AI 中文总结
研究任意域\(K\)上满足\(F(0)=0\)的平面多项式自同构\(F\)的分解，结合荣格 - 范德库尔克定理与帕斯卡有限类共轭不变性，得出分解形式及相关结论，回答了文献中二维相关问题，构成正特征下类似猜想。
AI 中文摘要
结合荣格 - 范德库尔克定理与帕斯卡有限类的共轭不变性，我们证明了在任意域\(K\)上满足\(F(0)=0\)的平面多项式自同构\(F\)可分解为\(F = \diag(\det J_F, 1) \circ P_1 \circ \dots \circ P_s\)的形式，其中所有\(P_i\)都是帕斯卡有限自同构。由于每个帕斯卡有限自同构的雅可比行列式等于\(1\)，对角因子是唯一障碍：\(F\)是帕斯卡有限映射的复合当且仅当\(\det J_F = 1\)。特别地，文献\(\cite{ABCH2}\)中的问题\(3.1\)在二维任意特征下有肯定答案，这构成了正特征下指数生成器猜想的一个类似物。在特征\(p\)下，这些因子的阶可被\(p^2\)整除。
英文摘要
Combining the Jung--van der Kulk theorem with the conjugacy invariance of the Pascal finite class, we show that every polynomial automorphism $F$ of the plane over an arbitrary field $K$, satisfying $F(0) = 0$, decomposes into the form $F = \diag(\det J_F, 1) \circ P_1 \circ \dots \circ P_s$, where all $P_i$ are Pascal finite automorphisms. Since every Pascal finite automorphism has Jacobian determinant equal to 1, the diagonal factor is the only obstacle: $F$ is a composition of Pascal finite maps if and only if $\det J_F = 1$. In particular, Question~3.1 from \cite{ABCH2} has a positive answer in dimension 2 in any characteristic, which constitutes an analogue of the Exponential Generators Conjecture in positive characteristic. In characteristic $p$, the factors can be chosen to have an order dividing $p^2$.
