The Tame Automorphism And Jacobian Conjucture

Abstract

Consider the notations for this paper as:  indicates a commutative ring, though in most result  are going to be domain, and  [X] the Polynomial ring in elements over.

Here assuming that subsequent subcategories of ::(,n) is equal  the affine subgroup of    including of every -  so that   = 1 . ,  the “de Jonquière’s” subgroup of    including the -  Of the arrangement.

 = () , . . . . ,) Where one of the   nd   [] for every d      The Elementary Automorphism generated, that is the form of the automorphisms is   = () for some   and  

  the tame subset of The subgroup generated by is (, n) and (,.

We get each part of (, n) is a multiplication of a part of,  and Elementary Automorphism finite in numbers. Therefore (, n)  (, n). Also, coupling the “de Jonquiere’s” automorphisms with appropriate one effectively verifies permutation maps that all Elementary Automorphism fit in to the subgroup of    created by  (, n) and J (, n). Hence, having (,n) =.

In this paper here we assume the condition  and consider the a domain. In this paper proving that the (, 2) is the free merged result of  (, 2) and (, 2) via their intersection. Moreover, we define an algorithm that determines if there is an endomorphism of polynomial of   is tame.

By means of this process, the paper demonstrate that if  it is not a field, so it    But, consider   may be a field then it seems that we have impartiality, that is each in dimension two, automorphism taken over a field is tame. This is the more popular “Jung-van der Kulk theorem” ().