Is it possible to generalize a result of Wang?

Question

Assume $A$ and $B$ are commutative algebras with $1$. There is a nice result of Wang, Corollary 8, which says the following: "Let $B = A[z] = A[Z]/(h(Z))$. Then $B$ is a separable algebra over $A$ if and only if $h'(z)$, the formal derivative of $h$ evaluated at $z$, is a unit of $B$".

By definition, an $A$-algebra $B$ is separable over $A$ if $B$ is a projective $B \otimes_A B$-module via $\mu: B\otimes_A B \to B$, $b_1 \otimes_A b_2 \mapsto b_1b_2$. In other words, $B$ is separable over $A$ if $pd_{B\otimes_A B}(B)=0$.

Hence Corollary 8 says: Let $B = A[z] = A[Z]/(h(Z))$. Then $pd_{B\otimes_A B}(B)=0$ if and only if the ideal of $B$ generated by $h'(z)$ is $B$.

My question: Can one generalize Corollary 8 to the following: Let $B = A[z] = A[Z]/(h(Z))$. Then $pd_{B \otimes_A B}(B) = n$ if and only if the ideal of $B$ generated by $h'(z),h''(z), \ldots, h^{(n+1)}$ is $B$.

(Remark: Maybe one should look at $n=1$ first).

Any comment will be appreciated.

Answer 1

I don't see any reason to expect the higher derivatives of $h$ to be important. For a simple counterexample to your conjecture, consider $h(Z)=Z^2$ (for, say, $A$ a field of characteristic $\neq2$). Then $h''$ is a unit, but $B$ has infinite projective dimension over $B\otimes_A B$. Another counterexample is $h(Z)=0$; then no number of iterated derivatives of $h$ can generate a nonzero ideal but the projective dimension of $B$ is $1$.