**Proof.**
As $B$ is essentially of finite type over $A$, so is $B \otimes _ A B$ and in particular $B \otimes _ A B$ is Noetherian. Hence the quotient $B$ of $B \otimes _ A B$ is pseudo-coherent over $B \otimes _ A B$ (More on Algebra, Lemma 15.64.17) which explains why perfectness of the ring map (More on Algebra, Definition 15.82.1) agrees with the condition of finite tor dimension.

We may write $B = R/K$ where $R$ is the localization of $A[x_1, \ldots , x_ n]$ at a prime ideal and $K \subset R$ is an ideal. Denote $\mathfrak m \subset R \otimes _ A R$ the maximal ideal which is the inverse image of the maximal ideal of $B$ via the surjection $R \otimes _ A R \to B \otimes _ A B \to B$. Then we have surjections

\[ (R \otimes _ A R)_\mathfrak m \to (B \otimes _ A B)_\mathfrak m \to B \]

and hence ideals $I \subset J \subset (R \otimes _ A R)_\mathfrak m$ as in Lemma 23.7.4. We conclude that $I/\mathfrak m I \to J/\mathfrak m J$ is injective.

Let $K = (f_1, \ldots , f_ r)$ with $r$ minimal. We may and do assume that $f_ i \in R$ is the image of an element of $A[x_1, \ldots , x_ n]$ which we also denote $f_ i$. Observe that $I$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and $1 \otimes f_1, \ldots , 1 \otimes f_ r$. We claim that this is a minimal set of generators of $I$. Namely, if $\kappa $ is the common residue field of $R$, $B$, $(R \otimes _ A R)_\mathfrak m$, and $(B \otimes _ A B)_\mathfrak m$ then we have a map $R \otimes _ A R \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ which factors through $(R \otimes _ A R)_\mathfrak m$. Since $B$ is flat over $A$ and since we have the short exact sequence $0 \to K \to R \to B \to 0$ we see that $K \otimes _ A \kappa \subset R \otimes _ A \kappa $, see Algebra, Lemma 10.39.12. Thus restricting the map $(R \otimes _ A R)_\mathfrak m \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ to $I$ we obtain a map

\[ I \to K \otimes _ A \kappa \oplus \kappa \otimes _ A K \to K \otimes _ B \kappa \oplus \kappa \otimes _ B K. \]

The elements $f_1 \otimes 1, \ldots , f_ r \otimes 1, 1 \otimes f_1, \ldots , 1 \otimes f_ r$ map to a basis of the target of this map, since by Nakayama's lemma (Algebra, Lemma 10.20.1) $f_1, \ldots , f_ r$ map to a basis of $K \otimes _ B \kappa $. This proves our claim.

The ideal $J$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and the elements $x_1 \otimes 1 - 1 \otimes x_1, \ldots , x_ n \otimes 1 - 1 \otimes x_ n$ (for the proof it suffices to see that these elements are contained in the ideal $J$). Now we can write

\[ f_ i \otimes 1 - 1 \otimes f_ i = \sum g_{ij} (x_ j \otimes 1 - 1 \otimes x_ j) \]

for some $g_{ij}$ in $(R \otimes _ A R)_\mathfrak m$. This is a general fact about elements of $A[x_1, \ldots , x_ n]$ whose proof we omit. Denote $a_{ij} \in \kappa $ the image of $g_{ij}$. Another computation shows that $a_{ij}$ is the image of $\partial f_ i / \partial x_ j$ in $\kappa $. The injectivity of $I/\mathfrak m I \to J/\mathfrak m J$ and the remarks made above forces the matrix $(a_{ij})$ to have maximal rank $r$. Set

\[ C = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ r) \]

and consider the naive cotangent complex $\mathop{N\! L}\nolimits _{C/A} \cong (C^{\oplus r} \to C^{\oplus n})$ where the map is given by the matrix of partial derivatives. Thus $\mathop{N\! L}\nolimits _{C/A} \otimes _ A B$ is isomorphic to a free $B$-module of rank $n - r$ placed in degree $0$. Hence $C_ g$ is smooth over $A$ for some $g \in C$ mapping to a unit in $B$, see Algebra, Lemma 10.137.12. This finishes the proof.
$\square$

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