# Partially commutative elements in powers of augmentation ideal

Let $$\vartheta$$ a relation of parcial commutation over a set $$X,$$ and consider the respective free parcially commutative group $$F(X, \vartheta).$$ Let $$K[F(X, \vartheta)]$$ the parcially commutative free group ring. Let $$\varepsilon \colon K[F(X, \vartheta)] \to K$$ $$\sum\limits_{\overline{x} \in F(X, \vartheta)} \alpha_{\overline{x}} \overline{x} \mapsto \sum\limits_{\overline{x} \in F(X, \vartheta)} \alpha_{\overline{x}}$$ We define in $$K[F(X, \vartheta)]$$ the augmentation ideal $$\mathfrak{X} = \ker(\varepsilon).$$

Denote by $$\frac{\partial f}{\partial \omega}$$ the Fox derivative, with $$\frac{\partial x_i}{\partial x_j} = \delta_{ij}$$ for $$x_i \in K[F(X)],$$ and

$$\frac{\partial^n f}{\partial x_{j_n}\partial x_{j_{n-1}} \cdots \partial x_{j_{1}} } = \frac{\partial}{\partial x_{j_n}} \left( \frac{\partial^{n-1} f}{\partial x_{j_{n-1}} \cdots \partial x_{j_{1}} } \right)$$

Also, $$\varphi \colon K[F(X)] \to K[F(X, \vartheta)]$$ is defined by $$\varphi(x_i) =\overline{x_i},$$ for $$x_i \in F(X).$$

We can define a derivation in $$K[F(X, \vartheta)],$$ letting $$D(\overline{x_i}) = \overline{x_1}-1$$ for $$\overline{x_i} \in F(X, \vartheta)$$ and

$$D(\overline{f}) = \sum\limits_i \varphi\left( \frac{\partial f}{\partial x_i}\right) (\overline{x_i} - 1)$$

How can I prove that $$\overline{f} \in R[F(X, \vartheta)]$$ is in $$\mathfrak{X}^n$$ iff $$\varepsilon\left( \varphi\left(\sum\limits_{\omega \in \overline{\omega}} \frac{\partial f}{\partial \omega}\right) \right) = 0 \quad \forall \overline{\omega} \in F(X, \vartheta), |\overline{\omega}|\le n-1?$$