An Algebraic Approach to Zariski Problem

Zariski Problem (Cancellation of indeterminates) is settled affirmatively, that is, it is proved that : Let k be an algebraically closed field of characteristic zero and let n, m ∈ N. If R[Y1, . . . , Ym] ∼=k k[X1, . . . , Xn+m] as k-algebras, where Y1, . . . , Ym, X1, . . . , Xn+m are indetermoinates, then R ∼=k k[X1, . . . , Xn]. Zariski Problem is the following : Zariski Problem. Let k be an algebraically closed field of characteristic zero and let n,m ∈ N. If R[Y1, . . . , Ym] ∼=k k[X1, . . . , Xn+m] as k-algebras, where Y1, . . . , Ym, X1, . . . , Xn+m are indetermoinates, is R ∼=k k[X1, . . . , Xn] ? We may assume that k = C. in the above problem. The answer is yes for n = 1 ([8]) or n = 2 ([4]). But the cases n ≥ 3 are open problems. Moreover, the known case n = 1, 2 are proved by some tools in algebraic geometry. The purely algebraic proof has not been known yet even in such cases. Our objective is to give an affirmative albegraic proof for all n ∈ N. Throughout this paper, all fields, rings and algebras are assumed to be commutative with unity. For a ring R, R denotes the set of units of R and K(R) the total quotient ring. Spec(R) denotes the affine scheme defined by R or merely 2000 Mathematics Subject Classification : Primary 13C25, Secondary 15A18