invariance of trace and inversion formula

Lemma 15   Let $k$ be a field of characteristic $p$. Let $\phi$ be a $k$-endomorphism of $A_n(k)$. Let $I,J$ be an element of $\{0,1,2,\dots,p-1\}^n$ (multi-index). Then for any $f \in Z(A_n(k))$, we have

$$\operatorname{trd}( \phi(\xi^I \eta^J f))= \begin{cases} -\... ... } I=J=(p-1,p-1,\dots,p-1) \\ 0 & \text{otherwise} \end{cases}$$

This is the same as Corollary 5. One may also prove this by using Corollary 2.

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For any index set $I\in \mathbb{N}^{2n}$, we denote by $I^c$ the index set $(p-1,p-1,\dots,p-1)-I$.

Corollary 7   Let $k$ be a field of characteristic $p$. Let $\phi$ be a $k$-endomorphism of $A_n(k)$. Assume we are given a set of $p^{2n}$ elements $\{f_{I J} \in Z(A_n(k)); I,J \subset \{0,1,2,\dots,p-1\}^n\}$. Let $M$ be the maximum of total degree of $\phi(\xi_1),\dots,\phi(\xi_n)$, $\phi(\eta_1),\dots,\phi(\eta_n)$. Then for any set $\{f_{I J}\}$ of $p^{2n}$ elements of $Z(A_n(k))$, we have

$$\operatorname{totaldeg}(\phi(\sum_{I,J}f_{I J} \xi^I \eta^J )) \geq \max_{I,J}(\operatorname{totaldeg}\phi(f_{I J}))-Mp^{2n}$$

Put $F=\phi(\sum_{I,J} f_{I J}\xi^I \eta^J )$. Then

$$\operatorname{trd}F \phi(\eta)^{J^c} \phi(\xi)^{I^c}= f_{I J}$$

Thus $\operatorname{totaldeg}(F\phi(\eta)^{J^c}\phi(\xi)^{I^c})\geq \operatorname{totaldeg}(f_{I J})$. Noting that total degree is additive ( $\operatorname{totaldeg}(F G)=\operatorname{totaldeg}(F)+\operatorname{totaldeg}(G)$), we complete the proof.

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Proposition 1 (inversion formula)   Let $k$ be a field of characteristic $p$. Assume we have an injective algebra endomorphism $\phi$ of $A_n(k)$. We use the notation $\overline{\xi_i}=\phi(\xi_i)$, $\overline{\eta_j}=\phi(\eta_j)$. Then for any element $x\in A_n(k)$, we have

$$x=-\sum_{I,J} \operatorname{trd}(x \overline{\eta}^{J^c} \overline{\xi}^{I^c}) \overline{\xi}^I\overline{\eta}^J$$

Corollary 8   Under the assumption of the Lemma above, if elements $\operatorname{trd}(\xi_i \overline{\eta}^{J^c} \overline{\xi}^{I^c})$, $\operatorname{trd}(\eta_i \overline{\eta}^{J^c} \overline{\xi}^{I^c})$ $(i=1,\dots,n)$ are all constants, then $\phi$ is invertible.

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