1 CONTRIBUTION TO THE JACOBIAN CONJECTURE : POLYNOMIAL MAPPING HAVING TWO ZEROS AT INFINITY

This article contains the theorems concerning the algebraic dependence of polynomial mappings with the constant Jacobian having two zeros at infinity. The work is related to the issues of the classical Jacobian Conjecture. This hypothesis affirm that the polynomial mapping of two complex variables with constant non-zero Jacobian is invertible. The Jacobian Conjecture is equivalent to the fact that polynomial mappings with constant non-zero Jacobian do not have two zeros at infinity, therefore it is equivalent to the two theorems given in the work. The proofs of these theorems proceeds by induction.


Introduction
The hypothesis formulated by O.H.Keller in 1939 [1] that a polynomial mapping of constant Jacobian is invertible is not up to now resolved.Even in the case of two variables.This classic case is the subject of this work.In 1974, T.T.Moh in work [2] formulated a known condition for the invertibility of the mapping that additionally having one branch at infinity.A certain contribution in this topic was given by the co-author of this article at work [3].Finally, in 1977 S.S. Abhyankar at work [4] proved that the polynomial mapping with constant Jacobian has at most two zeros at infinity.This work is particularly important for the authors of this article, because the theorems we prove excludes this possibility.It is worth mentioning about the work [5], of which Z.Charzyński was one of the co-authors.In the aforementioned work [4], Abhyankar showed that the mapping of the constant Jacobian having one zero at infinity is inveertible.The Jacobian Conjecture was also the subject of articles by D. Wright [6] and A. van den Essen [7].The article [8] by H. Bass, E. Connell and D. Wright is also highlighted.

Algebraic dependence of polynomial mappings
Let fi, hj be the complex forms of variables X, Y of degrees i, j respectively and ,1 ij  .
and   2 1 2 2 2 3 1 ... where Jac , f h const f h  then exist the form 1 ĥ for which and for some constants Proof.For q = 1 the theorem is true.Indeed, when and so 11 ĥh  .
For p = 1 the theorem is true, because then f = h.Let 2 p  .We assume that the formula (3) is true for exponents 1, ..., p -1, and we prove that it is true for p. Really, let and We have and back to the previous formula, we have Back to the coordinates f and h, consecutively we obtain so subtracting the above equations we get Using the induction assumption for exponent p -1 we have 12 12 ...
If A1= 0 we continue to calculate to the form f of the rank of 2p -4 and we get Next we proceed the same way as with the constant A1.This ends the first part of the proof.Let 2 p  .We assume that the formula (3) and (4) are true for exponents q = 1, ..., p -1.We will prove that for q = p the formulas are also true.Let's save again the formulas (1) and (2) We have consecutively 1) Jac , Jac , and 2) Jac , Jac , Jac , where Jac , Jac , 0 We insert (28), (33) to (24) and we get We subtract Now we apply the induction assumption for exponent p -1, converting f to h and h for f .Therefore .We have at this point and for some constants This allows us to determine the form 1 ĥ by the formula and If A1= 0 we continue to calculate to the form f of the rank of 2p -4 and we get Then we proceed in the same way as the constant A1 and for the next constants.This completes the second and the last part of the proof.
where kl  (k and l are relativity prim) and for which for some constants 1 In the proof we will use the following observation.
For p = 1 the theorem is true, because then f = h.In fact, for p = 1, we have ...
2) Jac , Jac , Jac , where  according to the formula (59).Consequently Therefore f = h.Now let 2 p  .We assume that the formulas (50) is true for exponents 1, ..., p -1, and we prove that it is true for p. Really, let We have 2) Jac , Jac , Jac , where and back to the previous formula, we have Continuing calculations up to form f(k+l ) (p-1) we obtain where e. g.
according to the above observation.
On the other hand we get Thus, the formulas (67), ( 71), (72) allow to write Using the induction assumption for exponent p -1 we have 12 12 ...
for some constants 12 ,..., If A1= 0 we continue to calculate to the form f of the rank of (k + l)(p -2) and we get With the constant A2 we proceed the same way as with the constant A1.What ends the first part of the proof.10 Let 2 p  .We assume that the formula (50) and (51) are true for exponents 11 qp    .We will prove that for q = p the formulas are also true.Then we have and and We proceed similarly to step k + l −1 and we also receive In the step k + l we have Now we subtract We receive the assumption for exponent q = p -1.When exist the forms and The first term in (96) after development give the formulas for successive homogeneous components of the polynomial h in (94).So

X Y h
  and we have the formula for the form Furthermore (101) and Next homogeneous component of the polynomial h in (94) is equal now and     We get the same After k+l−2 steps we receive The homogeneous component of the polynomial h of degree (k+l and the formula for the form 1 ĥ is ready   The formulas (113) and (115) give the result where 1 0 A  .If A1= 0 we continue to calculate to the form f of the rank of (k + l)(p -2) and we get With the constant A2 we proceed the same way as with the constant A1.What ends the final part of the proof.
Corollary 1.Under the assumptions of both theorems the polynomials f and h are algebraically depending.
Corollary 2.Not exist invertible polynomial mappings which have two zeros at infinity.
in step k + l − 1, we receive 11 hf 