0
$\begingroup$

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 $$\varepsilon\left( \varphi\left(\sum\limits_{\omega \in \overline{\omega}} \frac{\partial f}{\partial \omega}\right) \right) = 0 \quad \forall \overline{\omega} \in F(X, \vartheta), |\overline{\omega}|\le n-1?$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.