Is the smooth rational quartic curve in $\mathbb{P}^3$ a set-theoretic complete intersection?

The smooth rational quartic curve in projective three-space — canonically the monomial (Cremona) quartic parametrized by $(\lambda:\mu) \mapsto (\lambda^4:\lambda^3\mu:\lambda\mu^3:\mu^4)$ — is the standard open test case for Hartshorne's 1962 conjecture that every irreducible curve in $\mathbb{P}^3$ is a set-theoretic complete intersection, i.e., the common zero locus of two surfaces. Over an algebraically closed field of characteristic zero it is not known whether two polynomial equations can cut this curve out even set-theoretically, although partial results exclude surface pairs of degrees $(3,n)$ for every $n$ (for every smooth rational quartic) and $(4,4)$ (for the monomial quartic), and degree bounds reduce the question to a finite list of remaining degree pairs. In positive characteristic the answer for the monomial quartic is affirmative (Hartshorne 1979), making the problem genuinely characteristic-dependent.

Background

In projective algebraic geometry over an algebraically closed field, a subvariety $X \subset \mathbb{P}^n$ is a complete intersection when its homogeneous ideal $I(X)$ — the ideal of all polynomials vanishing on it — is generated by $\operatorname{codim}(X)$ homogeneous polynomials, so that $X$ is the scheme-theoretic intersection of that many hypersurfaces. A weaker, purely set-theoretic property asks only that the point set of $X$ be cut out by the right number of hypersurfaces: $X$ is a set-theoretic complete intersection (s.t.c.i.) when $\sqrt{(f_1,\dots,f_c)} = I(X)$ for some $c = \operatorname{codim}(X)$ homogeneous polynomials, where $\sqrt{(\,)}$ denotes the radical of an ideal. The minimum number of equations needed to generate an ideal up to radical is its arithmetical rank, so $X$ is an s.t.c.i. exactly when the arithmetical rank of $I(X)$ equals its height. Since passing to the radical discards all non-reduced structure (multiplicities, embedded components), set-theoretic generation is far more flexible than honest generation, and deciding it is notoriously difficult.

The question is classical: in the nineteenth century Kronecker and Cayley asked whether every variety in complex $\mathbb{P}^3$ is a set-theoretic complete intersection. Hartshorne (1962) proved this false for reducible varieties — in $\mathbb{C}^4$ the union of two 2-dimensional planes with one point in common, or in $\mathbb{P}^3$ the variety with defining ideal $\langle x_1,x_2\rangle \cap \langle x_3,x_4\rangle$, gives a connected variety that is not an s.t.c.i., the obstruction being the nonvanishing of a local cohomology module, a standard commutative-algebra tool that measures the failure of set-theoretic generation by few equations — and this led to what is now called Hartshorne's question, often stated as a conjecture: every irreducible curve in $\mathbb{P}^3$ is a set-theoretic complete intersection. The conjecture, still open, is one of the classical problems of projective algebraic geometry and commutative algebra; it is known for various classes of curves (for example, many projective monomial curves — those parametrized by monomials — are s.t.c.i. by work of Robbiano and Valla and their successors, and in a different, affine setting all local complete intersection curves in three-dimensional spectra were recently shown to be s.t.c.i.), but the known methods are piecemeal.

The standard open test case is the smooth rational quartic curve. Project the rational normal quartic — the image of $\mathbb{P}^1$ in $\mathbb{P}^4$ under all monomials of degree 4, the roundest embedding of the line — away from the $\lambda^2\mu^2$ coordinate; the image is the monomial or Cremona quartic $C$, parametrized by $(\lambda:\mu) \mapsto (\lambda^4:\lambda^3\mu:\lambda\mu^3:\mu^4)$. It is nonsingular, isomorphic to $\mathbb{P}^1$ (hence rational), has degree 4, and lies on the unique quadric surface $Q = \{xw = yz\}$ containing it — every smooth rational quartic in $\mathbb{P}^3$ lies on exactly one quadric, on which it has bidegree type $(1,3)$, meeting each line of one ruling (one of the two families of lines covering a quadric surface) in one point and each line of the other ruling in three. Its homogeneous ideal $I(C)$, the kernel of the map $k[x,y,z,w] \to k[\lambda,\mu]$ sending the variables to the four parametrizing monomials, is a toric ideal (an ideal generated by binomials): it has height 2 but requires more than two generators, so $C$ is not a scheme-theoretic complete intersection, and the question is only whether two equations suffice up to radical. The curve is the subject of Exercise 18.8 in Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, the reference customarily cited for the problem.

Substantial partial results exist, all in characteristic 0 and all excluding only special families of surface degrees. A series of papers from the Italian school culminating in 1989 (whose lineage is summarized in the Craighero–Gattazzo papers cited below) proved that no smooth rational quartic in $\mathbb{P}^3$ is the set-theoretic intersection of a cubic surface and a quartic surface (Stagnaro 1983), that the Cremona quartic is not the set-theoretic intersection of two quartic surfaces in characteristic different from 2 and 3 (Craighero–Gattazzo 1986), and — strongest — that no smooth rational quartic is the set-theoretic intersection of a cubic surface and a surface of any degree $n$ in characteristic 0 (Craighero–Gattazzo 1989). Jaffe (1995) converted the question into a finite problem: by iterated blowing up of the curve he bounded the degrees of two surfaces that cut out a smooth curve set-theoretically (assuming the surfaces have no common singular point and the curve is not a complete intersection), and for a rational quartic he reduced the possibilities to fifteen pairs of surface degrees, $(3,4), (3,8), (4,4), (4,7), (6,26), (9,48), (10,28), (12,18), (13,16), (17,220), (18,118), (19,84), (20,67), (22,50), (28,33)$ — of which he writes that 'very little is known about which of these degree pairs actually correspond to surface pairs'. His complementary linear-normality results force a curve cut out in this way to be linearly normal — embedded by a complete linear system, which a rational quartic in $\mathbb{P}^3$, being a projection of the rational normal quartic, is not — but only under additional hypotheses on the surfaces: both surfaces having only rational singularities, one surface being normal of degree below the curve's degree, or one surface being a quartic with only rational singularities. Configurations satisfying these hypotheses are thereby excluded; the case his analysis explicitly leaves open is two normal quartic surfaces, of which he writes 'It is not known if this is possible.' Since the Italian school excluded the $(3,n)$ family for every smooth rational quartic in characteristic 0 and the $(4,4)$ pair for the Cremona quartic, the characteristic-zero question survives, for the general statement over all smooth rational quartics, within Jaffe's no-common-singular-point framework, on the finite list $(4,7), (6,26), (9,48), \dots, (28,33)$ for surface pairs failing his singularity hypotheses (for instance pairs involving a non-normal surface), together with the two-normal-quartics case — and, beyond the framework altogether, on pairs of surfaces sharing singular points.

The problem is genuinely characteristic-dependent, which is why it cannot be settled by pure combinatorics of the ideal. Hartshorne (1979) proved that in positive characteristic the monomial rational quartic is a set-theoretic complete intersection — and Jaffe additionally cites one-page handwritten notes of Samuel from July 1964 as an independent record of the result: in characteristic 2 it is cut out set-theoretically by two explicit quartic surfaces, and for general $p$ the construction rests on Frobenius ($p$-th power) identities. The Frobenius map — raising to the $p$-th power, which is a ring homomorphism only in characteristic $p$ — has no characteristic-zero analogue, so the known positive technique fails exactly where the problem is open. The characteristic-zero question asked here is the one that Hartshorne's curve conjecture concerns.

Problem Statement

Let $k$ be an algebraically closed field of characteristic 0 and let $C \subset \mathbb{P}^3_k$ be a smooth rational quartic curve, with the monomial (Cremona) quartic — the image of $(\lambda:\mu) \mapsto (\lambda^4:\lambda^3\mu:\lambda\mu^3:\mu^4)$, the curve of Eisenbud's Exercise 18.8 — as the canonical instance. Determine whether $C$ is a set-theoretic complete intersection: do there exist two homogeneous polynomials $f, g \in k[x,y,z,w]$ with $\sqrt{(f,g)} = I(C)$, i.e., two surfaces of projective three-space whose common zero locus is exactly the curve $C$? Equivalently, is the arithmetical rank of the toric ideal $I(C)$, the kernel of $k[x,y,z,w] \to k[\lambda,\mu]$ sending $x \mapsto \lambda^4$, $y \mapsto \lambda^3\mu$, $z \mapsto \lambda\mu^3$, $w \mapsto \mu^4$, equal to its height 2? The question is asked both for the canonical monomial quartic and, more generally, for every smooth rational quartic curve in $\mathbb{P}^3_k$ in characteristic 0. The canonical monomial quartic is the weaker instance; the statement for every smooth rational quartic is the stronger form and implies the canonical case. A negative answer for any smooth rational quartic would refute Hartshorne's 1962 conjecture that every irreducible curve in $\mathbb{P}^3$ is a set-theoretic complete intersection; a positive answer would settle that conjecture's most famous open test case. The characteristic restriction is essential and part of the problem's content: in positive characteristic the monomial quartic is already known to be a set-theoretic complete intersection (Hartshorne 1979), so the open problem is specifically the characteristic-0 case.

References

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