An explicit failure of complex affine-space cancellation

Proof. Since H is a residual coordinate of P=\kk[p]\poly4, the theorem of Dutta–Lahiri [6, Theorem 3.4], applied with R=\kk[p] and four relative variables, gives G_2,G_3,G_4,G_5\in P[w] such that

Hence H is a coordinate of P[w] as a \kk-algebra. If H were a coordinate of P as a \kk-algebra, quotienting a polynomial coordinate system containing H by (H) would give A=P/(H)\simeq\kk\poly4, contrary to Theorem 1.1.□

Proof. Over \kk[p,p^{-1}], the elements x,y,z,u are polynomial coordinates on P[p^{-1}]. The inverse substitutions are

Substitution in either direction gives the identity, so in particular the new coordinates are algebraically independent. In these coordinates, Equation (2.2) reads

The coefficient of u is a unit. Replacing u by H is therefore another polynomial coordinate change over \kk[p,p^{-1}]. Consequently (H) is prime in P[p^{-1}], and its quotient is the Laurent polynomial ring asserted in the lemma, with u=p^{-3}(xy-z(z+1)).

Set x_0=s^2+u^3. Reduction modulo p gives

so p does not divide H in P. To descend irreducibility from the localization, factor H in the unique factorization domain P. Its image in P[p^{-1}] is irreducible, so all but one of its irreducible factors must become units there. An irreducible element of P that becomes a unit upon inverting p divides a power of p, and hence is associated to p. Nondivisibility by p excludes such factors. Thus H is irreducible in P, and is prime because P is a unique factorization domain. This proves that A is a domain. Its localization computed above is nonzero, which also proves that p\ne0 in A. This localization has fraction field \kk(p,x,y,z), of transcendence degree four over \kk. The same is true of A, so the dimension formula for finitely generated domains over a field [22, Tag 00P0] gives \dim A=4.□

Proof. We first change the equation by a polynomial automorphism, and then exhibit five polynomial coordinates on the resulting quotient.

Changing the equation. On P[p^{-1}]=\kk[p,p^{-1},x,y,z,u], define \Delta=-p^2\partial/\partial u, with p,x,y,z fixed. This derivation is locally nilpotent: every polynomial is killed by a sufficiently high iterate of \Delta. Applying it to Equation (2.3) gives

These values belong to P, so \Delta restricts to a locally nilpotent derivation of P. Extend it by \Delta w=0. The formula

is a finite sum for every f. The iterated Leibniz rule makes \Phi an algebra homomorphism, and the exponential identity gives its inverse \exp(-w\Delta). By Equation (2.4), \Delta H=p^3 and \Delta^2H=0. Hence

In particular, T is a domain by Lemma 2.1, and its element p is nonzero since \Phi fixes p.

Coordinates adapted to the fiber p=0. With x_0=s^2+u^3, put

The determinant of this linear change in (F,J), over \kk[p,s,u], is

Its inverse is

It follows that p,s,u,L,M are algebraically independent and that

For the rest of the proof, regard F,J,H as polynomials in L over B, using Equation (2.8). Expansion only in powers of p gives

At this point the transfer theorem of Edo–Vénéreau [7, Theorem 7] also applies, with coefficient ring B and parameter p^3: every coefficient of H(L)-L is divisible by p, hence nilpotent modulo p^3B. The following calculation gives a polynomial inverse in this instance.

A polynomial root modulo p^3. Define elements and a polynomial over B by

Since Q(L_*)-C=L_*Q_1(L_*), Equation (2.9) gives the exact identity

A polynomial representative for the fifth coordinate. The quotient (L-L_*)/p^3 suggests the remaining coordinate after localization. The next identity provides a polynomial representative for it before p is inverted. In B[L,w], put

Then H(L)+p^3w=L-ph. Using Q(L)-C=LQ_1(L), we obtain the polynomial identity

Thus the image e of e_0 in T satisfies

In particular, e belongs to the unlocalized ring T.

Generation and algebraic independence. Inverting p in the presentation T=B[L,w]/(H(L)+p^3w) allows us to eliminate w, giving the isomorphism

which is the identity on B and L. The composite B\to T\to T[p^{-1}] is therefore the natural injection B\hookrightarrow B[p^{-1}][L]. This proves that the displayed copy of B embeds in T. Moreover, Equation (2.14) identifies e in this localization with p^{-3}(L-L_*), an affine linear coordinate in the independent variable L. Consequently e is transcendental over B[p^{-1}], and hence over B.

It remains to prove that B and e generate T. By Equation (2.14), L=L_*+p^3e. The polynomial difference H(L_*+p^3Z)-H(L_*) is divisible by p^3 in B[Z]. Together with Equation (2.11), this shows that

The defining relation of T now gives p^3(w-W(e))=0. Since T is a domain and p\ne0, it follows that w=W(e). Thus L,w\in B[e], which proves

For completeness, the five independent coordinates in A[w] itself are the classes of

These are polynomials by the finite exponential formula. They are obtained by transporting the five generators in Equation (2.16) through the inverse of Equation (2.6), which proves the proposition.□

Proof. After inverting p, the defining polynomial of S is linear in u with invertible leading coefficient. It is therefore irreducible in \kk[p,p^{-1},x,y,z,u]. It is not divisible by p in \kk[p,x,y,z,u], so it is irreducible there as well: any factor becoming a unit after localization would be a scalar times a power of p. The latter polynomial ring is a unique factorization domain, so S is a domain and p is nonzero in S. The inverse formulas

show that S\to A becomes an isomorphism after inverting p. Its kernel is consequently p-torsion, hence zero in the domain S. This also proves equality of the fraction fields.

The polynomial xy-z(z+1) is primitive as a polynomial in y over \kk[x,z,u], since x and z(z+1) are coprime. It is linear and irreducible over \kk(x,z,u), so R is a domain. Its hypersurface is smooth by the Jacobian criterion [22, Tags 00T6 and 00T7]: simultaneous vanishing of the partial derivatives y, x, and -(2z+1) would give x=y=0, z=-1/2, which does not satisfy its equation.

Thus pS is a nonzero prime ideal, and the noetherian local domain T=S_{pS} has maximal ideal pT. To see directly that T is a discrete valuation ring, divide a nonzero nonunit repeatedly by p. This process must end: otherwise the principal ideals generated by the successive quotients form an ascending chain, and stabilization would imply that p is a unit, by cancellation in the domain T. Every nonzero element of T is therefore a power of p times a unit. This gives the stated valuation on \Frac(T).

If b\in S has positive order, then b\in pT, so tb\in pS for some t\in S\setminus pS. Primality of pS gives b\in pS. Dividing by p and repeating proves Equation (3.2); its converse is immediate. Finally, write h=p^{-N}b_0 with b_0\in S, N\geq 0, and apply Equation (3.2) to b_0 to divide out its full order. The asserted expression follows, and its uniqueness follows from the valuation and cancellation.□

Proof. The residues of s and u in R are the nonzero elements s_R and u. Also f and g are nonzero: specializing x=z=0 sends f to -y^2-u^3, and specializing x=0, z=-1 sends g to -1. The identities p^2F=x-s^2-u^3 and p^2J=z-sx therefore give

Here s_R is nonzero because its specialization at x=z=0 is y.

Let \theta:P\to R[\tau,\tau^{-1}] substitute the five initial forms in Equation (3.8). Its image contains the specified copy of R: the generators of that copy are recovered by

Consequently \theta(P) is exactly the subalgebra on the right of Equation (3.5). In particular this is an identification over the given coefficient ring R.

For the presentation weights, x has weight zero and H_{\mathrm{top}} is homogeneous of weight two, with

After inverting p, the coordinate change from (s,u,F,J) to (x,y,z,u) and its inverse in Lemma 3.1 identify \theta with the quotient map

Its kernel is (xy-z(z+1)), or equivalently (H_{\mathrm{top}}) in the original localized coordinates. This is a prime ideal. Moreover p does not divide H_{\mathrm{top}}, since its reduction modulo p is

which is nonzero. The same unique factorization argument as in Lemma 3.1 shows that H_{\mathrm{top}} is irreducible, hence prime, in P. Its principal ideal is thus saturated with respect to p: if p^NQ\in(H_{\mathrm{top}}) then Q\in(H_{\mathrm{top}}). Contracting the localized kernel yields

It remains to show that these initial forms generate all of \gr A. For a nonzero polynomial Q\in P, let \omega(Q) be its maximum monomial weight. If Q represents a nonzero h\in A, then \deg h\leq\omega(Q) by Equation (3.8) and the valuation inequalities. Suppose the inequality is strict, and write M=\omega(Q). The weight-M part Q_M has \theta(Q_M)=0, since the degree-M class of h is zero. By Equation (3.11),

where C may be taken homogeneous of weight M-2. To justify the homogeneity, decompose any quotient C into its finitely many homogeneous parts; multiplication by the nonzero homogeneous H_{\mathrm{top}} cannot cancel contributions of different weights in the domain P. The polynomial

represents the same element h and has maximum weight strictly less than M, because puC has weight M-3. Repeating this operation terminates: the maximum weights are strictly decreasing integers bounded below by the fixed finite number \deg h. Thus it ends with a representative of weight exactly \deg h. No lower bound on the weights of all monomials in P is needed.

The initial form of h is now the image under \theta of the top part of such an exact representative. This proves \gr A=\theta(P) and, together with Equation (3.11), proves both Equation (3.5) and Equation (3.7).

Finally, G is a domain by its embedding in R[\tau,\tau^{-1}]. On the open set where p, or equivalently \tau, is invertible, the presentation gives the smooth algebra R[\tau,\tau^{-1}]. On p=0, the partial derivatives of H_{\mathrm{top}} with respect to F and J reduce to x_0^2 and -(1+2sx_0). They cannot vanish simultaneously, as

The Jacobian criterion proves smoothness also at every point of p=0, and hence on all of \Spec G.□

Proof. Let (h_1,\ldots,h_5)=(p,s,u,F,J) be the generators of A and set

The maximum is over a finite nonempty set, since D is nonzero. We claim that

By Proposition 3.2, choose a representative of h of maximum weight exactly \deg h. Applying the product rule to each monomial, every nonzero summand in its derivative has degree at most the monomial’s weight plus j: differentiating a factor h_i changes its degree by at most j. Summing these bounds gives Equation (3.13).

It follows that D(\mathcal F_dA)\subseteq\mathcal F_{d+j}A, so

defines a homogeneous derivation of degree shift j on G. It is well defined because D(\mathcal F_{d-1}A) lies in \mathcal F_{d+j-1}A, and the product rule descends to the graded algebra. For a generator attaining the maximum in Equation (3.12), the right side of Equation (3.14) applied to its initial form is nonzero. Thus D_0\ne0.

For every h\in\mathcal F_dA and m\geq 0, induction gives the identity in the indicated graded component

It includes the case when an intermediate class vanishes: the shift bound ensures that a drop below the indicated degree persists in all later iterates. Local nilpotence of D therefore kills each homogeneous class after finitely many iterates of D_0. Every element of G is a finite sum of homogeneous classes, so a maximum of their nilpotence bounds kills that element. This proves local nilpotence on all of G, without a sign restriction on j. Finally, if Dh=0, Equation (3.14) gives D_0\inif(h)=0.□

Proof. Under the hypothesis choose polynomial coordinates q_1,q_2,q_3,q_4 of A. The subset \mathcal F_0A is a \kk-subalgebra. Since F has degree two by Equation (3.8), this subalgebra is proper. Therefore at least one coordinate q_r has positive degree; otherwise the coordinates would generate A inside \mathcal F_0A. Choose i\ne r. The coordinate partial derivative D=\partial/\partial q_i is a nonzero LND of A and satisfies Dq_r=0. By Lemma 3.3, its induced derivation D_0 on G is nonzero and homogeneous, and it kills the nonzero homogeneous element \inif(q_r) of degree \deg q_r>0.□

Proof. First, \Rt is a domain: the polynomial ac-(bd+1) is primitive as a polynomial in c over the unique factorization domain \kk[a,d,b,u], and is irreducible over its fraction field because it is linear. The element v is nonzero, since u is an indeterminate over the determinant ring and its coefficient in degree three in v is -a^2\ne0.

The principal opens D(z+1) and D(z) cover \Spec R. Over the first open, ac=z+1 is invertible, and the formulas

give mutually inverse identifications

Indeed, they give ac=z+1, ab=x, dc=y, and db=xy/(z+1)=z. Over the second open, the analogous formulas are

Here db=z, dc=y, ab=x, and ac=xy/z=z+1. On the overlap the fiber parameters satisfy

Thus their transition function is a base unit. Gluing affine lines by this same transition function gives a line bundle whose complement of the zero section is \Spec\Rt. The indicated weights are exactly scalar multiplication on the two fiber parameters. Each Laurent polynomial chart is faithfully flat over its base chart: as a module it is free with basis the integral powers of its fiber parameter. Since the base charts cover, R\to\Rt is faithfully flat.

We prove principalization by global identities. In \Rt put

Since s_R=dc-abh and ac-bd=1, we have

Expansion also gives

Finally, with q=bd the coefficient identity is

Consequently

which proves I\Rt=(v).

Flatness permits us to tensor the injection G\hookrightarrow R[\tau,\tau^{-1}] with \Rt. The resulting injection has image generated by \Rt, \tau, f/\tau^2, and g/\tau^2. By Equation (4.7), this image is precisely \Rt[\tau,v/\tau^2]. In particular it is a domain. The map G\to G\otimes_R\Rt is faithfully flat by base change, and hence injective.

For completeness, the single displayed relation is the entire presentation. Let W be an indeterminate and set Q=\Rt[\tau,W]/(\tau^2W-v). If \tau F=(\tau^2W-v)B in \Rt[\tau,W], reduction modulo \tau gives vB(0,W)=0. Since \Rt[W] is a domain and v\ne0, we have B=\tau B_1; cancellation then gives F=(\tau^2W-v)B_1. Thus \tau is a nonzerodivisor in Q. After inverting \tau, the relation eliminates W and identifies Q[\tau^{-1}] with \Rt[\tau,\tau^{-1}]. It follows that Q injects into that Laurent ring with image \Rt[\tau,v/\tau^2], proving Equation (4.4). The two gradings follow from \deg(v)=0 and \wt(v)=2; the relation \tau^2V=v is homogeneous for each.□

Proof. Put C=\Gamma(X,\mathcal O_X) and K=\Frac C. Both X and \Aff^1\times X are regular. Their local rings are factorial, so every Weil divisor is Cartier and their Picard groups identify with their divisor class groups [22, Tags 056S, 0AG0, and 0BE9]. Let D be a Weil divisor on \Spec C[t]. Its restriction to \Spec K[t] is principal, since K[t] is a principal ideal domain. Choose r\in K(t)^* giving that divisor. Subtracting \operatorname{div}(r) from D removes every prime component whose defining prime ideal has zero contraction to C.

Each remaining component has a height-one prime ideal \mathfrak P\subset C[t] with nonzero contraction. Choose 0\ne c\in\mathfrak P\cap C, and choose a prime \mathfrak q minimal over (c) and contained in \mathfrak P\cap C. The principal ideal theorem [22, Tag 00KV] and the fact that C is a domain give \operatorname{ht}\mathfrak q=1. Now \mathfrak qC[t] is a nonzero prime contained in the height-one prime \mathfrak P, so \mathfrak P=\mathfrak qC[t]. The corresponding divisor is the pullback of the prime divisor defined by \mathfrak q, with multiplicity one: a uniformizer of the discrete valuation ring C_{\mathfrak q} is also a uniformizer at the generic point of that pulled-back divisor. Thus every divisor class on \Aff^1\times X is pulled back from X, proving surjectivity of \pi^*. Restriction to i_0\colon X\to\Aff^1\times X, x\mapsto(0,x), is a left inverse to \pi^* on Picard groups, proving injectivity as well.

Since \alpha\circ i_0=\pi\circ i_0=\mathrm{id}_X, the line bundles \alpha^*L and \pi^*L have the same restriction to the identity section. The Picard isomorphism just proved implies that they are isomorphic. Take an isomorphism \phi_1 between them, with the orientation in Equation (4.8). Its restriction i_0^*\phi_1 is an automorphism of L. Composing on the source with \pi^*((i_0^*\phi_1)^{-1}) produces an isomorphism \phi whose restriction is the identity.

An automorphism of any line bundle on a scheme is multiplication by a global unit: its endomorphism sheaf is canonically the structure sheaf. Because C is a domain,

Indeed, total degree of a product of nonzero polynomials over a domain is the sum of their total degrees, so a polynomial unit has degree zero. The ratio of two normalized isomorphisms in Equation (4.8) is a unit of C[t] whose value at t=0 is one. By Equation (4.9), that ratio is one. This proves uniqueness, without a restriction on the units of C.

To check the cocycle, denote the fiber map by \phi_{t,x}\colon L_x\to L_{\alpha(t,x)}. Over \Aff^2\times X, the two compositions to compare are

Their sources and targets agree by the action law for \alpha. Thus

is an automorphism of the pullback of L from X, hence a unit of C[s,t]. At s=0 it is the identity. By Equation (4.9), it is the identity everywhere. This proves the cocycle. The family of fiber maps consequently defines an algebraic additive action on the total space of L. Linearity preserves zero and nonzero vectors and commutes with scalar multiplication, giving the asserted action on L^\times.

Finally, suppose L^\times=\Spec T is affine. Write its action on functions as

where the sum is finite for each r. The identity and action laws give \delta_0=\mathrm{id} and

The coefficient of t in multiplicativity of \rho says that \delta_1 is a derivation. Since the characteristic is zero, the displayed composition identity gives \delta_n=\delta_1^n/n!. Finiteness of \rho(r) now says that \delta_1 is locally nilpotent. Commutation with fiber scaling says precisely that this derivation preserves the fiber weights.□

Proof. By Proposition 3.2, Z=\Spec G is smooth and integral. Exponentiating D_0 gives the algebraic action on functions

Apply Proposition 4.2 to this action and the line bundle L on Z whose complement is \Spec\Gt. The resulting locally nilpotent derivation \Dt restricts to D_0 because the action covers the given action on Z, and has fiber weight zero because the lift commutes with fiber scaling. Since G\hookrightarrow\Gt, a nonzero D_0 gives a nonzero \Dt.

It remains to establish its valuation degree. For \lambda\in\kk^*, let \sigma_\lambda be the automorphism of G multiplying a homogeneous element of degree h by \lambda^h. It fixes the coefficient subring R pointwise. Therefore it has the pullback lift \widetilde\sigma_\lambda=\sigma_\lambda\otimes\mathrm{id}_{\Rt} on \Gt. Explicitly,

Geometrically this lift acts on L=Z\times_Y L_Y by the degree action on Z and the identity on L_Y. In particular it extends to a linear bundle automorphism, and its action on the complement agrees with Equation (4.10).

On the coordinate ring of the base, homogeneity of D_0 gives

Conjugate the lifted additive action by \widetilde\sigma_\lambda, using the expression on coordinate rings

This is a normalized linear lift of the base action \exp(\lambda^jtD_0), by Equation (4.11). The action \exp(\lambda^jt\Dt) is another normalized linear lift of that same base action. Uniqueness in Proposition 4.2 therefore gives equality of these two families. Taking their coefficients of t yields

For a homogeneous r of degree h, this says \widetilde\sigma_\lambda(\Dt r)=\lambda^{h+j}\Dt r for every \lambda. The distinct integral characters of \Gm are linearly independent, so \Dt r is homogeneous of degree h+j whenever it is nonzero. This proves the required degree shift.□

Proof. Local nilpotence makes the defining maximum finite. Set r=\nu_\delta(f) and s=\nu_\delta(g). The iterated Leibniz rule gives

For n>r+s every term is zero. For n=r+s only the term with i=r can be nonzero, and it equals

Here we use both characteristic zero and the domain property. This proves Equation (5.2). The assertion about \delta(h) follows directly from the definition. Since a nonzero element has order zero exactly when it is invariant, product additivity proves factorial closure. Finally, if 0\ne\delta(h)=hq, then q\ne0 and

which is impossible.□

Proof. By Proposition 3.4, polynomiality of A supplies a nonzero homogeneous LND D_0 of G and a nonzero homogeneous h_0\in\ker D_0 of positive degree. Apply Proposition 4.3 to obtain an LND \Dt of \Gt lifting D_0, homogeneous of some degree shift j\in\mathbb Z and of fiber weight shift zero. The injection G\hookrightarrow\Gt in Proposition 4.1 ensures both that \Dt\ne0 and that the image h of h_0 is nonzero. This image is a homogeneous invariant of the same positive degree. By Equation (5.3), h is divisible by V in \Gt. Factorial closure in Lemma 5.1 therefore gives

We distinguish whether \tau is also invariant. If it is, the derivation restricts after removing a power of \tau. If it is not, the homogeneous pieces force a positive factor of V into its values on \Rt; removing this factor produces a derivation of negative fiber weight. The LND order proves local nilpotence in this second case without assuming that \Dt preserves \Rt.

Case 1: \Dt(\tau)=0. In this case \Dt extends to a locally nilpotent derivation of

Indeed, for f\in\Gt and m\ge0, all iterates on f\tau^{-m} equal \tau^{-m}\Dt^n(f). The degree-j component of this Laurent polynomial ring is exactly \tau^{-j}\Rt. Since every r\in\Rt has degree zero, there is thus a unique E(r)\in\Rt satisfying

Linearity and the Leibniz rule for \Dt, followed by cancellation of \tau^{-j}, show that E:\Rt\to\Rt is a \kk-derivation. It has fiber weight shift zero because \tau has fiber weight zero and \Dt preserves fiber weight. Since \Dt(\tau)=0, induction in Equation (5.5) gives

Local nilpotence of \Dt therefore proves local nilpotence of E. By Equation (5.4),

so E(v)=0. Finally, E=0 would make \Dt vanish on \Rt, \tau, and V, which generate \Gt; hence E\ne0. This proves the desired conclusion with q=0.

Case 2: \Dt(\tau)\ne0. Write \nu=\nu_{\Dt} on the nonzero elements of \Gt, and put m=\nu(\tau)\ge1. By Lemma 5.1, \Dt(\tau)\notin\tau\Gt. On the other hand, this element is homogeneous of degree j-1. Equation (5.3) shows that every component of negative degree, and every component of positive odd degree, is contained in \tau\Gt. We must therefore have

For r\in\Rt, the value \Dt(r) belongs to the degree-j piece. By Equation (5.3) there is a unique element E(r)\in\Rt such that

Thus the values of E lie in \Rt before any iteration or nilpotence argument is used. Cancellation of the common nonzero multiplier in the linearity and Leibniz identities again proves that E is a \kk-derivation of \Rt. Since that multiplier has fiber weight 2(\ell+1), the fiber weight shift of E is

We prove local nilpotence inside \Gt using its LND order. Whenever E(r)\ne0, Equation (5.7), Equation (5.4), and Lemma 5.1 imply

In fact, the order of the left side of Equation (5.7) is \nu(r)-1, whereas that of its right side is m+(\ell+1)\nu(V)+\nu(E(r))=m+\nu(E(r)). Every iterate E^n(r) still lies in \Rt. Consequently, as long as it is nonzero, successive applications of Equation (5.9) give

This nonnegative integer cannot decrease indefinitely, proving that E is locally nilpotent.

Differentiating the relation in Equation (5.1) gives

The last inequality uses characteristic zero and the domain property. Thus E(v)\ne0, and in particular E\ne0. Product additivity and Equation (5.4) give \nu(v)=\nu(\tau^2V)=2m; hence Equation (5.9) yields

If E^2(v) were nonzero, a second application would yield

an impossibility. Therefore E^2(v)=0, as required.□

Proof. Primes in this proof denote differentiation with respect to q. Consider the Wronskian

It is nonzero. Indeed, if W=0, relative primality implies that X divides X'. In characteristic zero this is impossible for nonconstant X, so X is a nonzero constant. Then Y'=0, hence Y and Z are constant as well, contrary to the hypothesis.

The same Wronskian, up to sign, is obtained from either of the other two pairs, because Z=X+Y. Work over an algebraic closure only to count roots. A root of multiplicity h of one of the three polynomials is not a root of either of the others. Computing W from a pair containing that polynomial shows that W vanishes at the root to order at least h-1. These roots are distinct across the three polynomials, so, on writing T=\deg X+\deg Y+\deg Z, we obtain

On the other hand, compute W from the two polynomials other than one of maximum degree M. Differentiation lowers the degree of every nonconstant polynomial by one, and therefore

Together these inequalities give M\le r(XYZ)-1.

The nonconstant hypothesis is necessary: for three nonzero constants the right side would be -1. The nonzero hypothesis also causes no missing nonconstant case: if one member of a pairwise relatively prime triple X+Y=Z is zero, the other two must be nonzero constants. The all-zero triple is not pairwise relatively prime.□

Proof. Write \delta=P(t)\,d/dt. There is nothing to prove if P=0. Otherwise let m=\deg P and let \gamma\ne0 be its leading coefficient. If m\ge1, induction shows that, for every n\ge1, the polynomial \delta^n(t) has degree 1+n(m-1) and leading coefficient

Every factor is nonzero in characteristic zero, including when m=1. Thus no iterate vanishes, contradicting local nilpotence. Consequently m=0.□

Proof. Suppose such an E exists. We introduce an auxiliary grading, distinct from both the valuation degree and the fiber weight:

The relation ac-bd-1 is homogeneous of auxiliary degree zero, so this is an actual \mathbb Z-grading on \Rt. Its components need not vanish in negative degrees. The generators are homogeneous for both this grading and the fiber grading, and the relation is homogeneous for both; thus the two gradings commute.

The highest auxiliary component. For an auxiliary-homogeneous r\in\Rt of degree i, define E_j(r) to be the degree-(i+j) component of E(r), and extend linearly to \Rt. Projecting the product rule to each degree shows that every E_j is a derivation. Only finitely many of these derivations are nonzero: their values on the five generators are obtained from the finitely many homogeneous components of their images under E, and a derivation vanishing on all five generators is zero. We therefore have a finite decomposition

Let \lambda be its greatest nonzero shift and set E'=E_\lambda. Then E'\ne0. For every auxiliary-homogeneous r of degree i and every n\ge1, the component of E^n(r) of degree i+n\lambda equals (E')^n(r): in the expansion of the iterate, each chosen shift is at most \lambda, and their sum can equal n\lambda only when every chosen shift equals \lambda. Local nilpotence of E therefore implies local nilpotence of E' on every homogeneous r, and hence on every element, which is a finite sum of such elements. This reasoning uses finite shift sets and finite sums, so the negative degrees in Equation (6.1) cause no difficulty. Projection to auxiliary degree preserves fiber weight, and consequently E' still has fiber shift e.

Set

Both its summands have auxiliary degree 6, whereas a^3b has auxiliary degree -3. In E^2(v)=E^2(a^3b-N), the first term has auxiliary degrees at most -3+2\lambda. The degree 6+2\lambda component of the second term is (E')^2(N). Taking that component of E^2(v)=0 gives

The orbit polynomials fix a,d,u. The localization obtained by solving for c is

In particular, a,d,b,u are algebraically independent, N\ne0, and the fraction field is

Since E' is locally nilpotent, exponentiating it gives polynomials

Each is nonzero, since its value at q=0 is the corresponding nonzero element of \Rt. The exponential is multiplicative by the iterated product rule. Hence Equation (6.2) implies

Here N(q)\ne0 because N(0)=N\ne0. Thus neither summand nor their sum is the zero polynomial.

The three polynomials in Equation (6.5) are pairwise relatively prime in K[q]. To see this, a common root of the two summands over \overline K would have multiplicity at least two in each summand. For the second summand this follows whether the root belongs to a(q), to u(q), or to both. Their sum would then have a root of multiplicity at least two, which a nonzero polynomial of degree at most one cannot have. Thus the summands are relatively prime, and the identity X+Y=Z implies relative primality of the other two pairs. No relative primality between a(q) and u(q) is needed.

Write

Suppose at least one summand in Equation (6.5) is nonconstant. A nonconstant summand has degree at least two. Since the sum has degree at most one, both summands must then have the same degree

The number of distinct roots of their product with N(q) is at most D_d+D_a+D_u+\deg_q N(q). Applying Lemma 6.1 therefore yields

This also covers constant N(q); in that case the first bound is stronger. The last inequality gives D_d\le D_a+D_u. Combining it with 2D_d=2D_a+3D_u gives 3D_u\le2D_u, so D_u=0. Consequently u(q)=u and D_d=D_a>0.

Let \alpha,\delta\in K^* be the leading coefficients of a(q) and d(q), respectively. Cancellation of the degree-M terms in Equation (6.5) requires

This equality would hold in the original field K of Equation (6.4), not merely in \overline K. It is impossible: put K_0=\kk(a,d,b), so that K=K_0(u). For a nonzero rational function of u, the multiplicity of u in its numerator minus its multiplicity in its denominator defines an integer-valued order. A square has even order, while -u^3 has order three. Passing to \overline K above was used only for counting roots and has not changed the field of the leading coefficients.

We conclude that both summands in Equation (6.5) are constant. Their nonvanishing gives 2D_d=0 and 2D_a+3D_u=0. All three degrees are nonnegative, so D_a=D_d=D_u=0. Taking coefficients of q in their exponentials now gives

The remaining coefficient has an impossible fiber weight. Differentiating ac-bd=1 and using Equation (6.7) gives aE'(c)=dE'(b). Define

Using both determinant identities, without dividing by either a or d, gives

Let C_0=\kk[a,d,u] and Q=\Frac(C_0)=\kk(a,d,u). By Equation (6.7), every nonzero element of C_0 is invariant. Thus E' extends to a locally nilpotent Q-derivation of

where the equality follows by solving for c as in Equation (6.3). Local nilpotence is preserved because (E')^n(r/s)=(E')^n(r)/s for s\in C_0\setminus\{0\}. By Lemma 6.2, E'(b)\in Q; hence Equation (6.8) implies h\in Q.

We spell out the coefficient intersection that improves this to h\in C_0. Besides Equation (6.3), solving for b gives

Since b and c are each transcendental over Q, comparison of polynomial coefficients in these two presentations gives

The last intersection is taken in Q: in the unique factorization domain C_0, the reduced denominator of an element belonging to both rings must divide a power of a and a power of d. As a and d are relatively prime, that denominator is a unit. Conversely C_0\subset\Rt\cap Q, so the intersection is exactly C_0. In particular h\in C_0.

The derivation E' is nonzero and fixes a,d,u; therefore Equation (6.8) implies h\ne0. Since E' has fiber shift e, the definition of h shows that it is homogeneous of fiber weight e-2<0. But every monomial in C_0=\kk[a,d,u] has nonnegative fiber weight, because the weights of a,d,u are 1,1,0. A nonzero element of C_0 cannot have negative fiber weight. This contradiction proves the proposition.□

Proof of Theorem 1.1. The explicitly defined algebra A=P/(H) is a finitely generated unital \kk-algebra. By Lemma 2.1 it is integral, with fraction field \kk(p,x,y,z), and hence has dimension four. Moreover, Proposition 2.2 gives a \kk-algebra isomorphism

Suppose that A\cong\kk\poly{4}. By Propositions 3.2 and 3.4, its associated graded ring G admits a nonzero homogeneous locally nilpotent derivation D_0 with a nonzero homogeneous invariant of positive valuation degree. The principal bundle and its coordinate ring \Gt are given by Proposition 4.1. By Proposition 4.3, D_0 lifts to a locally nilpotent derivation \Dt on \Gt with the same valuation degree shift and fiber shift zero. The reduction of this lift and invariant in Proposition 5.2 produces a nonzero locally nilpotent derivation E of \Rt with a fixed nonpositive fiber shift and E^2(v)=0. This contradicts Proposition 6.3. Thus A\not\cong\kk\poly{4}, while its cylinder is \kk\poly{5}. Taking n=4 proves the claimed counterexample over \kk.□

Proof. Set \Lambda=\kk[p] and write \mathbf z=(s,u,F,J). For every prime \mathfrak p of \Lambda, let H_{\mathfrak p} denote the image of H in \kappa(\mathfrak p)[\mathbf z]. The coordinate changes in Equations (2.3) and (2.4) show that H is a coordinate over \kk[p,p^{-1}]. At \mathfrak p=(p), the determinant-one change in Equations (2.7) and (2.8) gives H_{\mathfrak p}=L\bmod p. Every other prime of \Lambda avoids p. It follows that for every \mathfrak p there are three polynomials G_1,G_2,G_3 over \kappa(\mathfrak p) such that

is a polynomial coordinate system.

Let y be any point of \Aff^2_{p,q}, set \mathfrak p=\ker(\Lambda\to\kappa(y)), and let b be the image of q in \kappa(y). The field \kappa(y) is an extension of \kappa(\mathfrak p). Tensoring (7.1) with \kappa(y) and then setting the coordinate H_{\mathfrak p} equal to b gives

Here H_{\mathfrak p} in the quotient denotes its image after this field extension. Thus every fiber of \rho is affine three-space over its residue field, and the same is true after extension to an algebraic closure. In particular, \rho is surjective. Moreover,

so \pi is the base change of \rho along the closed immersion \Aff^1_p\hookrightarrow\Aff^2_{p,q} defined by q=0. The fiber assertions and surjectivity follow for \pi as well.

We next check smoothness without assuming local triviality. Let

For every \mathfrak p, differentiation with respect to the coordinate H_{\mathfrak p} in (7.1) gives a \kappa(\mathfrak p)-derivation \delta with \delta(H_{\mathfrak p})=1. The chain rule expresses 1 as a linear combination of the four partial derivatives of H_{\mathfrak p}. Consequently the image of \mathfrak j generates the unit ideal in \kappa(\mathfrak p)[\mathbf z]. If a prime of P contained \mathfrak j, localization and reduction over its contraction to \Lambda would give a proper prime of this fiber algebra containing that unit ideal. Hence \mathfrak j=P.

As a \Lambda[q]-algebra, P has the presentation

The principal open sets on which one of the four displayed partial derivatives is invertible cover \Spec P. On each such open set this presentation has one relation with an invertible Jacobian minor, so it is standard smooth of relative dimension 4-1=3. Thus \rho is smooth of relative dimension three. Smoothness and relative dimension are preserved by base change, giving the same assertion for \pi.

Suppose now that one of the two morphisms is Zariski-locally trivial, and put

respectively, with the algebra structures specified in the statement. In either case, B is a finitely presented R_0-algebra, and local triviality implies B_{\mathfrak m}\simeq R_{0,\mathfrak m}\poly3 for every maximal ideal \mathfrak m of R_0. Bass–Connell–Wright’s theorem [3, Theorem 3(a)], in the formulation recalled by Gupta [14, Theorem 4.1], therefore gives an R_0-algebra isomorphism

for a projective R_0-module M. This conclusion allows arbitrary polynomial transition automorphisms in the assumed local trivializations; no linear structure is assumed beforehand.

The module M is finitely generated. Indeed, if \mathfrak a is the augmentation ideal of \operatorname{Sym}_{R_0}(M), then M\simeq\mathfrak a/\mathfrak a^2; finite generation follows because B is a finite-type augmented R_0-algebra. Tensoring with each residue field and comparing the dimensions of the resulting polynomial rings shows that M has rank three. For the freeness step, Suslin’s Theorem 2(b) [24] states that a finitely generated projective module of rank greater than \dim C over C[t_1,\ldots,t_m] is extended from C when C is regular and noetherian. Here C=\kk has dimension zero, giving the needed Quillen–Suslin statement: M is extended from a three-dimensional complex vector space. Thus M is free and

For \pi this would give A\simeq\kk[p]\poly3\simeq\kk\poly4, contrary to Theorem 1.1. For \rho an R_0-algebra isomorphism would give a polynomial coordinate system P=\kk[p,H,G_1,G_2,G_3] for suitable G_1,G_2,G_3\in P. In particular, H would be a coordinate of P, contrary to Corollary 1.2. Neither morphism is therefore Zariski-locally trivial.□