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Let $I = \langle f_1, \ldots f_r \rangle$ be an ideal in $R=k[x_1,\ldots,x_n]$ where $k$ is a field, and put $A = R/I$.

(If $k$ is algebraically closed and $I$ is radical then $A$ is the coordinate ring of an affine variety.)

Let $\mathfrak{X}^p(A) = \operatorname{Der}_k(\wedge^p A, A) $ be skew-symmetric $p$-derivations (derivation in each argument) of $A$.

Question: Is there an algorithm to calculate $\mathfrak{X}^p(A)$ in terms of $f_1,\ldots,f_r$? I am interested in $p=1,2,3$.

What I have tried/observed so far:

  1. The condition that $P \in \mathfrak{X}^p(R)$ descends to (is well-defined on) $\mathfrak{X}^p(A)$ is a system of (ideal membership) equations $$P(x_{i_1}, \ldots, x_{i_{p-1}}, f_j) \in I \quad \text{for each } 1\leq i_1<\cdots<i_{p-1}\leq n \text{ and } 1\leq j\leq r.$$ (Note also every element of $\mathfrak{X}^p(A)$ has a lift to $\mathfrak{X}^p(R)$.)

  2. Letting $P_{i_1,\ldots,i_p} = P(x_{i_1},\ldots,x_{i_p}) \in R$ be the (sought for) coefficients of $P$, the system becomes $$\sum_k P_{i_1,\ldots,i_{p-1},k} \frac{\partial f_j}{\partial x_k} \in I \quad \text{for each } 1\leq i_1<\cdots<i_{p-1}\leq n \text{ and } 1\leq j\leq r.$$

  3. Here one can restrict to the sum over $k \in \{1,\ldots,n\} \setminus \{i_1,\ldots i_{p-1}\}$. Note the equations are not independent, due to skew-symmetry.

  4. The equations above for fixed $j$ suggest to consider the intersection $I \cap \langle \frac{\partial f_j}{\partial x_1}, \ldots \frac{\partial f_j}{\partial x_n} \rangle$ which can be computed using Groebner bases. I am not sure what we can conclude about the coefficients $P_{i_1,\ldots,i_{p-1},k}$. Furthermore, I do not see how one would combine these results for different $(i_1,\ldots,i_{p-1})$'s and $j$'s.

  5. $\mathfrak{X}^p(A)$ is an $A$-module. I think it is finitely generated. (I am interested in the case where it is.)

  6. $\mathfrak{X}^p(A) \cong \operatorname{Hom}_A(\Omega^p(A), A)$ where $\Omega^p(A)$ are Kähler $p$-forms.

  7. $\mathfrak{X}^2(A) \neq \wedge^2 \mathfrak{X}^1(A)$ in general: if $I = \langle yx, yz, y^2 \rangle$ in $R=\mathbb{C}[x,y,z]$ then $0 \neq y \frac{\partial}{\partial y} \wedge \frac{\partial}{\partial z}$ is in $\mathfrak{X}^2(A)$ and not in $\wedge^2 \mathfrak{X}^1(A)$.


I would also be interested in (classes of) examples where one can calculate $\mathfrak{X}^p(A)$ explicitly for $p=1,2,3$.

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Put $N_p =\binom{n}{p}$ and consider the sought-for $R$-submodule $M_p \subset R^{N_p}$ consisting of $(P_{i_1,\ldots,i_p})$ with $1 \leq i_1<\ldots< i_p \leq n$ such that $$\sum_{1 \leq i_1<\ldots< i_p \leq n} P_{i_1,\ldots,i_p} \frac{\partial}{\partial x_{i_1}} \wedge \cdots \wedge \frac{\partial}{\partial x_{i_p}} \in \mathfrak{X}^p(A).$$

Define $P_{i_1,\ldots,i_p}$ for unsorted $(i_1,\ldots,i_p)$ in the usual way, by using the appropriate sign.

The defining equations are (as in the question): for every $1 \leq j_1 < \ldots < j_{p-1} \leq n$ and $1 \leq j \leq r$, $$\sum_{k=1}^n P_{j_1,\ldots,j_{p-1},k}\frac{\partial f_j}{\partial x_k} \in I. \tag{1}$$

Here the coefficients are unsorted in general, so we define a $R$-linear map $\Phi^p_{j_1,\ldots,j_{p-1}}:R^{N_p}\to R^n$ by $$\Phi^p_{j_1,\ldots,j_{p-1}}((P_{i_1,\ldots,i_p}))_k = \begin{cases} (-)^\sigma P_{i_1,\ldots,i_{p}} & \text{ if } \exists \sigma \in S_{p} \text{ such that } \sigma(i_1,\ldots,i_p) = (j_1,\ldots,j_{p-1},k)\\ 0 & \text { otherwise. }\end{cases}$$ which gives the coefficients of the equation $(1)$ in terms of the sorted ones.

(For example, $\Phi_{1}^2((P_{12},P_{13},P_{23})) = (0, P_{12}, P_{13})$ and $\Phi_{2}^2((P_{12},P_{13},P_{23})) = (-P_{12},0,P_{23})$.)

Put $T_j = \{ c \in R^n | \sum c_k \frac{\partial f_j}{\partial x_k} \in I\}$ and let $\pi_j : R^n \to R^n/T_j$ be the quotient map. Note $T_j$ can be computed.

Then we have constructed the $R$-module as $$M_p = \bigcap_{\substack{1\leq j_1 < \ldots < j_{p-1} \leq n\\1 \leq j \leq r}} \ker(\pi_j \circ \Phi^p_{j_1,\ldots,j_{p-1}})$$ which can be computed because kernels of $R$-linear maps and intersections of $R$-modules can be computed.

(For example in Singular using modulo and intersect respectively.)

Of course the coefficients can be taken modulo $I$, so we can consider $M_p/IR^{N_p}$. In practice it means we can throw away some of the generators of $M_p$ which we obtain, in the same way as in the answer that was just linked.

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