We shall use the notation R[n] for a polynomial ring in n variables over a commutative ring R. Thus, E = R[n] will mean that E = R[t1, . . . , tn] for some elements t1, . . . , tnin E which are algebraically independent over R.
Let k be an algebraically closed field. The Zariski Cancellation Problem for Affine Spaces asks whether the affine space An
k is cancellative, i.e., if V is an affine k-variety
N. Gupta / Advances in Mathematics 264 (2014) 296–307 297
such that V × A1k∼= An+1
k , does it follow that V ∼= Ank? Equivalently, if A is an affine
k-algebra such that A[1] = k[n+1], does it follow that A = k[n]?The affine line A1
k was shown to be cancellative by S.S. Abhyankar, P. Eakin and W.J. Heinzer [1] and the affine plane A2
k was shown to be cancellative by T. Fujita, M. Miyanishi and T. Sugie [5,10] in characteristic zero and by P. Russell [11] in positive characteristic. However, in [8], the author showed that when ch. k = p > 0, the affine space A3
k is not cancellative. She showed that the threefold R = k[X, Y, Z, T ]/(XmY +Zpe + T + T sp), where m, e, s are positive integers such that pe � sp and sp � pe, con-structed by Asanuma in [2], is not isomorphic to k[3] for m ≥ 2. It was already shown by Asanuma in [2] that the ring R is a stably polynomial ring, i.e., R[1] = k[4]. Note that the polynomial f(Z, T ) := Zpe +T +T sp, with pe � sp and sp � pe, satisfies k[Z, T ]/(f) = k[1]
but k[Z, T ] �= k[f ][1] [7, p. 62]. Such a polynomial f is called a non-trivial line. In a subsequent paper [9], the author proved that, for any non-trivial line f ∈ k[Z, T ], the threefold k[X, Y, Z, T ]/(XmY − f(Z, T )), m ≥ 2, is not isomorphic to the polynomial ring k[X1, X2, X3] but stably isomorphic to a polynomial ring.
In this paper, we shall show (Theorem 3.7) that for any non-trivial line f(Z, T ) ∈k[Z, T ] the ring
A = k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY − f(Z, T )
),
where ri > 1 for each i, 1 ≤ i ≤ m, is not a polynomial ring but is a stably polynomial ring; i.e., A �= k[m+2] but A[1] = k[m+3]. Since non-trivial lines exist over any field of positive characteristic, we obtain that, when ch. k > 0, the affine space An
k is not cancellative for any n ≥ 3 (Corollary 3.8). Thus, this result completely settles Zariski’s Cancellation Problem for any affine n-space over any field of positive characteristic.
2. Preliminaries
For any ring R, R∗ will denote the group of units of R.We shall use the following term from affine algebraic geometry.
Definition. Let k be a field. An element f ∈ k[Z, T ] is called a line if k[Z, T ]/(f) = k[1]. A line f is called a non-trivial line if k[Z, T ] �= k[f ][1].
We recall below the definition of an exponential map and the Derksen invariant.
Definition. Let k be a field, A a k-algebra, φ : A → A[1] a k-algebra homomorphism and Aφ = {a ∈ A | φ(a) = a}, the ring of invariants of φ. For an indeterminate U over A, let φU denote the map φ : A → A[U ]. φ is said to be an exponential map on A if φ satisfies the following two properties:
(i) ε0φU is identity on A, where ε0 : A[U ] → A is the evaluation at U = 0.
298 N. Gupta / Advances in Mathematics 264 (2014) 296–307
(ii) φV φU = φV +U , where φV : A → A[V ] is extended to a homomorphism φV : A[U ] →A[V, U ] by setting φV (U) = U .
An exponential map φ is said to be non-trivial if Aφ �= A. The Derksen invariant of A is the subring of A generated by the ring of invariants of all non-trivial exponential maps on A, i.e.,
DK(A) = k[f∣∣ f ∈ Aφ, φ a non-trivial exponential map
].
We state below a few properties of an exponential map (cf. [3, pp. 1291–1292] and [8, Lemma 2.1]).
Lemma 2.1. Let A be an affine domain over a field k. Suppose that there exists a non-trivial exponential map φ on A. Then the following statements hold:
(i) Aφ is factorially closed in A, i.e., for any non-zero a, b ∈ A, the condition ab ∈ Aφ
implies that both a ∈ Aφ and b ∈ Aφ. In particular, Aφ is algebraically closed in A.(ii) tr.degk(Aφ) = tr.degk(A) − 1.
Below, we recall the concept of an admissible proper Z-filtration on an affine domain.
Definition. Let A be an affine domain over a field k. A collection of k-linear subspaces {An}n∈Z of A is said to be a proper Z-filtration if it satisfies the following conditions:
(i) An ⊆ An+1 for all n ∈ Z,(ii) A =
⋃n∈Z
An,(iii)
⋂n∈Z
An = (0), and(iv) (An \An−1) · (Am \Am−1) ⊆ An+m \An+m−1 for all n, m ∈ Z.
We shall call a proper Z-filtration {An}n∈Z of A admissible if there exists a finite generating set Γ of A such that, for any n ∈ Z and a ∈ An, a can be written as a finite sum of monomials in elements of Γ and each of these monomials is an element of An.
Any proper Z-filtration on A determines the following Z-graded integral domain
gr(A) :=⊕i
Ai/Ai−1,
and a map
ρ : A → gr(A) defined by ρ(a) = a + An−1, if a ∈ An \An−1.
Moreover, gr(A) is generated by ρ(Γ ) where Γ is a finite generating set of A that makes the filtration admissible.
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An exponential map φ on a graded ring A is said to be homogeneous if φ : A → A[U ]becomes homogeneous when A[U ] is given a grading induced from A such that U is a homogeneous element.
We quote below a result on homogenization of exponential maps due to H. Derksen, O. Hadas and L. Makar-Limanov [4]; the following version is presented in [3, Theo-rem 2.6].
Theorem 2.2. Let A be an affine domain over a field k with an admissible proper Z-filtration and gr(A) the induced Z-graded domain. Let φ be a non-trivial exponen-tial map on A. Then φ induces a non-trivial homogeneous exponential map φ on gr(A)such that ρ(Aφ) ⊆ gr(A)φ.
We recall below three other results involving exponential maps and the Derksen in-variant from [8, Lemma 2.4], [8, Lemma 3.3] and [9, Theorem 3.7] respectively.
Lemma 2.3. Let k be a field and A = k[n], where n > 1. Then DK(A) = A.
Lemma 2.4. Let B be an affine domain over an infinite field k. Let f ∈ B be such that f−λ is a prime element of B for infinitely many λ ∈ k. Let φ be a non-trivial exponential map on B such that f ∈ Bφ. Then there exist infinitely many β ∈ k such that each f −β
is a prime element of B and φ induces a non-trivial exponential map φ on B/(f − β). Moreover, Bφ/(f − β)Bφ is contained in (B/(f − β))φ.
Theorem 2.5. Let k be a field and A be an integral domain defined by
A = k[X,Y, Z, T ]/(XrY − F (X,Z, T )
), where r > 1.
Set f(Z, T ) := F (0, Z, T ). Let x, y, z and t denote, respectively, the images of X, Y , Zand T in A. Suppose that DK(A) �= k[x, z, t]. Then the following statements hold.
(i) There exist Z1, T1 ∈ k[Z, T ] and a0, a1 ∈ k[1] such that k[Z, T ] = k[Z1, T1] and f(Z, T ) = a0(Z1) + a1(Z1)T1.
(ii) If k[Z, T ]/(f) = k[1], then k[Z, T ] = k[f ][1].
3. Main theorem
Throughout the paper, k will denote a field (of any characteristic unless otherwise specified) and A an integral domain defined by
A = k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY − F (X1, . . . , Xm, Z, T )
),
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where ri > 1 for each i, 1 ≤ i ≤ m. The images of X1, . . . , Xm, Y , Z and T in A will be denoted by x1, . . . , xm, y, z and t respectively. Let B denote the k-subalgebra of Adefined by
B := k[x1, . . . , xm, z, t] ↪→ A ↪→ B[(x1 · · ·xm)−1].
Note that B = k[m+2]. For each m-tuple (q1, . . . , qm) ∈ Zm, consider the Z-grading on B[(x1 · · ·xm)−1] given by
B[(x1 · · ·xm)−1] =
⊕i∈Z
Bi,
where Bi =⊕
(i1,...,im)∈Zm, q1i1+q2i2+···+qmim=i
k[z, t]x1i1x2
i2 . . . xmim .
Each non-zero element H ∈ B[(x1 · · ·xm)−1] can be uniquely written as H =∑eH≤j≤dH
Hj , where Hj ∈ Bj , HeH �= 0 and HdH�= 0 for some integers eH and dH .
(Note that if H ∈ B then Hj ∈ B for each j.) We call dH the degree of H. Thus, HdHis
the leading homogeneous summand of H.For each integer n ∈ Z, set An := A ∩
⊕i≤n Bi. Then xj ∈ Aqj \ Aqj−1, 1 ≤ j ≤ m
and z, t ∈ A0 \ A−1. Since A is an integral domain, F (X1, . . . , Xm, Z, T ) �= 0. Thus, F (x1, . . . , xm, z, t) is a non-zero element of B. For simplicity, we shall use the notation F
for both the polynomial in k[m+2] and its image in B and denote the degree dF of F
by d. Thus, Fd �= 0 and y ∈ Ac \Ac−1, where c = d − (q1r1 + · · · + qmrm).
Lemma 3.1. The k-linear subspaces {An}n∈Z define an admissible proper Z-filtration Fon A with the generating set Γ := {x1, . . . , xm, y, z, t}, and the induced graded ring grA :=
⊕n∈Z
An/An−1 is generated by the image of Γ in grA.
Proof. Clearly, the linear subspaces {An}n∈Z define a proper Z-filtration on A. Note that each element h ∈ A can be uniquely written as a sum of monomials of the form
x1i1 · · ·xm
imzj1tj2y�, (1)
where i1, i2, . . . , im, j1, j2, are all non-negative integers and satisfy the condition:
is < rs for at least one s, 1 ≤ s ≤ m, whenever > 0.
Therefore, it can be seen that, if h ∈ An \An−1 then h can be uniquely written as a sum of monomials of the form as in Eq. (1) and that each of these monomials lies in An. Thus, the filtration defined on A is admissible with the generating set Γ := {x1, . . . , xm, y, z, t}. Therefore, grA is generated by the image of Γ in grA (cf. [8, Remark 2.2 (2)]). �
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We shall call F defined in Lemma 3.1 the (admissible proper) Z-filtration with respect to (q1, . . . , qm) ∈ Zm. We now exhibit a structure of grA when Fd is not divisible in Bby xj for any j.
Lemma 3.2. Let d = degF . Suppose that Fd /∈ xjB for any j, 1 ≤ j ≤ m. Then grA is isomorphic to
D = k[X1, . . . , Xm, Y, Z, T ](X1
r1 · · ·XmrmY − Fd(X1, . . . , Xm, Z, T ))
as k-algebras.
Proof. For a ∈ A, let gr(a) denote the image of a in grA. Then by Lemma 3.1, grAis generated by gr(x1), . . . , gr(xm), gr(y), gr(z) and gr(t). Since F = x1
Since Fd is not divisible by xj for any j, 1 ≤ j ≤ m, D is an integral domain. As grAcan be identified with a subring of gr(B[(x1 · · ·xm)−1]) ∼= B[(x1 · · ·xm)−1], we see that the elements gr(x1), . . . , gr(xm), gr(z), gr(t) of grA are algebraically independent over k. Hence grA ∼=k D. �Lemma 3.3. B ⊆ DK(A).
Proof. Define k-algebra homomorphisms φ1 and φ2 from A to A[U ] by φ1(xj) = xj for each j, 1 ≤ j ≤ m, φ1(z) = z, φ1(t) = t + x1
r1 · · ·xmrmU ,
φ1(y) = F (x1, . . . , xm, z, t + x1r1 · · ·xm
rmU)x1r1 · · ·xm
rm= y + Uα(x1, . . . , xm, z, t, U)
and φ2(xj) = xj for each j, 1 ≤ j ≤ m, φ2(t) = t, φ2(z) = z + x1r1 · · ·xm
rmU ,
φ2(y) = F (x1, . . . , xm, z + x1r1 · · ·xm
rmU, t)x1r1 · · ·xm
rm= y + Uβ(x1, . . . , xm, z, t, U),
where α(x1, . . . , xm, z, t, U), β(x1, . . . , xm, z, t, U) ∈ k[x1, . . . , xm, z, t, U ]. It is easy to see that φ1 and φ2 are nontrivial exponential maps on A such that k[x1, . . . , xm, z] ⊆ Aφ1
and k[x1, . . . , xm, t] ⊆ Aφ2 . Hence k[x1, . . . , xm, z, t] ⊆ DK(A). �We now prove a generalization of Theorem 2.5.
Proposition 3.4. Let k be an infinite field. Suppose that f(Z, T ) := F (0, . . . , 0, Z, T ) �= 0and that DK(A) = A. Then the following statements hold.
302 N. Gupta / Advances in Mathematics 264 (2014) 296–307
(i) There exist Z1, T1 ∈ k[Z, T ] and a0, a1 ∈ k[1] such that k[Z, T ] = k[Z1, T1] and f(Z, T ) = a0(Z1) + a1(Z1)T1.
(ii) Suppose that k[Z, T ]/(f) = k[1]. Then k[Z, T ] = k[f ][1].
Proof. (i) We prove the result by induction on m. The result is true for m = 1 by Theorem 2.5. Suppose that m > 1. Since DK(A) = A, there exists an exponential map φ
on A such that Aφ � B(= k[x1, x2, . . . , xm, z, t]). Let g ∈ Aφ\B. Since g /∈ B, by Eq. (1), there exists a monomial in g which is of the form x1
i1 . . . xmimzj1tj2y�, where > 0 and
is < rs for some s, 1 ≤ s ≤ m. Without loss of generality, we may assume that s = 1.Consider the proper Z-filtration F1 on A with respect to (−1, 0, . . . , 0) ∈ Zm and let
A denote the induced graded ring. Let d = degF . For h ∈ A, let h denote the image of h in A. Since A is an integral domain, we have F (0, x2, . . . , xm, z, t) �= 0 and so, Fd = F (0, x2, . . . , xm, z, t). Since f(Z, T ) �= 0, Fd is not divisible by any xj for any j, 1 ≤ j ≤ m and hence by Lemma 3.2, we have
A ∼=k k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY − Fd
).
By Theorem 2.2, φ induces a non-trivial homogeneous exponential map φ on A such that g ∈ Aφ. By the choice of g and the filtration F1 defined on A, we have y | g. Since Aφ is factorially closed in A (cf. Lemma 2.1 (i)), it follows that y ∈ Aφ. Since x1 is a homogeneous element of A and Aφ is a homogeneous subring of A, we see that either x1 ∈ Aφ or Aφ ⊆ k[x2, . . . , xm, z, t, y]. We consider these two cases separately.
Case 1: Suppose that x1 ∈ Aφ. Since x1 − λ is a prime element in A for every λ ∈ k∗, by Lemma 2.4 there exists β ∈ k∗ for which φ induces a non-trivial exponential map ψon the ring C := A/(x1 − β)A such that the image of y lies in Cψ. Now
C = A/(x1 − β)A ∼=kk[X2, X3, . . . , Xm, Y, Z, T ]
(X2r2X3
r3 · · ·Xmrmβr1Y − Fd)
∼=kk[X2, X3, . . . , Xm, Y, Z, T ]
(X2r2X3
r3 · · ·XmrmY − Fd)
.
Therefore, using Lemma 3.3, we have DK(C) = C.Thus by induction on m, f(Z, T ) satisfies the required condition.
Case 2: Now suppose that Aφ ⊆ k[x2, . . . , xm, z, t, y]. Consider the proper Z-filtration F2
on A with respect to (−1, −1, . . . , −1) ∈ Zm and let A denote the induced graded ring. For h ∈ A, let h denote the image of h in A. Since f(Z, T ) �= 0, the leading homogeneous summand of Fd is f(z, t) and hence, by Lemma 3.2,
A ∼=k k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY − f(Z, T )
).
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Again by Theorem 2.2, φ induces a non-trivial homogeneous exponential map φ on Asuch that y ∈ Aφ. Since tr.degk Aφ = m + 1, there exist m algebraically independent elements in Aφ over k[y ].
If Aφ ⊆ k[y, z, t ], then since m ≥ 2 and Aφ is algebraically closed in A, we have Aφ = k[y, z, t ]. Since x1
r1 · · · xmrm y(= f(z, t )) ∈ k[y, z, t ], we have x1, . . . , xm, y ∈ Aφ
as Aφ is factorially closed in A. This would contradict the fact that φ is non-trivial. Thus, there exists a homogeneous element h ∈ Aφ \ k[y, z, t ]. Hence, h contains a monomial which is divisible by xi for some i, 2 ≤ i ≤ m. Without loss of generality we may assume that x2 divides a monomial of h.
Now again consider the proper Z-filtration F3 on A with respect to the m-tuple (0, 1, 0, . . . , 0) ∈ Zm and let A denote the induced graded ring. Then, it is easy to see that A ∼= A. For a ∈ A, let a denote the image of a in A. Then x2 | h in A. Again by Theorem 2.2, φ induces a non-trivial homogeneous exponential map φ on A such that yand h ∈ Aφ. Since x2 | h in A, we have x2 ∈ Aφ (cf. Lemma 2.1 (i)).
Now, arguing as in Case 1, we get β ∈ k∗ for which the ring
C := A/(x2 − β)A ∼=kk[X1, X3, . . . , Xm, Y, Z, T ]
(X1r1X3
r3 · · ·Xmrmβr2Y − f(Z, T ))
∼=kk[X1, X3, . . . , Xm, Y, Z, T ]
(X1r1X3
r3 · · ·XmrmY − f(Z, T ))
has the property that DK(C) = C. We are thus through by induction on m.(ii) Suppose that k[Z, T ]/(f) = k[1]. By (i) above, there exist Z1, T1 ∈ k[Z, T ] and
a0, a1 ∈ k[1] such that k[Z, T ] = k[Z1, T1] and f(Z, T ) = a0(Z1) +a1(Z1)T1. If a1(Z1) = 0, then we have k[Z1,T1]
(f) = k[Z1](a0(Z1)) [T1] = k[1] and hence k[Z1]
(a0(Z1)) = k, showing that f(Z, T ) = a0(Z1) must be a linear polynomial in Z1 and hence a variable in k[Z, T ]. Now consider the case a1(Z1) �= 0. As f(Z, T ) is a line in k[Z, T ], a0(Z1) and a1(Z1) are coprime in k[Z1]. Hence k[Z, T ]/(f) = k[Z1, T1]/(a0(Z1) +a1(Z1)T1) ∼=k k[Z1, −a0(Z1)
a1(Z1) ] =k[Z1, 1
a1(Z1) ] and, since (k[Z, T ]/(f))∗ = k∗, we have a1(Z1) ∈ k∗. This again implies that f(Z, T ) is a variable in k[Z, T ]. �Lemma 3.5. Let R be an integral domain, π1, π2, . . . , πn ∈ R and π = π1π2 · · ·πn. Let G(Z, T ) ∈ R[Z, T ] be such that R[Z, T ]/(π, G(Z, T )) ∼=R (R/(π))[1]. Then
R[Z, T ]/(π1
r1π2r2 · · ·πm
rm , G(Z, T )) ∼=R
(R/
(π1
r1π2r2 · · ·πm
rm))[1]
for every m-tuple of positive integers (r1, r2, . . . , rm).
Proof. Let h ∈ R[Z, T ] be such that R[Z, T ] = R[h] +πR[Z, T ] +G(Z, T )R[Z, T ]. There-fore, for any i, 1 ≤ i ≤ m,
R[Z, T ] = R[h] + πR[Z, T ] + G(Z, T )R[Z, T ]
⊆ R[h] + πiR[Z, T ] + G(Z, T )R[Z, T ] ⊆ R[Z, T ].
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Thus, R[Z, T ] = R[h] + πiR[Z, T ] + G(Z, T )R[Z, T ]. Now, for any i, we have
R[Z, T ] = R[h] + πR[Z, T ] + G(Z, T )R[Z, T ]
= R[h] + π(R[h] + πiR[Z, T ] + G(Z, T )R[Z, T ]
)+ G(Z, T )R[Z, T ]
= R[h] + ππiR[Z, T ] + G(Z, T )R[Z, T ].
Since, π1r1π2
r2 · · ·πmrm = π(π1
r1−1π2r2−1 · · ·πm
rm−1), repeating the above process, we see that
R[Z, T ] = R[h] + π1r1π2
r2 · · ·πmrmR[Z, T ] + G(Z, T )R[Z, T ].
Hence the result. �Proposition 3.6. Let R be an integral domain, π1, π2, . . . , πn ∈ R and π = π1π2 · · ·πn. Let G(Z, T ) ∈ R[Z, T ] be such that R[Z, T ]/(π, G(Z, T )) ∼=R (R/π)[1]. Let r1, . . . , rm be a set of positive integers and set
D := R[Z, T, Y ]/(π1
r1π2r2 · · ·πm
rmY −G(Z, T )).
Then D[1] = R[3].
Proof. We may assume that πi �= 0 for each i, 1 ≤ i ≤ n. Let V be an indetermi-nate over R and ψ : R[V ] → R[V ]/(π1
r1π2r2 · · ·πm
rm) be the canonical surjection. By Lemma 3.5, there exists a surjective R-algebra homomorphism φ : R[Z, T ] →R[V ]/(π1
r1π2r2 · · ·πm
rm) such that Kerφ = (π1r1π2
r2 · · ·πmrm , G(Z, T )). Let H(Z, T ) ∈
R[Z, T ], P (V ) ∈ R[V ] and Q(V ) ∈ R[V ] be such that
φ(H(Z, T )
)= ψ(V ), φ(Z) = ψ
(P (V )
)and φ(T ) = ψ
(Q(V )
).
Let W be an indeterminate over D and set
W1 := π1r1 · · ·πm
rmW + H(Z, T ),
Z1 :=(Z − P (W1)
)/π1
r1 · · ·πmrm ,
T1 :=(T −Q(W1)
)/π1
r1 · · ·πmrm .
Let y be the image of Y in D so that D = R[Z, T, y]. We show that D[W ] = R[Z1, T1, W1]. Set C := R[Z1, T1, W1]. We have
Z = P (W1) + π1r1 · · ·πm
rmZ1, (2)
T = Q(W1) + π1r1 · · ·πm
rmT1, (3)
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y = G(Z, T )π1r1 · · ·πm
rm= G(π1
r1 · · ·πmrmZ1 + P (W1), π1
r1 · · ·πmrmT1 + Q(W1))
π1r1 · · ·πmrm
= G(P (W1), Q(W1))π1r1 · · ·πm
rm+ α(Z1, T1,W1), (4)
W = W1 −H(Z, T )π1r1 · · ·πm
rm= W1 −H(π1
r1 · · ·πmrmZ1 + P (W1), π1
r1 · · ·πmrmT1 + Q(W1))
π1r1 · · ·πmrm
= W1 −H(P (W1), Q(W1))π1r1 · · ·πm
rm+ β(Z1, T1,W1) (5)
for some α, β ∈ C. Since
ψ(G(P (V ), Q(V )
))= G
(ψ(P (V )
), ψ
(Q(V )
))= G
(φ(Z), φ(T )
)= φ
(G(Z, T )
)= 0
and
ψ(V −H
(P (V ), Q(V )
))= ψ(V ) −H
(ψ(P (V )
), ψ
(Q(V )
))= φ
(H(Z, T )
)−H
(φ(Z), φ(T )
)= 0,
it follows that
G(P (W1), Q(W1)
)∈ π1
r1π2r2 · · ·πm
rmR[W1]
and
W1 −H(P (W1), Q(W1)
)∈ π1
r1π2r2 · · ·πm
rmR[W1].
Hence, by the relations (2), (3), (4) and (5), we see that D[W ] ⊆ C. We now show that C ⊆ D[W ]. Clearly, W1 ∈ D[W ], so it suffices to show that Z1, T1 ∈ D[W ]. Now
Z1 = Z − P (W1)π1r1π2r2 · · ·πm
rm= Z − P (π1
r1π2r2 · · ·πm
rmW + H(Z, T ))π1r1π2r2 · · ·πm
rm
= Z − P (H(Z, T ))π1r1π2r2 · · ·πm
rm+ γ(Z, T,W )
and
T1 = T −Q(W1)π1r1π2r2 · · ·πm
rm= T −Q(π1
r1π2r2 · · ·πm
rmW + H(Z, T ))π1r1π2r2 · · ·πm
rm
= T −Q(H(Z, T ))π1r1π2r2 · · ·πm
rm+ δ(Z, T,W )
for some γ, δ ∈ D[W ]. Since
φ(Z − P
(H(Z, T )
))= ψ
(P (V )
)− ψ
(P (V )
)= 0
306 N. Gupta / Advances in Mathematics 264 (2014) 296–307
and
φ(T −Q
(H(Z, T )
))= ψ
(Q(V )
)− ψ
(Q(V )
)= 0
we see that
Z − P(H(Z, T )
)∈(π1
r1π2r2 · · ·πm
rm , G(Z, T ))R[Z, T ]
and
T −Q(H(Z, T )
)∈(π1
r1π2r2 · · ·πm
rm , G(Z, T ))R[Z, T ].
Hence there exist α1, α2, β1, β2 ∈ R[Z, T ] such that
Z − P(H(Z, T )
)= α1π1
r1π2r2 · · ·πm
rm + β1G(Z, T )
and
T −Q(H(Z, T )
)= α2π1
r1π2r2 · · ·πm
rm + β2G(Z, T ).
Therefore, since y = G(Z, T )/π1r1π2
r2 · · ·πmrm ∈ D, we see that
Z1 = Z − P (H(Z, T ))π1r1π2r2 · · ·πm
rm+ γ(Z, T,W ) = α1 + β1y + γ(Z, T,W )
and
T1 = T −Q(H(Z, T ))π1r1π2r2 · · ·πm
rm+ δ(Z, T,W ) = α2 + β2y + δ(Z, T,W ).
Since R[Z, T ] ⊆ D, we have Z1, T1 ∈ D[W ]. Hence C ⊆ D[W ]. Since C = R[3], the result follows. �
We now prove our main theorem.
Theorem 3.7. Let k be a field of any characteristic and A an integral domain defined by
A = k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY − F (X1, X2, . . . , Xm, Z, T )
),
where ri ≥ 2 for each i and let f(Z, T ) := F (0, 0, . . . , 0, Z, T ). Suppose that
k[X1, . . . , Xm, Z, T ]/(X1 · · ·Xm, F (X1, X2, . . . , Xm, Z, T )
)∼=
(k[X1, . . . , Xm]/(X1 · · ·Xm)
)[1]
N. Gupta / Advances in Mathematics 264 (2014) 296–307 307
as k[X1, . . . , Xm]-algebras. Then
A[1] = k[X1, . . . , Xm][3] = k[m+3].
Moreover, if ch. k > 0 and f(Z, T ) is a non-trivial line in k[Z, T ] then
A �= k[m+2].
Proof. The isomorphism A[1] = k[X1, . . . , Xm][3] = k[m+3] follows from Proposition 3.6. Now let f(Z, T ) be a non-trivial line in k[Z, T ]. Suppose, if possible, that A = k[m+2]. Let k be an algebraic closure of k. We note that by a result of Ganong [6, Proposition 1.16], f(Z, T ) is a non-trivial line in k[Z, T ]. Now A ⊗k k = k[m+2] and so by Lemma 2.3, DK(A ⊗k k) = A ⊗k k. But this contradicts Proposition 3.4 (ii). �Corollary 3.8. When k is a field of positive characteristic, Zariski’s Cancellation Con-jecture does not hold for the affine n-space An
k for any n ≥ 3.
Proof. When ch. k = p > 0, there exist non-trivial lines in k[Z, T ]; for instance, we may take the line f(Z, T ) = Zp2 + T + T qp, where q is a prime other than p [7, p. 62]. Now consider A = k[X1, . . . , Xm, Y, Z, T ]/(X1
r1 · · ·XmrmY −f(Z, T )) where ri ≥ 2 for each i.
Then, by Theorem 3.7, A[1] = k[m+3] but A �= k[m+2]. �Acknowledgments
The author thanks Professor Amartya K. Dutta for carefully going through the earlier draft and suggesting improvements. The author also acknowledges Department of Science and Technology for their INSPIRE Research Grant.
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