Proof. Proposition 5 identifies each fiber with \(\mathbb A^3\); 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. ◻

Proof. Over \(U_0\), the identity \(ac=z+1\) makes \(a\) invertible. The formulas \[ c=\frac{z+1}{a},\qquad b=\frac{x}{a},\qquad d=\frac{ay}{z+1} \tag{26}\] 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 \[ b=\frac{z}{d},\qquad c=\frac{y}{d},\qquad a=\frac{xd}{z}, \tag{27}\] 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 \[ d=\frac{y}{z+1}\,a, \tag{28}\] 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 \(\mathop{\mathrm{Spec}}C\) above. Both fiber coordinates have weight one, so the scaling agrees with (25). ◻

Proof. In \(C[a^{-1}]\), put \(\xi=d/a\) and \(j=x+u^2-\xi^2\). The determinant relation and (24) give \[z=\xi x,\qquad y=\xi+\xi^2x,\qquad s_B=y-x(x+u^2)=\xi-xj.\] Consequently \[ f_0=(1+2\xi x-x^2j)j,\qquad g_0=x^2j,\qquad v=a^2j. \tag{30}\] The two coefficients of \(j\) generate the unit ideal, as the identity \[(1-2\xi x)(1+2\xi x-x^2j) +x^2(4\xi^2+j-2\xi xj)=1\] 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 9 have coordinate rings \[\begin{align*} B[(z+1)^{-1}]&=k[x,t,u,(xt+1)^{-1}], &t&=\frac{y}{z+1},\quad z=xt,\\ B[z^{-1}]&=k[x,t,u,(xt-1)^{-1}], &t&=\frac{y}{z},\quad z=xt-1. \end{align*}\] 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 \(\mathbb A^1\times Y\). Choose trivializations on \(\mathbb A^1\times U_0\) and \(\mathbb A^1\times U_1\). 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 \(\alpha:\mathbb A^1\times X\longrightarrow X\) be the action. If \(\pi:\mathbb A^1\times X\longrightarrow X\) is projection, then \(\alpha^*\mathcal M\) and \(\pi^*\mathcal M\) restrict to \(\mathcal M\) at parameter zero. The hypothesis therefore supplies an isomorphism \[\eta:\pi^*\mathcal M\longrightarrow\alpha^*\mathcal M.\] 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 \(\mathbb A^2\times X\), 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 \[\exp(T\delta)(r)\,\exp(T\delta)(s)=rs.\] 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 11 and 12, it lifts to \(\mathcal L\) by fiberwise linear maps. Restrict to the complement of the zero section, namely \(\mathop{\mathrm{Spec}}C\). 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 (25).

Finally choose \(0\ne h\in I\cap\ker E\). The inclusion \(B\subset C\) and Lemma 10 give a factorization \(h=vq\) in \(C\), with both factors nonzero. Since \(\widetilde E(h)=0\), Lemma 13 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: \[ \begin{array}{c|rrrrr} &a&d&b&c&u\\ \hline \text{fiber weight}&1&1&-1&-1&0\\ \text{auxiliary degree}&0&1&-1&0&1 \end{array} \tag{31}\] 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 \[\deg_{\mathrm{aux}}(a^3b)=-1,\qquad \deg_{\mathrm{aux}}(a^2u^2-d^2)=2.\]

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 (31) are compatible, \(E'\) still preserves fiber weights.

Taking the component of degree \(m+2\) in \(\widetilde E(v)=0\) gives \[E'(a^2u^2-d^2)=0.\] The factorization \[a^2u^2-d^2=(au-d)(au+d)\] and Lemma 13 imply that \(E'\) annihilates both \(au-d\) and \(au+d\). Thus \(E'(d)=E'(au)=0\), and another application of that lemma gives \[ E'(a)=E'(d)=E'(u)=0. \tag{32}\] All the factors used here are nonzero; for example, after inverting \(a\) one has \[ C[a^{-1}]=k[a,a^{-1},d,u][b],\qquad c=\frac{1+bd}{a}, \tag{33}\] where their nonvanishing is immediate.

It remains to control the possible values of \(E'\) on \(b,c\). Differentiating \(ac-bd=1\) and using (32) yields \(aE'(c)=dE'(b)\). Define \[q=cE'(b)-bE'(c)\in C.\] Multiplication by \(a\) and \(d\), respectively, then gives \[ E'(b)=aq,\qquad E'(c)=dq. \tag{34}\] 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 (33). 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 (34) therefore gives \(q\in K\).

To compute \(C\cap K\) inside the fraction field of \(C\), use both coordinate charts: \[C[a^{-1}]=k[a,a^{-1},d,u][b],\qquad C[d^{-1}]=k[a,d,d^{-1},u][c].\] 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 \[C\cap K\subset k[a,a^{-1},d,u]\cap k[a,d,d^{-1},u]=k[a,d,u].\] 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 (34), 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. By Proposition 3, the algebra \(A\) is a polynomial ring in four variables and its element \(p\) becomes the polynomial \(f\) stated in the theorem. Proposition 4 makes \(p\) a coordinate of \(A[w]\). If \(p\) were a coordinate of \(A\), Proposition 8 would produce a nonzero locally nilpotent derivation \(E\) of \(B\) with \(I\cap\ker E\ne0\). Proposition 14 would then supply a derivation forbidden by Proposition 15. Thus \(p\) is not a coordinate of \(A\). Transporting these two properties through the polynomial presentation proves the assertion for \(f\). ◻