A stable coordinate that is not a coordinate in four variables

Proof. Proposition 2.3 identifies each fiber with \A^3; in particular, (f-\lambda) is its prime defining ideal. If an ambient automorphism takes f^{-1}(\lambda) to a coordinate hyperplane, it takes its defining principal ideal to the ideal of a coordinate. The two generators therefore differ by a nonzero constant. This would make f-\lambda, and hence f, a coordinate, contrary to Theorem 1.1.□

Proof. The relation H becomes linear after a change of coordinates preserving x. To exhibit that change, write

Then x=-\det M, e^{\mathsf t}n=0, and

Keep J fixed and define F',s',u',p' by

Since (ne^{\mathsf t})^2=0 and \det(I_2-2ne^{\mathsf t})=1, we have -\det M'=x. Consequently x and J, and hence e and n, are unchanged by the substitution. Its inverse is the polynomial substitution

so p',s',u',F',J form a coordinate system on P.

Taking half the difference of the off-diagonal entries in Equation (2.3) gives

Eliminating u' therefore identifies A with k[p',s',F',J]. In this quotient u'=-J, so the determinant identity gives the first formula in Equation (2.2). The lower-right entry of the inverse matrix identity gives

which becomes the second formula after substituting u'=-J.□

Proof. A derivation \delta of a k-algebra is locally nilpotent if, for every element g, some iterate \delta^r(g) is zero. We use such a derivation to replace the equation H=0 by H+pw=0.

In the coordinates of Equation (2.6), take

This locally nilpotent derivation of P[p^{-1}] preserves P, because Equation (2.7) yields

Its restriction to P is therefore locally nilpotent as well. Applying \Delta to Equation (2.5) gives \Delta(H)=p. Extend \Delta to P[w] by \Delta(w)=0. The exponential

is a polynomial automorphism: each sum is finite, and its inverse is \exp(-w\Delta). It fixes p and sends H to H+pw, inducing a k[p]-algebra isomorphism

It remains to eliminate one variable from the equation on the right. Put x_0=s^2-u^2 and introduce

The coefficient matrix of this substitution has determinant 4s^2x_0^2+(1+2sx_0)(1-2sx_0)=1; explicitly,

Thus p,s,u,L,N are polynomial coordinates on P. Expanding x=x_0+pF in H gives

Since Q_0\in P, the substitution w'=w+Q_0 is another polynomial change of coordinates, and

Together with Equation (2.9), this proves the assertion with p fixed throughout. Composing with the identification in Proposition 2.1 and relabeling the target variables so that p maps to x_1 gives an automorphism of R[w] sending f to x_1.□

Proof. At \lambda=0, Equations (2.10)–(2.12) give

For \lambda\ne0, Equations (2.6) and (2.5) give

by elimination of u.□

Proof. Exhaustiveness follows because every element of A has a polynomial representative. For r>0, every monomial of weight at most -r is divisible by p^r, so

By Proposition 2.1, A is a polynomial ring, and Equation (2.2) shows that p is a nonzero nonunit. Therefore \bigcap_{r>0}p^rA=0: in a unique factorization domain, a nonzero element cannot be divisible by arbitrarily high powers of an irreducible factor of p. This proves separation. In particular, every nonzero a\in A has a well-defined degree

and a nonzero initial form in \mathcal F_dA/\mathcal F_{d-1}A.

The element x is homogeneous of weight zero in P, and H=H_*-u has highest weight part H_*, of weight one. For every nonzero q\in P, the highest weight part of qH is (\operatorname{in}q)H_*: the polynomial ring is a graded domain, so the product of the highest parts does not vanish. Consequently the ideal of initial relations is exactly (H_*). For completeness, the natural graded map P\to\gr A sends a homogeneous polynomial of weight d to its class in \mathcal F_dA/\mathcal F_{d-1}A, which may at this stage be zero. This map is surjective. A homogeneous polynomial of weight d lies in its kernel exactly when its image in A has a representative of weight at most d-1; subtracting that representative gives a relation with the original polynomial as its highest part. Thus the kernel is the ideal of initial relations, proving (3.2).

To prove that this quotient is a domain, use the polynomial coordinates x,y,z,u of P[p^{-1}] from the preceding section. The identity xy-z(z+1)=p(H+u) becomes

The polynomial on the right is irreducible over k[p,p^{-1},x,z,u]: as a polynomial in y, it is primitive, because x and z(z+1) are coprime, and it has degree one over the fraction field. Thus H_* is irreducible after inverting p. It is not divisible by p before localization, since its reduction modulo p is

Any factorization of H_* in P has a factor that becomes a unit after inverting p, hence is a scalar times a power of p. Since p\nmid H_*, this factor is already a unit in P. Thus H_* is irreducible in P, and hence prime by unique factorization. Finally, H_* is not associated to any of the five variables, so their classes in G are nonzero. These classes are the initial forms of the corresponding elements of A, proving the asserted degrees.□

Proof. A monomial in the generators in (3.6), with coefficient in B, has the form

It has degree \ell exactly when i+j-r=\ell. In that case it equals \tau^{-\ell}b f_0^i g_0^j, whose numerator belongs to I^{\ell+r}\subseteq I^\ell. Since the presentation is homogeneous, such monomials span G_\ell. Conversely, every element of I^\ell is a B-linear combination of the products f_0^i g_0^j with i+j=\ell, and dividing these products by \tau^\ell gives elements of G_\ell.□

Proof. Suppose A=k[p,q_1,q_2,q_3] for a polynomial coordinate system. At least one q_i has strictly positive filtration degree. Indeed, if all three had degree at most zero, then so would every polynomial in p,q_1,q_2,q_3, contrary to \deg F=1. Relabel so that \deg q_1>0, and take

This is a nonzero locally nilpotent derivation of A satisfying D(p)=D(q_1)=0.

We give the details of its passage to G, because the filtration has both positive and negative indices. Write \omega(t) for the assigned weight of a generator t\in\{p,s,u,F,J\}, and put

This maximum is defined: a nonzero derivation cannot kill all algebra generators. The product rule shows that a monomial of weight at most d has derivative in \mathcal F_{d+m}A. Every element of \mathcal F_dA has a representative that is a sum of such monomials, by the definition of the quotient filtration. Thus

It follows that D induces a homogeneous derivation D_0 of G of degree m. For a\in\mathcal F_dA, write [a]_d for its class in \mathcal F_dA/\mathcal F_{d-1}A. Explicitly,

The first formula is well-defined because D also sends \mathcal F_{d-1}A into \mathcal F_{d+m-1}A; the second follows by induction. It proves that D_0 is locally nilpotent, first on homogeneous elements and then on finite sums of them. Moreover, equality is attained in (3.7) for some generator t, whose initial form has degree \omega(t) by Proposition 3.1. Its image under D_0 is the nonzero initial form of D(t), so D_0\ne0. Finally, (3.8) shows that D_0 kills both \tau and the nonzero initial form \eta of q_1. The latter is homogeneous of degree \ell=\deg q_1>0.

Because D_0(\tau)=0, extending D_0 to G[\tau^{-1}]=B[\tau,\tau^{-1}] preserves local nilpotence. Homogeneity implies that there is a derivation E:B\to B with

Indeed, every degree-m element of the Laurent ring is uniquely \tau^{-m} times an element of B. The derivation E is nonzero: otherwise D_0 would kill both B and \tau, hence the whole Laurent ring and its subring G. It is locally nilpotent because

Here, as in (3.8), the formula follows by induction, using that D_0 fixes \tau.

By Lemma 3.2, write \eta=\tau^{-\ell}h with 0\ne h\in I^\ell. Equations (3.9) and D_0(\eta)=0 give

Thus E(h)=0, and I^\ell\subseteq I because \ell>0.□

Proof. Over U_0, the identity ac=z+1 makes a invertible. The formulas

give an isomorphism C[(z+1)^{-1}]=B[(z+1)^{-1}][a,a^{-1}]. In particular the localized map from B is injective, and so is the original map because B is a domain. Over U_1, the corresponding formulas are

and give C[z^{-1}]=B[z^{-1}][d,d^{-1}]. The opens cover Y because z and z+1 generate the unit ideal. On their overlap the fiber coordinates satisfy

where y/(z+1) is a unit with inverse x/z. Gluing the trivial line bundles by this transition gives \mathcal L; removing their zero sections gives precisely the two charts of \Spec C above. Both fiber coordinates have weight one, so the scaling agrees with (4.3).□

Proof. In C[a^{-1}], put \xi=d/a and j=x+u^2-\xi^2. The determinant relation and (4.2) give

Consequently

The two coefficients of j generate the unit ideal, as the identity

shows. Thus IC[a^{-1}]=(j)=(v).

It remains to check primes of C containing a. Modulo such a prime, bd=-1, while x=0 and z=-1. Hence g_0=z-s_Bx has nonzero residue -1, and v has nonzero residue -d^2. Both are units in the corresponding local ring. The ideals IC and (v) therefore agree at every prime, which proves the assertion.□

Proof. The two charts of Lemma 4.1 have coordinate rings

Here t denotes a separate coordinate on each chart. Each ring, and its polynomial extension by the affine-line coordinate T, is a noetherian unique factorization domain. A line bundle on the spectrum of such a ring is trivial: its divisor class is represented by a Cartier divisor, and unique factorization makes every divisor principal.

Let \mathcal M be a line bundle on \A^1\times Y. Choose trivializations on \A^1\times U_0 and \A^1\times U_1. Its transition function belongs to B[z^{-1},(z+1)^{-1}][T]^*. A polynomial over a domain is a unit only when it is a constant unit, so this transition function is independent of T. Its specialization at T=0 is therefore the same transition function. The chosen trivializations identify \mathcal M with the pullback of its restriction at zero.□

Proof. Write R=k[X], let \mathcal M be the line bundle, and let \alpha:\A^1\times X\longrightarrow X be the action. If \pi:\A^1\times X\longrightarrow X is projection, then \alpha^*\mathcal M and \pi^*\mathcal M restrict to \mathcal M at parameter zero. The hypothesis therefore supplies an isomorphism

At zero, \eta is an automorphism of \mathcal M, hence multiplication by a unit of R. Multiplying \eta by the inverse pullback of that unit normalizes it to the identity at zero.

The isomorphism \eta gives a fiberwise linear lift of \alpha. To check the group law, compare, over \A^2\times X, the map on fibers obtained by applying parameters S and then T with the map obtained by applying T+S. Both are global isomorphisms of the same source and target line bundles, with fiber maps from \mathcal M_x to \mathcal M_{\alpha(T+S,x)}. Their ratio is an automorphism of the source line bundle. Its endomorphism sheaf is canonically \mathcal O, so this automorphism is multiplication by a global unit in R[T,S]. Since R is a domain, this unit is independent of T,S. At (T,S)=(0,0) it equals one by the normalization of \eta. Thus the group law holds. The resulting action is linear on fibers, so it commutes with fiber scaling and preserves the complement of the zero section.□

Proof. The exponential \exp(T\delta) takes values in R[T]. If rs is fixed, then

Degrees in T add under multiplication of nonzero polynomials over a domain. Both factors on the left therefore have degree zero, and their coefficients of T give the assertion.□

Proof. The exponential of E defines an additive-group action on Y. By Lemmas 4.3 and 4.4, it lifts to \mathcal L by fiberwise linear maps. Restrict to the complement of the zero section, namely \Spec C. Differentiating this action at the identity gives a derivation \widetilde E of C. The action law expresses its pullback on each function as \sum_{r\ge0}T^r\widetilde E^{\,r}/r!; because the action is algebraic, the sum is finite for each function. Hence \widetilde E is locally nilpotent. It extends E and is nonzero, and commutation with fiber scaling says exactly that it preserves the weights (4.3).

Finally choose 0\ne h\in I\cap\ker E. The inclusion B\subset C and Lemma 4.2 give a factorization h=vq in C, with both factors nonzero. Since \widetilde E(h)=0, Lemma 4.5 yields \widetilde E(v)=0.□

Proof. Suppose that \widetilde E is such a derivation. Give C the auxiliary grading in the second row of the following table:

The two gradings are compatible because ac-bd-1 is homogeneous of degree zero for each. The auxiliary degrees of the two parts of v are

Decompose \widetilde E into its homogeneous components for auxiliary degree shift, and let E' be the nonzero component with largest shift, say m. There are only finitely many components: a derivation is determined by its values on the five homogeneous generators, and each value is a finite sum of homogeneous terms. Each component is a derivation by the product rule. Moreover, E' is locally nilpotent. Indeed, for homogeneous r\in C of auxiliary degree e, the component of auxiliary degree e+nm in \widetilde E^{\,n}(r) is (E')^n(r). For sufficiently large n the former expression vanishes, so the latter vanishes as well. Homogeneous elements span C, proving local nilpotence on every element. Since the gradings in (5.1) are compatible, E' still preserves fiber weights.

Taking the component of degree m+2 in \widetilde E(v)=0 gives

The factorization

and Lemma 4.5 imply that E' annihilates both au-d and au+d. Thus E'(d)=E'(au)=0, and another application of that lemma gives

All the factors used here are nonzero; for example, after inverting a one has

where their nonvanishing is immediate.

It remains to control the possible values of E' on b,c. Differentiating ac-bd=1 and using (5.2) yields aE'(c)=dE'(b). Define

Multiplication by a and d, respectively, then gives

The element q is nonzero, since otherwise E' would kill every generator. We claim that q belongs to the polynomial subring k[a,d,u].

The elements a,d,u are algebraically independent by (5.3). Put K=k(a,d,u) and localize C at the nonzero elements of k[a,d,u]. Because these elements lie in \ker E', the localized derivation remains locally nilpotent. The resulting algebra is K[b], with c=(1+bd)/a. A locally nilpotent K-derivation of K[b] sends b into K: if its value P(b) had degree at least one, iterating P(b)\,d/db on b would never give zero. More explicitly, differentiation of a nonconstant polynomial multiplies its nonzero leading coefficient by its positive degree, and multiplication by P then leaves the result nonconstant. Equation (5.4) therefore gives q\in K.

To compute C\cap K inside the fraction field of C, use both coordinate charts:

An element of the first polynomial ring lying in K has no b terms, and an element of the second lying in K has no c terms. Hence

The last equality follows from unique factorization in k[a,d,u]: a reduced denominator that is both a power of a and a power of d is a unit. The reverse inclusion is immediate, proving the claim.

Finally, preservation of fiber weights makes E'(b) have weight -1. By (5.4), the nonzero polynomial q has weight -2. This is impossible in k[a,d,u], whose generators have weights 1,1,0. The contradiction proves the proposition.□

Proof of Theorem 1.1. By Proposition 2.1, the algebra A is a polynomial ring in four variables and its element p becomes the polynomial f stated in the theorem. Proposition 2.2 makes p a coordinate of A[w]. If p were a coordinate of A, Proposition 3.3 would produce a nonzero locally nilpotent derivation E of B with I\cap\ker E\ne0. Proposition 4.6 would then supply a derivation forbidden by Proposition 5.1. Thus p is not a coordinate of A. Transporting these two properties through the polynomial presentation proves the assertion for f.□