ which generate $M/M'$. Then $x_1, \ldots, x_k$ generate $M_f = (M/M')_f$. Hence, if there is a relation $\sum a_ix_i = 0$ in $M$, then we see that $a_1, \ldots, a_k$ map to zero in $R_f$ since otherwise $\text{Fit}_{k - 1}(M) R_f = \text{Fit}_{k - 1}(M_f)$ would be nonzero. Since $f$ is a nonzerodivisor, we conclude $a_1 = \ldots = a_k = 0$. Thus $M \cong R^{\oplus k} \oplus M'$. After a change of basis in our presentation above, we may assume the first $n - k$ basis vectors of $R^{\oplus n}$ map into the summand $M'$ of $M$ and the last $k$-basis vectors of $R^{\oplus n}$ map to basis elements of the summand $R^{\oplus k}$ of $M$. Having done so, the last $k$ rows of the matrix $A$ vanish. In this way we see that, replacing $M$ by $M'$, $k$ by $0$, $n$ by $n - k$, and $A$ by the submatrix where we delete the last $k$ rows, we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $R$ is local, $k = 0$, and $M$ annihilated by $f$. Now the $0$th Fitting ideal of $M$ is $(f)$ and is generated by the $n \times n$ minors of the matrix $A$ of size $n \times m$. (This in particular implies $m \geq n$.) Since $R$ is local, some $n \times n$ minor of $A$ is $uf$ for a unit $u \in R$. After renumbering we may assume this minor is the first one. Moreover, we know all other $n \times n$ minors of $A$ are divisible by $f$. Write $A = (A_1 A_2)$ in block form where $A_1$ is an $n \times n$ matrix and $A_2$ is an $n \times (m - n)$ matrix. By Algebra, Lemma \ref{algebra-lemma-matrix-right-inverse} applied to the transpose of $A$ (!) we find there exists an $n \times n$ matrix $B$ such that $$ BA = B(A_1 A_2) = f \left( \begin{matrix} u 1_{n \times n} & C \end{matrix} \right) $$ for some $n \times (m - n)$ matrix $C$ with coefficients in $R$. Then we first conclude $BA_1 = fu 1_{n \times n}$. Thus $$ BA_2 = fC = u^{-1}fuC = u^{-1}BA_1C $$ Since the determinant of $B$ is a nonzerodivisor we conclude that $A_2 = u^{-1}A_1C$. Therefore the image of $A$ is equal to the image of $A_1$ which is isomorphic to $R^{\oplus n}$ because the determinant of $A_1$ is a nonzerodivisor. This finishes the proof.
The set of points of $\Spec(B)$ where $A \to B$ is not \'etale is a closed subset of $\Spec(B)$, see Algebra, Definition \ref{algebra-definition-etale}. Write this as $V(J')$ for some ideal $J' \subset B$. Then $V(J') \cap V(J) = \emptyset$ hence $J + J' = B$ by Algebra, Lemma \ref{algebra-lemma-Zariski-topology}. Write $1 = f + g$ with $f \in J$ and $g \in J'$. Then $g$ works.
We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say $\deg(\overline{g}) = n$ and $\deg(\overline{h}) = m$ so that $\deg(f) = n + m$. Write $f = x^{n + m} + \sum \alpha_i x^{n + m - i}$ for some $\alpha_1, \ldots, \alpha_{n + m} \in A$. Consider the ring map $$ R = \mathbf{Z}[a_1, \ldots, a_{n + m}] \longrightarrow S = \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m] $$ of Algebra, Example \ref{algebra-example-factor-polynomials-etale}. Let $R \to A$ be the ring map which sends $a_i$ to $\alpha_i$. Set $$ B = A \otimes_R S $$ By construction the image $f_B$ of $f$ in $B[x]$ factors, say $f_B = g_B h_B$ with $g_B = x^n + \sum (1 \otimes b_i) x^{n - i}$ and similarly for $h_B$. Write $\overline{g} = x^n + \sum \overline{\beta}_i x^{n - i}$ and $\overline{h} = x^m + \sum \overline{\gamma}_i x^{m - i}$. The $A$-algebra map $$ B \longrightarrow A/I, \quad 1 \otimes b_i \mapsto \overline{\beta}_i, \quad 1 \otimes c_i \mapsto \overline{\gamma}_i $$ maps $g_B$ and $h_B$ to $\overline{g}$ and $\overline{h}$ in $A/I[x]$. The displayed map is surjective; denote $J \subset B$ its kernel. From the discussion in  Algebra, Example \ref{algebra-example-factor-polynomials-etale} it is clear that $A \to B$ is etale at all points of $V(J) \subset \Spec(B)$. Choose $g \in B$ as in Lemma \ref{lemma-localize-upstairs} and consider the $A$-algebra $B_g$. Since $g$ maps to a unit in $B/J = A/I$ we obtain also a map $B_g/I B_g \to A/I$ of $A/I$-algebras. Since $A/I \to B_g/I B_g$ is \'etale, also $B_g/IB_g \to A/I$ is \'etale (Algebra, Lemma \ref{algebra-lemma-map-between-etale}). Hence there exists an idempotent $e \in B_g/I B_g$ such that $A/I = (B_g/I B_g)_e$ (Algebra, Lemma \ref{algebra-lemma-surjective-flat-finitely-presented}). Choose a lift $h \in B_g$ of $e$. Then $A \to A' = (B_g)_h$ with factorization given by the image of the factorization $f_B = g_B h_B$ in $A'$ is a solution to the problem posed by the lemma.
Applying Lemma \ref{lemma-lift-invertible-element} we may assume that the leading coefficient of $\overline{g}$ is the reduction of an invertible element $u \in A$. Then we may replace $\overline{g}$ by $\overline{u}^{-1}\overline{g}$ and $\overline{h}$ by $\overline{u}\overline{h}$. Thus we may assume that $\overline{g}$ is monic. Since $f$ is monic we conclude that $\overline{h}$ is monic too. In this case the result follows from Lemma \ref{lemma-lift-factorization-monic}.
Say $f = x^d + a_1 x^{d - 1} + \ldots + a_d$ has degree $d$. Write $\overline{g} = \sum \overline{b}_j x^j$ and $\overline{h} = \sum \overline{c}_j x^j$. Then we see that $1 = \sum \overline{b}_j \overline{c}_{d - j}$. It follows that $\Spec(A/I)$ is covered by the standard opens $D(\overline{b}_j \overline{c}_{d - j})$. However, each point $\mathfrak p$ of $\Spec(A/I)$ is contained in at most one of these as by looking at the induced factorization of $f$ over the field $\kappa(\mathfrak p)$ we see that $\deg(\overline{g} \bmod \mathfrak p) + \deg(\overline{h} \bmod \mathfrak p) = d$. Hence our open covering is a disjoint open covering. Applying Lemma \ref{lemma-lift-open-covering} (and replacing $A$ by $A'$) we see that we may assume there is a corresponding disjoint open covering of $\Spec(A)$. This disjoint open covering corresponds to a product decomposition of $A$, see Algebra, Lemma \ref{algebra-lemma-disjoint-implies-product}. It follows that $$ A = A_0 \times \ldots \times A_d, \quad I = I_0 \times \ldots \times I_d, $$ where the image of $\overline{g}$, resp.\ $\overline{h}$ in $A_j/I_j$ has degree $j$, resp.\ $d - j$ with invertible leading coefficient. Clearly, it suffices to prove the result for each factor $A_j$ separatedly. Hence the lemma follows from Lemma \ref{lemma-lift-factorization-easy}.
Let $I' \subset R$ be an ideal such that $V(I')$ is the closure of the image of $V(J)$. Then $V(I) \cap V(I') = \emptyset$ by assumption and hence $I + I' = R$ by Algebra, Lemma \ref{algebra-lemma-Zariski-topology}. Write $1 = g + f$ with $g \in I$ and $f \in I'$. We have $V(f') \supset V(J)$ where $f'$ is the image of $f$ in $S$. Hence $(f')^n \in J$ for some $n$, see Algebra, Lemma \ref{algebra-lemma-Zariski-topology}. Replacing $f$ by $f^n$ we win.
Observe that $z = b^2 - b$ is an element of $IB$. By Algebra, Lemma \ref{algebra-lemma-integral-integral-over-ideal} there exist a monic polynomial $g(x) = x^d + \sum a_j x^j$ of degree $d$ with $a_j \in I$ such that $g(z) = 0$ in $B$. Hence $f(x) = g(x^2 - x) \in A[x]$ is a monic polynomial such that $f(x) \equiv x^d(x - 1)^d \bmod I$ and such that $f(b) = 0$ in $B$.
Choose an element $y \in B$ lifting $\overline{e}$. Choose $f \in A[x]$ as in Lemma \ref{lemma-helper-integral} for $y$. By Lemma \ref{lemma-lift-factorization-easy} we can find an \'etale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and such that $f = gh$ in $A[x]$ with $g(x) = x^d \bmod IA'$ and $h(x) = (x - 1)^d \bmod IA'$. After replacing $A$ by $A'$ we may assume that the factorization is defined over $A$. In that case we see that $b_1 = g(y) \in B$ is a lift of $\overline{e}^d = \overline{e}$ and $b_2 = h(y) \in B$ is a lift of $(\overline{e} - 1)^d = (-1)^d (1 - \overline{e})^d = (-1)^d(1 - \overline{e})$ and moreover $b_1b_2 = 0$. Thus $(b_1, b_2)B/IB = B/IB$ and $V(b_1, b_2) \subset \Spec(B)$ is disjoint from $V(IB)$. Since $\Spec(B) \to \Spec(A)$ is closed (see Algebra, Lemmas \ref{algebra-lemma-integral-going-up} and \ref{algebra-lemma-going-up-closed}) we can find an $a \in A$ which maps to an invertible element of $A/I$ whose image in $B$ lies in $(b_1, b_2)$, see Lemma \ref{lemma-separate-image-closed-from-closed}. After replacing $A$ by the localization $A_a$ we get that $(b_1, b_2) = B$. Then $\Spec(B) = D(b_1) \amalg D(b_2)$; disjoint union because $b_1b_2 = 0$ and covers $\Spec(B)$ because $(b_1, b_2) = B$. Let $e \in B$ be the idempotent corresponding to the open and closed subset $D(b_1)$, see Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}. Since $b_1$ is a lift of $\overline{e}$ and $b_2$ is a lift of $\pm (1 - \overline{e})$ we conclude that $e$ is a lift of $\overline{e}$ by the uniqueness statement in Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}.
We can choose an integer $n$ and a direct sum decomposition $(A/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $R/I$-module $\overline{K}$. Choose a lift $\varphi : A^{\oplus n} \to A^{\oplus n}$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$. Let $f \in A[x]$ be the characteristic polynomial of $\varphi$. Set $B = A[x]/(f)$. By Cayley-Hamilton (Algebra, Lemma \ref{algebra-lemma-charpoly}) there is a map $B \to \text{End}_A(A^{\oplus n})$ mapping $x$ to $\varphi$. For every prime $\mathfrak p \supset I$ the image of $f$ in $\kappa(\mathfrak p)$ is $(x - 1)^rx^{n - r}$ where $r$ is the dimension of $\overline{P} \otimes_{A/I} \kappa(\mathfrak p)$. Hence $(x - 1)^nx^n$ maps to zero in $B \otimes_A \kappa(\mathfrak p)$ for all $\mathfrak p \supset I$. Thus $x(1 - x)$ is contained in every prime ideal of $B/IB$. Hence $x^N(1 - x)^N$ is contained in $IB$ for some $N \geq 1$. It follows that $x^N + (1 - x)^N$ is a unit in $B/IB$ and that $$ \overline{e} = \text{image of }\frac{x^N}{x^N + (1 - x)^N}\text{ in }B/IB $$ is an idempotent as both assertions hold in $\mathbf{Z}[x]/(x^N(x - 1)^N)$. The image of $\overline{e}$ in $\text{End}_{A/I}((A/I)^{\oplus n})$ is $$ \frac{\overline{p}^N}{\overline{p}^N + (1 - \overline{p})^N} = \overline{p} $$ as $\overline{p}$ is an idempotent. After replacing $A$ by an \'etale extension $A'$ as in the lemma, we may assume there exists an idempotent $e \in B$ which maps to $\overline{e}$ in $B/IB$, see Lemma \ref{lemma-lift-idempotent-upstairs}. Then the image of $e$ under the map $$ B = A[x]/(f) \longrightarrow \text{End}_A(A^{\oplus n}). $$ is an idempotent element $p$ which lifts $\overline{p}$. Setting $P = \Im(p)$ we win.
Let $J = \Ker(\alpha)$. The lemma asserts that $J/J^2 \cong K \otimes_A C$. Note that $\alpha$ is a homomorphism of graded algebras. We will prove that in degree $d$ we have $(J/J^2)_d = K \otimes_A C_{d - 1}$. Note that $$ J_d = \Ker(\text{Sym}^d_A(A^{\oplus m}) \to \text{Sym}^d_A(M)) = \Im(K \otimes_A \text{Sym}^{d - 1}_A(A^{\oplus m}) \to \text{Sym}^d_A(A^{\oplus m})), $$ see Algebra, Lemma \ref{algebra-lemma-presentation-sym-exterior}. It follows that $(J^2)_d = \sum_{a + b = d} J_a \cdot J_b$ is the image of $$ K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^d_A(A^{\oplus m}). $$ The cokernel of the map $K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^{d - 1}_A(A^{\oplus m})$ is $\text{Sym}^{d - 1}_A(M)$ by the lemma referenced above. Hence it is clear that $(J/J^2)_d = J_d/(J^2)_d$ is equal to \begin{align*} \Coker( K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to K \otimes_A \text{Sym}^{d - 1}_A(A^{\otimes m})) & = K \otimes_A \text{Sym}^{d - 1}_A(M) \\ & = K \otimes_A C_{d -1} \end{align*} as desired.
Let $\sigma : C \to A$ be the projection onto the degree $0$ part of $C$. Then $J = \Ker(\sigma)$ is the part of degree $> 0$ and we see that $J/J^2 = M$ as an $A$-module. Hence if $A \to C$ is smooth then $M$ is a finite projective $A$-module by Algebra, Lemma \ref{algebra-lemma-section-smooth}. \medskip\noindent Conversely, assume that $M$ is finite projective and choose a surjection $A^{\oplus n} \to M$ with kernel $K$. Of course the sequence $0 \to K \to A^{\oplus n} \to M \to 0$ is split as $M$ is projective. In particular we see that $K$ is a finite $A$-module and hence $C$ is of finite presentation over $A$ as $C$ is a quotient of $A[x_1, \ldots, x_n]$ by the ideal generated by $K \subset \bigoplus Ax_i$. The computation of Lemma \ref{lemma-cotangent-complex-symmetric-algebra} shows that $\NL_{C/A}$ is homotopy equivalent to $(K \to M) \otimes_A C$. Hence $\NL_{C/A}$ is quasi-isomorphic to $C \otimes_A M$ placed in degree $0$ which means that $C$ is smooth over $A$ by Algebra, Definition \ref{algebra-definition-smooth}.
The ring map $A \to B$ is finite by Algebra, Lemma \ref{algebra-lemma-characterize-finite-in-terms-of-integral}. Hence $B$ is finitely presented as an $A$-module by Algebra, Lemma \ref{algebra-lemma-finite-finitely-presented-extension}. Hence $B$ is a finite locally free $A$-module by Algebra, Lemma \ref{algebra-lemma-finite-projective}. Since the module $B$ has rank $1$ along $V(I)$ (see rank function described in Algebra, Lemma \ref{algebra-lemma-finite-projective}), and as $(A, I)$ is a Zariski pair, we conclude that the rank is $1$ everywhere. It follows that $A \to B$ is an isomorphism: it is a pleasant exercise to show that a ring map $R \to S$ such that $S$ is a locally free $R$-module of rank $1$ is an isomorphism (hint: look at local rings).
By Lemma \ref{lemma-lift-idempotent-upstairs} we can find an \'etale ring map $A \to A'$ inducing an isomorphism $A/I \to A'/IA'$ such that $B' = B \otimes_A A'$ contains an idempotent $e'$ lifting the image of $b$ in $B'/IB'$. Consider the corresponding $A'$-algebra decomposition $$ B' = B'_1 \times B'_2 $$ which is compatible with the one given in the lemma upon reduction modulo $I$. The map $A' \to B'_1$ is surjective modulo $IA'$. By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we can find $i \in IA'$ such that after replacing $A'$ by $A'_{1 + i}$ the map $A' \to B'_1$ is surjective. Observe that the image $b'_1 \in B'_1$ of $b$  satisfies $b'_1 - 1 \in IB'_1$. Thus we may pick $a' \in IA'$ mapping to $b'_1 - 1$. On the other hand, the image $b'_2 \in B'_2$ of $b$ is in $IB'_2$. By Algebra, Lemma \ref{algebra-lemma-integral-integral-over-ideal} there exist a monic polynomial $g(x) = x^d + \sum a'_j x^j$ of degree $d$ with $a'_j \in IA'$ such that $g(b'_2) = 0$ in $B'_2$. Thus the image $b' = (b'_1, b'_2) \in B'$ of $b$ is a root of the polynomial $(x - 1 - a')g(x)$. We conclude that $$ (b' - 1)(b')^d \in \sum\nolimits_{j = 0, \ldots, d} IA' \cdot (b')^j $$ We claim that this implies $$ (b - 1)b^d \in \sum\nolimits_{j = 0, \ldots, d} I \cdot b^j $$ in $B$. For this it is enough to see that the ring map $A \to A'$ is faithfully flat, because the condition is that the image of $(b - 1)b^d$ is zero in $B/\sum_{j = 0, \ldots, d} Ib^j$ (use Algebra, Lemma \ref{algebra-lemma-faithfully-flat-universally-injective}). The map $A \to A'$ flat because it is \'etale (Algebra, Lemma \ref{algebra-lemma-etale}). On the other hand, the induced map on spectra is open (see Algebra, Proposition \ref{algebra-proposition-fppf-open} and use previous lemma referenced) and the image contains $V(I)$. Since $I$ is contained in the Jacobson radical of $A$ we conclude.