Dimension Two

Abstract

This chapter examines locally nilpotent R-derivations of R[x, y] = R ^[2] for certain rings R containing $\mathbb{Q}$. This set is denoted LND _R (R[x, y]).

Appendix: Newton Polytopes

It is natural to define higher dimensional analogues of Newton polygons and investigate their properties relative to locally nilpotent derivations of polynomial rings. Such investigation is the subject of papers by Hadas, Makar-Limanov, and Derksen [97, 196, 197].

Let f ∈ B = k ^[n] be given. Relative to a coordinate system B = k[x ₁, …, x _n ], we define the Newton polytope of f, denoted Newt( f), as follows. Write:

$$\displaystyle{\,f =\sum \alpha _{e}x_{1}^{e_{1} }\cdots x_{n}^{e_{n} }\,\,,\quad e = (e_{1},\ldots,e_{n}) \in \mathbb{Z}^{n}\,\,,\,\,e_{ i} \geq 0\,\,,\,\,\alpha _{e} \in k}$$

The support of f, denoted Supp( f), equals $\{e \in \mathbb{Z}^{n}\vert \alpha _{e}\neq 0\}$; and Newt( f) equals the convex hull of Supp( f) ∪{ 0} in $\mathbb{R}^{n}$. Note that this definition depends on the choice of coordinates for B.

For such polynomials, the following special property emerges.

Proposition 4.34 (Thm. 3.2 of [197])

Suppose B = k[x ₁, …, x _n ] = k ^[n] , D ∈ LND(B), D ≠ 0, and f ∈ ker D. Then every vertex of Newt( f) is contained in a coordinate hyperplane. More precisely, if $q = (q_{1},\ldots,q_{n}) \in \mathbb{Q}^{n}$ is a vertex of Newt( f), then q _i = 0 for at least one i.

Proof

Suppose q ≠ 0. By convexity, there exists a hyperplane $H \subset \mathbb{R}^{n}$ such that H ∩ Newt( f) = {q}. We may suppose that H is defined by the equation:

$$\displaystyle{\sum _{1\leq i\leq n}a_{i}y_{i} = d\,\,,\quad y = (\,y_{1},\ldots,y_{n}) \in \mathbb{R}^{n}\,\,,\,\,a_{ i},d \in \mathbb{Z}\,\,,\,\,d> 0}$$

Since 0 ∈ Newt( f), we conclude that ∑a _i y _i ≤ d for y ∈ Newt( f), with equality only at q.

Note that H determines a $\mathbb{Z}$-grading on B, namely, the degree of the monomial $x_{1}^{t_{1}}\cdots x_{n}^{t_{n}}$ is ∑ _{i = 1} ⁿ a _i t _i . For this grading, write $B = \oplus _{i\in \mathbb{Z}}B_{i}$; then f = ∑ _{i = 0} ^d f _i for f _i ∈ B _i , and f _d is the monomial supported at q. In addition, if D′ is the highest-degree homogeneous summand of D relative to this grading, then according to Corollary 1.31 , D′ ∈ LND(B) and f _d ∈ ker D′. Since D ≠ 0, D′ ≠ 0 as well. If each q _i were strictly positive, then since ker D′ is factorially closed and f _d is a monomial, we would have that x _i ∈ ker D′ for each i, implying D′ = 0. Therefore, at least one of the q _i equals 0. □

Remark 4.35

Any variable f of B = k ^[n] is in the kernel of some nonzero D ∈ LND(B), for example, D = ∂ _g where ( f, g) is a partial system of variables for B. Thus, any variable possesses the property described in the theorem above. In fact, it is shown in [97] that, regardless of the characteristic of the field k, the Newton polytope of an invariant of an algebraic $\mathbb{G}_{a}$-action on $\mathbb{A}^{n}$ has all its vertices on coordinate hyperplanes (Thm. 3.1). Thus, this property applies to any variable, regardless of characteristic.

A second property of Newton polytopes involves its edges. Continuing the assumptions and notations above, an edge E of Newt( f) is called an intrusive edge if it is contained in no coordinate hyperplane of $\mathbb{R}^{n}$. (Such edges are called trespassers in [97].) If E is the intrusive edge joining vertices p and q, then e(E), or just e, will denote the vector e = p − q (or q − p, it doesn’t matter).

Proposition 4.36 (Thm. 4.2 of [197])

Suppose f ∈ ker D for D ∈ LND(B) and D ≠ 0. Let E be an intrusive edge of Newt( f), where e = (e ₁, …, e _n ) = p − q for vertices p = ( p ₁, …, p _n ) and q = (q ₁, …, q _n ) of Newt( f).

• (a) There exist integers r ≠ s such that p _r = 0, q _s = 0, and either $e \in q_{r} \cdot \mathbb{Z}^{n}$ or $e \in p_{s} \cdot \mathbb{Z}^{n}$

• (b) min_{1 ≤ i ≤ n} {e _i } < 0 and max_{1 ≤ i ≤ n} {e _i } > 0

Proof (Following [197])

We first make several simplifying assumptions and observations.

• Given i, set $B_{i} = k[x_{1},\ldots,\hat{x_{i}},\ldots,x_{n}]$. Note that f ∉ ∪ _{i = 1} ⁿ B _i , since the Newton polytope of an element of B _i has no intrusive edges.

• Suppose this result holds for algebraically closed fields of characteristic zero, and let L denote the algebraic closure of k. If D _L denotes the extension of D to B _L = L[x ₁, …, x _n ], then D _L ∈ LND(B _L ) and f ∈ ker (D _L ). Note that the Newton polytope of f relative to D or D _L is the same. So from now on we assume k is algebraically closed.

• To simplify notation, for $u \in \mathbb{Z}^{n}$, x ^u will denote the monomial $x_{1}^{u_{1}}\cdots x_{n}^{u_{n}}$.

• After a permutation of the variables, we may assume that x ₁, …, x _γ ∈ ker D and x _γ+1, …, x _n ∉ ker D, where γ is an integer with 0 ≤ γ ≤ n.

We now proceed with the proof.

To show (a), first choose a $\mathbb{Q}$-linear function $\lambda: \mathbb{Q}^{n} \rightarrow \mathbb{R}$ such that ker (λ) = 〈e〉. (For example, choose a projection $\mathbb{Q}^{n} \rightarrow \mathbb{Q}^{n-1}$ mapping E to a single point, and then pick your favorite copy of $\mathbb{Q}^{n-1}$ in $\mathbb{R}$, such as $\mathbb{Q}[\zeta ]$ for a real algebraic number ζ of degree n − 1.) Thus, λ is constant on E, and we may further assume that λ(E) = max λ(Newt( f)) ≥ 0. Such λ induces a grading $B = \oplus _{\rho \in \mathbb{R}}B_{\rho }$ in which the degree of the monomial x ^u equals λ(u). Elements of B _ρ will be called λ-homogeneous of λ-degree ρ.

As before, if D′ is the highest λ-homogeneous summand of D, and if f′ is the highest λ-homogeneous summand of f, then D′ is locally nilpotent, D′( f′) = 0, and Newt( f′) is the convex hull of 0, p and q. So we might just as well assume D = D′ and f = f′.

Choose $\epsilon \in \mathbb{Q}^{n}$ which spans the kernel of λ. We may assume that $\epsilon \in \mathbb{Z}^{n}$, and that the entries of ε have no common factor. It follows that every λ-homogeneous element of B can be written as x ^α P(x ^ε), where $\alpha \in \mathbb{Z}^{n}$ has non-negative entries, and P is a univariate polynomial over k. (Note that some entries of ε can be negative.) In particular, f = x ^α P(x ^ε) for some α and some P. This implies that we may write

$$\displaystyle{\,f = x^{\alpha }\prod _{i}(x^{\epsilon } - t_{i}) = x^{w}\prod _{ i}(x^{u} - t_{ i}x^{v})}$$

where t _i ∈ k, $u,v,w \in \mathbb{Z}^{n}$ have non-negative entries, and u _i v _i = 0 for each i. Moreover, since every x _i appears in f, it follows that u _i + v _i + w _i > 0 for each i.

Since e is an intrusive edge, f is divisible by at least one factor of the form x ^u − tx ^v, t ∈ k, meaning x ^u − tx ^v ∈ ker D. If also x ^u − sx ^v divides f for s ≠ t, then we have x ^u, x ^v ∈ ker D. Since ker D is factorially closed, x ^w ∈ ker D as well. But then since u _i + v _i + w _i > 0, it would follow that x _i ∈ ker D for every i (again since ker D is factorially closed), which is absurd. Therefore ker D contains x ^u − tx ^v for exactly one value of t, and t ≠ 0. Altogether, this implies f = cx ^w(x ^u − tx ^v)^m for some c ∈ k ^∗ and positive integer m.

In the same way, if g ∈ ker D is any other λ-homogeneous element, then g = dx ^ω(x ^u − tx ^v)^μ, where d ∈ k, $\omega \in \mathbb{Z}^{n}$ has non-negative entries, and μ is a non-negative integer. Since ker D is factorially closed, x ^ω ∈ k[x ₁, …, x _γ ]. It follows that ker D = k[x ₁, …, x _γ , x ^u − tx ^v], and since the transcendence degree of ker D is n − 1, we have γ + 1 = n − 1, or γ = n − 2.

Let K = k(x ₁, …, x _n−2). Then D extends to a locally nilpotent K-derivation of R = K[x _n−1, x _n ], so by Rentschler’s Theorem x ^u − tx ^v is a K-variable of R. As such, it must have a degree-one term in either x _n−1 or x _n over K.

Let us assume u _n = 1 (so v _n = 0). Recalling that ε = u − v, we have that ε _n = 1. Now f = x ^w(x ^u − tx ^v)^m, where w _n = 0. It follows that p = mu + w and q = mv + w. Since v _n = 0, we conclude that q _n = 0 as well. Since e = p − q is an integral multiple of ε, and e _n = p _n , we conclude that e = p _n ε. The other cases are similar. This completes the proof of (a).

By the preceding result, both p and q have at least one 0 entry, and since e is an intrusive edge, we can find r and s with p _r = 0, q _s = 0, and r ≠ s. Thus, e _r = p _r − q _r = −q _r ≤ 0 and e _s = p _s − q _s = p _s ≥ 0, and (b) follows. □

As noted in [197], these two results provide a quick way to determine that certain polynomials cannot be annihilated by a nonzero locally nilpotent derivation. For example, if $a,b,c \in \mathbb{Z}$ have a, b, c > 1 and gcd(a, b, c) = 1, then f = x ^a + y ^b + z ^c, g = x ^a y ^b + z ^c, and h = x ^a+1 y ^b + x ^a z ^c + 1 are elements of k[x, y, z] not in the kernel of any nonzero D ∈ LND(k[x, y, z]). In the first case, Newt( f) has the intrusive edge (a, −b, 0). In the second case, Newt(g) has the intrusive edge (a, b, −c). An in the third case, Newt(h) has vertices p = (a + 1, b, 0) and q = (a, 0, c), so that e = (1, b, −c) is an intrusive edge which belongs to neither $b\mathbb{Z}^{3}$ nor $c\mathbb{Z}^{3}$. Consequently, the surfaces in $\mathbb{A}^{3}$ defined by f, g and h are not stabilized by any $\mathbb{G}_{a}$-action on $\mathbb{A}^{3}$.

Example 4.37

Let B = k[x, y, z] = k ^[3], and define G ∈ B by:

$$\displaystyle{G = x^{2}z^{3} - 2xy^{2}z^{2} + y^{4}z + 2x^{3}yz - 2x^{2}y^{3} + x^{5}}$$

In Chap.  5 it is shown that G is irreducible, and that there exist nonzero D ∈ LND(B) with DG = 0. Since G is homogeneous in the standard sense, of degree 5, its support is contained in the plane H defined by x ₁ + x ₂ + x ₃ = 5. Let Q be the convex hull of Supp(G), which is contained in H. It is easy to see that the vertex set of Q is

$$\displaystyle{\{(2,0,3),(0,4,1),(2,3,0),(5,0,0)\}}$$

so Q is a quadrilateral, and Newt(G) consists of a cone over Q. Thus, Newt(G) has 5 faces. Two of these faces lie in coordinate planes, and 3 do not (call these intrusive faces). There are two intrusive edges, namely, E ₁ joining (0, 4, 1) and (2, 0, 3), and E ₂ joining (0, 4, 1) and (2, 3, 0). They define vectors e ₁ = (−2, 4, −2) and e ₂ = (−2, 1, 1).

One can also consider higher-dimensional faces of Newton polytopes for polynomials f annihilated by locally nilpotent derivations, and thereby get further conditions on Newt( f). The last section of [97] gives some results in this direction.

Remark 4.38

Define $f = x^{2} + y^{2} + z^{2} \in \mathbb{C}[x,y,z]$, and $D \in \mathrm{ LND}(\mathbb{C}[x,y,z])$ by:

$$\displaystyle{Dx = -z\,\,,\,\,Dy = -iz\,\,,\,\,Dz = x + iy}$$

Then Df = 0. Likewise, $f \in \mathbb{R}[x,y,z]$, but it is shown in Chap.  9 below that there is no nonzero $\delta \in \mathrm{ LND}(\mathbb{R}[x,y,z])$ with δf = 0. This example points out the limitations of the information provided by Newt( f).