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2 struck out the claim about k[x]/x^2

Let $k$ be a commutative ring, $A$ a commutative $k$-algebra, and for some other part of why I'm asking this question I only care about the case when $k \supseteq \mathbb Q$. Recall the following notion, I think originally due to Grothendieck:

Definition (differential operator): Let $D : A\to A$ be $k$-linear. Define $s_nD : A^{\otimes n} \to A$ by: $$s_nD(a_1,\dots,a_n) = \sum_{I \subseteq \lbrace 1,\dots,n\rbrace} (-1)^{|I|}\; \left( \prod_{i\not\in I} a_i\right) \;D\left(\prod_{i\in I}a_i\right)$$ One says that $D$ is an $n$th order differential operator if $s_{n+1}D = 0$.

Examples: $s_0D = D(1) \in A$. $s_1D(a) = D(a) - aD(1)$. $s_2D(a,b) = D(ab) - aD(b) - bD(a) + abD(1)$.

Remark: $s_nD$ is symmetric in the $a_i$s. If $D$ is an $n$th order differential operator, then $s_nD(-,a_2,\dots,a_n)$ is a derivation, for $a_2,\dots,a_n$ fixed. Thus if $D$ is an $n$th order differential operator, $s_nD$ is a symmetric polyderivation. It deserves the be called the principal symbol of $D$. It measures the failure of $D$ to be an $(n-1)$th order differential operator.

Question: For which algebras $A$ (i.e. what are natural, checkable conditions) does the following "PBW theorem" hold: $$s_n: \lbrace n\text{th order differential operators}\rbrace \to \lbrace \text{symmetric }n\text{-polyderivations} \rbrace$$ surjective for every $n$?

Examples: This PBW theorem holds for $A = k[x_1,\dots,x_n]$ and $A = k [\![ x_1,\dots,x_n ]\!]$ and $A = \mathscr C^\infty(M)$ where $M$ is a smooth manifold. This PBW theorem failes for $A = k[x]/x^2$, as the space of symmetric biderivations is one-dimensional spanned by $x \frac{\partial}{\partial x}\otimes \frac{\partial}{\partial x}$, whereas every second-order differential operator is also a first-order differential operator. Edit: I don't know a characteristic-$0$ example for which PBW theorem fails, but I don't expect it to always hold.

Remark: I would expect that algebras $A$ for which the PBW theorem holds are those for which $\operatorname{spec}(A)$ is "smooth" in the appropriate sense, but I don't know if this is "smooth" in other usual senses of the word.

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# For which algebras does \{Differential Operators\} satisfy a PBW-like theorem?

Let $k$ be a commutative ring, $A$ a commutative $k$-algebra, and for some other part of why I'm asking this question I only care about the case when $k \supseteq \mathbb Q$. Recall the following notion, I think originally due to Grothendieck:

Definition (differential operator): Let $D : A\to A$ be $k$-linear. Define $s_nD : A^{\otimes n} \to A$ by: $$s_nD(a_1,\dots,a_n) = \sum_{I \subseteq \lbrace 1,\dots,n\rbrace} (-1)^{|I|}\; \left( \prod_{i\not\in I} a_i\right) \;D\left(\prod_{i\in I}a_i\right)$$ One says that $D$ is an $n$th order differential operator if $s_{n+1}D = 0$.

Examples: $s_0D = D(1) \in A$. $s_1D(a) = D(a) - aD(1)$. $s_2D(a,b) = D(ab) - aD(b) - bD(a) + abD(1)$.

Remark: $s_nD$ is symmetric in the $a_i$s. If $D$ is an $n$th order differential operator, then $s_nD(-,a_2,\dots,a_n)$ is a derivation, for $a_2,\dots,a_n$ fixed. Thus if $D$ is an $n$th order differential operator, $s_nD$ is a symmetric polyderivation. It deserves the be called the principal symbol of $D$. It measures the failure of $D$ to be an $(n-1)$th order differential operator.

Question: For which algebras $A$ (i.e. what are natural, checkable conditions) does the following "PBW theorem" hold: $$s_n: \lbrace n\text{th order differential operators}\rbrace \to \lbrace \text{symmetric }n\text{-polyderivations} \rbrace$$ surjective for every $n$?

Examples: This PBW theorem holds for $A = k[x_1,\dots,x_n]$ and $A = k [\![ x_1,\dots,x_n ]\!]$ and $A = \mathscr C^\infty(M)$ where $M$ is a smooth manifold. This PBW theorem failes for $A = k[x]/x^2$, as the space of symmetric biderivations is one-dimensional spanned by $x \frac{\partial}{\partial x}\otimes \frac{\partial}{\partial x}$, whereas every second-order differential operator is also a first-order differential operator.

Remark: I would expect that algebras $A$ for which the PBW theorem holds are those for which $\operatorname{spec}(A)$ is "smooth" in the appropriate sense, but I don't know if this is "smooth" in other usual senses of the word.