gpt 6 astra ultra attempt on Math-369

Nagata Conjecture on Plane Curve Linear Systems

Definitions and mathematical statement

Choose $r\ge10$ very general points $p_1,\ldots,p_r\in{\mathbb{P}}^2_{{\mathbb{C}}}$, meaning outside a countable union of proper closed subsets of the configuration space. For a plane curve $C$ of positive degree $d$, its multiplicity at $p_i$ is the vanishing order of a local defining equation. The selected Nagata inequality asserts that, whenever $\operatorname{mult}_{p_i}C\ge m_i$ for integers $m_i\ge0$, \[ d>\frac1{\sqrt r}\sum_{i=1}^r m_i. \] The configuration is very general, not every arrangement of distinct points. Unequal multiplicities and reducible curves are included in the statement. The restriction to actual positive-degree curves excludes the empty degree-zero divisor, for which a strict $0>0$ inequality would be meaningless.

Formulation and concept references: [S1].

Short English statement

For sufficiently many very general points in the projective plane, must a curve's degree strictly exceed the stated average multiplicity bound?

Research attempt: the nef boundary, a square specialization, and the strictness gap

Scope and source audit.

This is an attempt at the plane-curve statement above, with arbitrary nonnegative integer multiplicities. It is not an assertion about every configuration, or about infinitely near points. The introduction of the 2025 primary paper [S1], inspected on 27 September 2026, states the strict inequality in this form, recalls Nagata's theorem for square $r$, and describes the remaining cases as open. Its new results concern the valuative version and equivalent conditions; they are not a proof for every nonsquare number of ordinary points. The elementary arguments below recover a weaker square bound and explicit low-degree cases, then locate the obstruction to improving them. No novelty is claimed for these reductions.

Put $S=\sum_i m_i$. The case $S=0$ is immediate because $d>0$. When proving bounds it suffices to use the actual multiplicities $\mu_i=\operatorname{mult}_{p_i}C$: a bound for $\sum\mu_i$ implies the corresponding bound for every prescribed $m_i\le\mu_i$. This distinction matters when a component is removed during a specialization.

1. The exact numerical target on a blowup.

Let $\pi:Y\to\mathbb P^2$ be the blowup of the distinct points, write $H=\pi^*\mathcal O_{\mathbb P^2}(1)$, and denote the exceptional curves by $E_i$. Their intersections are \[ H^2=1,\qquad H E_i=0,\qquad E_iE_j=-\delta_{ij}. \] For the real divisor class \[ B_r=\sqrt r\,H-\sum_iE_i \] we have $B_r^2=0$ and $B_rE_i=1$. If $C$ is an integral plane curve of degree $d$ with actual multiplicities $\mu_i$, its strict transform satisfies \[ B_r\widetilde C=\sqrt r\,d-\sum_i\mu_i. \tag{1} \] Every irreducible curve on $Y$ is either exceptional or such a strict transform. Consequently, $B_r$ is nef if and only if the nonstrict inequality $\sqrt r\,d\ge\sum_i\mu_i$ holds for all integral plane curves. Additivity extends that nonstrict inequality to arbitrary effective curves, including repeated components.

When $r$ is nonsquare and $d>0$, equality in $1$ is impossible: the left term $\sqrt r\,d$ is irrational and the sum of multiplicities is an integer. Thus in the nonsquare cases the desired strict inequality follows from this nefness assertion alone. For square $r=s^2$, however, nefness permits a nonexceptional curve orthogonal to $B_r$. Strictness then contains an extra exclusion. In particular, the calculation $B_r^2=0$ does not prove nefness, and a nef isotropic class need not intersect every curve strictly positively.

For instance, on the blowup of $s^2$ transverse intersection points of two curves of degree $s$, the strict transform of either defining curve has intersection zero with $B_{s^2}$. This is a special configuration, so it does not contradict Nagata's theorem. It is a useful test against an argument that silently upgrades a weak bound on a special fiber to a strict bound on general fibers.

2. Why fixed numerical data may be specialized.

Fix $r,d$ and $(m_1,\ldots,m_r)$. On an affine chart of the configuration space, the conditions that a homogeneous form of degree $d$ vanish to the specified orders are linear equations in its coefficients. Taylor coefficients of order less than $m_i$ at $p_i$ give the equations. Their matrix entries are polynomial functions of the point coordinates after choosing the chart. Hence the existence of a nonzero such form is a closed rank-drop condition. The same assertion is intrinsic, using the evaluation map to the finite jet spaces at the points.

It follows that if one distinct-point configuration admits no such form, configurations admitting a form lie in a proper closed subset. There are only countably many choices of the integer data $d,m_1,\ldots,m_r$. Therefore a separate nonexistence certificate for every forbidden numerical type suffices for a very-general assertion. There is no requirement to find one special configuration simultaneously excluding all types, though it is useful when possible.

The direction is important. Existence on a special configuration proves nothing about existence on a very general one. Nonexistence on a special configuration does propagate to an open set for those fixed numerical data. Ranks may increase when the points become general, so dimensions of the spaces of forms may decrease.

3. An elementary weak square bound.

Here is a complete specialization argument giving \[ d\ge {1\over s}\sum_{i=1}^{s^2}m_i \tag{2} \] for very general $s^2$ points, for every positive integer $s$. Choose a smooth plane curve $A$ of degree $s$ and a degree-$s$ curve $D$ meeting $A$ transversely in exactly $s^2$ distinct points. These intersection points form our special configuration. We claim $2$ for every effective curve at this configuration.

First suppose $C$ has no component $A$. Bézout gives $A\cdot C=sd$. Since $A$ is smooth at each marked point, the local intersection multiplicity of $A$ and $C$ there is at least $\mu_i$. Thus \[ sd\ge\sum_i\mu_i\ge\sum_i m_i. \] If $C$ contains $A$ with coefficient $q\ge1$, write $C=qA+C_0$ where $A$ is not a component of $C_0$. Multiplicity of a product of local defining equations is additive, so \[ \deg C=qs+\deg C_0,\qquad \sum_i\operatorname{mult}_{p_i}C=qs^2+ \sum_i\operatorname{mult}_{p_i}C_0. \] Bézout applied to $C_0$ proves the same inequality. The case $C_0=0$ is allowed and gives equality. Therefore no curve with $sd<S$ exists on this configuration. The rank argument of paragraph 2 proves $2$ for very general points, simultaneously for all integer multiplicities and degrees.

This elementary proof is deliberately weaker than the externally known strict square theorem. Its equality example $C=qA$ is present for every $q$, and prevents it from establishing strictness. A nonstrict argument valid even for $s=3$ is consistent with the fact that nine general points lie on a cubic: $d=3$ and all $m_i=1$ give equality. The restriction $r\ge10$ in the problem is essential for the strict formulation.

4. Bounds between consecutive squares.

Let $s=\lfloor\sqrt r\rfloor$ and $t=\lceil\sqrt r\rceil$. Every subconfiguration of $s^2$ points in a very general $r$-tuple is itself sufficiently general for $2$. For each subset $I\subset\{1,\ldots,r\}$ of cardinality $s^2$, therefore, \[ \sum_{i\in I}m_i\le sd. \] Sum these inequalities over all such subsets. Each $m_i$ occurs $\binom{r-1}{s^2-1}$ times, while the right side occurs $\binom r{s^2}$ times. Division gives \[ S\le {r\over s}\,d,\qquad d\ge {s\over r}S. \tag{3} \] An independent bound comes from enlarging a configuration to $t^2$ points and giving the extra points multiplicity zero. The nonexistence assertion for those numerical data does not depend on the extra coordinates. Since it holds for very general $t^2$-tuples, it holds for very general $r$-tuples. Thus \[ d\ge {1\over t}S. \tag{4} \] Together these prove the explicit coefficient \[ d\ge \max\left\{{s\over r},{1\over t}\right\}S. \tag{5} \] For nonsquare $r$, both displayed coefficients are strictly less than $1/\sqrt r$. For example, $r=10$ gives $d\ge3S/10$, while the target coefficient is $1/\sqrt{10}$. Neither averaging nor adding zero multiplicities closes this numerical interval. Integer rounding can strengthen a fixed instance of $5$, but cannot replace its coefficient uniformly as the multiplicities grow.

The subconfiguration argument uses finitely many subsets for each fixed $r$. The extension argument can also be stated purely in terms of the closed rank-drop locus: a proper locus independent of the last $t^2-r$ coordinates must come from a proper locus downstairs. No probabilistic assertion about independently chosen "very general" coordinates is required.

5. Complete degree-one and degree-two cases.

Very general points have no three collinear and no six on a conic. For completeness, the latter is a proper closed condition on any six chosen points: it is the vanishing of the determinant of their evaluations on the six-dimensional vector space of quadratic forms, and there exist six points with independent evaluations. There are finitely many triples and six-element subsets to exclude for a fixed $r$.

An integral line is smooth and contains at most two marked points, each with multiplicity one. Therefore its total multiplicity is at most two, and \[ 1>{2\over\sqrt r}\qquad(r\ge10). \] An integral conic over $\mathbb C$ is smooth and contains at most five marked points. Thus \[ 2>{5\over\sqrt r}\qquad(r\ge10), \] because $4r>25$. This proves the target for integral curves of degrees one and two. A reducible conic is a sum of two lines, allowing equality of the two components. Add the strict degree-one inequalities, with the coefficient of a repeated line included. This proves the target for every effective positive-degree curve of degree at most two, with every permitted choice of the $m_i$.

More generally, if every integral component of a given effective curve satisfies the target with its actual multiplicities, the curve does too. Indeed, degree and multiplicity are both additive, and at least one positive-degree component occurs. Reducibility is therefore a reduction to integral curves, not a separate source of counterexamples once the integral statement is known.

6. The genus bound is a useful but incomplete obstruction.

For an integral degree-$d$ plane curve, the arithmetic genus is $(d-1)(d-2)/2$. The normalization formula and the standard local bound $\delta_p\ge\mu_p(\mu_p-1)/2$ give \[ \sum_i\mu_i(\mu_i-1)\le(d-1)(d-2). \tag{6} \] One can also obtain the same inequality from the arithmetic genus of the strict transform after blowing up the marked points: it equals the original arithmetic genus minus the indicated sum divided by two, and remains nonnegative for an integral projective curve. Cauchy--Schwarz therefore implies, for $S=\sum_i\mu_i$, \[ {S^2\over r}-S\le d^2-3d+2. \tag{7} \] This is necessary for an integral curve, but is not the desired bound. In the equal-multiplicity case it reads \[ r m(m-1)\le(d-1)(d-2). \tag{8} \] For example, the numerical data \[ r=10,\qquad m=6,\qquad d=19 \] satisfy $10\cdot6\cdot5=300\le18\cdot17=306$, while they are not a violating type because $19^2=361>10\cdot6^2=360$. This near miss illustrates why decimal comparisons must be replaced by exact squares. An actual numerical type not excluded by the genus test and lying on the forbidden side is \[ r=10,\qquad m=19,\qquad d=60: \quad 3420\le3422,\qquad 3600<3610. \tag{9} \] These numbers do not assert existence of a curve. They show that $6$, even with integrality, is logically insufficient to rule out all violations. In fact the raw number of imposed multiplicity conditions in $9$ is $10\binom{20}{2}=1900$, while the space of degree-60 forms has dimension $\binom{62}{2}=1891$. Independence would give nonexistence, but that independence is the very issue being studied.

The saved exact diagnostic verifies $3$--$5$, the square tests in $9$, and jet-evaluation ranks for the low-degree general-position claims. No finite search over degrees is used as a substitute for the unbounded statement.

7. Why the obvious dimension count is not a proof.

At one point, multiplicity at least $m$ imposes at most $\binom{m+1}{2}$ linear conditions. Thus \[ h^0\left(\mathbb P^2,\mathcal I_{\sum m_ip_i}(d)\right) \ge \binom{d+2}{2}-\sum_i\binom{m_i+1}{2}. \tag{10} \] The right side is a lower bound, not an upper bound. If it is positive, existence follows; if it is negative, no conclusion follows. To turn negativity into nonexistence would require maximal rank of the jet evaluation matrix for those data. General points do not by definition make every conceivable interpolation problem independent.

For a basic diagnostic, five general points lie on a conic $Q$. The quartic $Q^2$ has multiplicity two at all five points. Here the degree-four space has dimension $15$, and the five double-point conditions number $5\cdot3=15$, so the naive expected vector-space dimension is zero while a nonzero form exists. This example has $r<10$ and is not a counterexample to the target. It refutes the blanket linear-algebra assertion that point generality alone guarantees independence in every degree and multiplicity.

Nor does dividing a repeated component by its multiplicity prove the desired uniform statement. Special linear systems can contain forced curves, and recognizing all possible fixed components is itself difficult on blowups at many general points. The elementary Bézout proof succeeded only because it supplied a specific divisor against which every curve could be tested.

8. The first missing step and the outcome.

For nonsquare $r$, a sufficient missing statement is: the class $B_r$ in paragraph 1 is nef at a very general configuration. Equivalently, every fixed integer type with $rd^2<S^2$ has a jet-evaluation matrix of full column rank on some configuration. The arguments above establish this only for the weaker coefficients in $5$, for low degrees, and for assorted individual numerical types. They do not provide the needed certificate for all forbidden types.

For square $r$, the weak specialization is complete, but excluding the equality types requires more than this argument; the known strict square theorem is attributed to Nagata rather than claimed as a consequence of our calculation. Moving the complete-intersection configuration off its special locus may destroy the equality curves, but saying that it does so for all powers and all unequal multiplicities is an additional theorem, not semicontinuity alone.

Outcome: UNRESOLVED. The attempt proves a weak square bound, explicit interpolating bounds for every number of points, complete degree-one and degree-two cases, and a precise nefness reduction for nonsquare $r$. The genus and interpolation stress tests identify the exact failure of two tempting general proofs. None of these steps settles the remaining nonsquare cases of the selected conjecture.

Sources

• S1 Carlos Galindo et al., On the valuative Nagata conjecture, Research in the Mathematical Sciences 12, 18 $2025$. https://link.springer.com/article/10.1007/s40687-025-00500-2. Formulation source.