gpt 6 astra ultra attempt on Math-367

Fujita Freeness Conjecture

Definitions and mathematical statement

Let $X$ be a smooth complex projective variety of dimension $n$, $K_X=\bigwedge^n\Omega_X^1$ its canonical line bundle, and $L$ an ample line bundle. Ampleness means that some positive tensor power defines a projective embedding. A line bundle $A$ is basepoint-free if the evaluation map $H^0(X,A)\otimes\mathcal O_X\to A$ is surjective, equivalently every point has a global section nonvanishing there. Fujita freeness asserts \[ K_X\otimes L^{\otimes(n+1)}\text{ is basepoint-free}. \] In additive divisor notation this is $K_X+(n+1)L$. The assertion requires no extra positivity such as very ampleness of $L$, and it does not claim that this adjoint bundle itself gives an embedding; the very-ampleness target is listed separately.

Formulation and concept references: [S1].

Short English statement

Does adding $(n+1)$ copies of any ample divisor to the canonical divisor of a smooth projective $n$-fold always remove every base point?

Research attempt: point evaluation, a Koszul proof, and the cost of blowing up

Scope and current-source check.

All varieties here are smooth and projective over $\mathbb C$, unless explicitly stated otherwise. The target is generation at each point by $K_X+(n+1)L$, not separation of tangent vectors. The primary formulation [S1], version 2 dated 23 March 2026, was inspected on 27 September 2026. Its introduction records the characteristic-zero freeness theorem through dimension five and the case of ample globally generated $L$. Its new results treat a different adjoint variant; positive-characteristic counterexamples discussed there do not refute the present complex statement.

A newer primary preprint [E1], version 2 dated 11 September 2026, reports the bound $K_X+mL$ for all $m\ge\lceil C_0n\rceil$, with $C_0=1.77629\ldots$, and hence the bound $2n$. This source check inspected its abstract and version record, not the complete proof. The reported coefficient is asymptotically larger than the conjectured one. It is relevant current progress, not a resolution at $n+1$ in arbitrary dimension. The arguments below are independent elementary reductions using classical vanishing and intersection theory; known subcases are not claimed as new.

1. The evaluation problem and an exact sufficient vanishing.

Set $M=K_X+(n+1)L$. For a closed point $p\in X$, tensor the ideal sequence by $M$: \[ 0\longrightarrow M\otimes\mathcal I_p\longrightarrow M \longrightarrow M|_p\longrightarrow0. \] The relevant portion of the cohomology sequence is \[ H^0(X,M)\longrightarrow M|_p\longrightarrow H^1(X,M\otimes\mathcal I_p). \tag{1} \] Thus $H^1(X,M\otimes\mathcal I_p)=0$ is sufficient for generation at $p$. It is stronger than needed: evaluation may be onto even if this cohomology group is nonzero, because the connecting map can vanish. We will use the sufficient condition only where it is actually established.

Kodaira vanishing gives $H^i(X,K_X+aL)=0$ for $i>0$ and $a>0$. The ideal $\mathcal I_p$ in $1$ is not an ample line bundle. Deleting it and applying Kodaira vanishing to $M$ does not prove $1$: the two cohomology groups are different. Resolving that ideal while preserving enough positivity is the main issue in the methods below.

2. The complete curve case, including small genera.

Let $n=1$, let $C$ have genus $g$, and write $d=\deg L\ge1$. For every point $p$, \[ H^1(C,K_C+2L-p)^{\vee}=H^0(C,-2L+p)=0, \tag{2} \] because $\deg(-2L+p)=1-2d<0$. Equations $1$--$2$ show that $K_C+2L$ is globally generated. This proof works for $g=0$ and $g=1$ as well as for higher genus; it uses Serre duality and negativity of the displayed degree, not an implicit assumption that $K_C$ is positive.

There is a sharpness test already on $\mathbb P^1$ with $L=\mathcal O(1)$: $K_C+L=\mathcal O(-1)$ has no nonzero global section, whereas $K_C+2L=\mathcal O$ is generated. On an elliptic curve, a line bundle of degree one has exactly one independent section, whose divisor is a single point. It is therefore ample but not globally generated. Its square has degree two and is globally generated by the same duality calculation. This example will prevent us from identifying ampleness with generation or from descending generation through tensor roots.

3. A complete higher-dimensional subcase: $L$ globally generated.

Assume now that $L$ is both ample and globally generated. We prove the conjectured conclusion for this case in all dimensions. The argument deliberately allows the morphism defined by $L$ to be ramified; it does not ask for a smooth member of $|L|$ through every point.

Let $\varphi:X\to\mathbb P^N$ be the morphism defined by the complete linear system of $L$. It is finite onto its image. Indeed, a positive-dimensional projective fiber would contain an integral curve $C$. Since $L=\varphi^*\mathcal O(1)$, that curve would have $L\cdot C=0$, contradicting ampleness. Properness plus finite fibers makes the morphism finite.

Fix $p\in X$ and put $q=\varphi(p)$. Choose $n$ hyperplanes through $q$ whose intersection with the $n$-dimensional image $\varphi(X)$ is zero-dimensional. Such choices can be made inductively: the linear system of hyperplanes through $q$ has base locus exactly $q$ in projective space, so a general hyperplane does not contain any positive-dimensional irreducible component of the previous intersection. Intersecting lowers each positive dimension by one; after $n$ steps only finitely many points remain.

Pull back their equations to sections $s_1,\ldots,s_n\in H^0(X,L)$. Their common zero scheme $Z$ is zero-dimensional, is nonempty, and contains $p$. Since $X$ is smooth and hence Cohen--Macaulay, these $n$ equations of codimension $n$ form a regular sequence locally wherever they all vanish. Away from their common zero locus one is a unit. Consequently the Koszul complex is exact globally, giving a resolution \[ 0\longrightarrow L^{-n}\longrightarrow (L^{-(n-1)})^{\binom n{n-1}}\longrightarrow\cdots \longrightarrow (L^{-1})^{\oplus n} \longrightarrow\mathcal I_Z\longrightarrow0. \tag{3} \] No reducedness or smoothness of $Z$ is asserted or needed.

Tensor $3$ by $M$ and write \[ F_j=(K_X+(n+1-j)L)^{\oplus\binom nj},\quad1\le j\le n. \] Each $n+1-j$ is a positive integer. Kodaira vanishing therefore gives \[ H^i(X,F_j)=0\quad\text{for every }i>0 \text{ and }1\le j\le n. \tag{4} \] In particular, the long exact sequences for the successive kernels in $3$ imply $H^1(X,M\otimes\mathcal I_Z)=0$. To make the dimension shift explicit, if $n\ge2$ put $Q_1=\ker(F_1\to M\otimes\mathcal I_Z)$ and continue with $0\to Q_j\to F_j\to Q_{j-1}\to0$. Then $H^1(M\otimes\mathcal I_Z)$ injects into $H^2(Q_1)$, which injects into $H^3(Q_2)$, and so on, ending in $H^n(F_n)=0$. For $n=1$, the ideal is simply $L^{-1}$ and the result is the first vanishing in $4$.

It follows that \[ H^0(X,M)\longrightarrow H^0(Z,M|_Z) \] is surjective. Because $Z$ is a finite scheme, its global sections surject onto the fiber at $p$: on the Artinian local factor at $p$, a generator of the invertible module maps to a nonzero element of that fiber. Composing yields generation of $M$ at $p$. Since $p$ was arbitrary, $M$ is globally generated.

Thus the target is proved when the polarization is already globally generated, with no very-ampleness assumption. The proof is also a direct explanation of the usual regularity argument in this complex setting: the crucial positive exponents in $4$ range from $n$ down to $1$.

4. Why smooth hyperplane induction needs an additional check.

A tempting alternative is to choose a smooth divisor $D\in|L|$ through $p$, use adjunction on $D$, and lift sections. For a very ample $L$ such an induction is available by Bertini with a point condition. For a merely globally generated $L$, its associated finite morphism can be ramified. At a point where its differential vanishes, every hyperplane section through the image can have singular pullback at $p$. Global generation alone does not supply the smooth divisor required by that proof.

The Koszul argument avoids that assumption by using a zero-dimensional complete intersection, possibly nonreduced. As a local model, in a smooth $n$-dimensional affine chart the functions \[ z_1^{a_1},\ldots,z_n^{a_n},\qquad a_i\ge1, \] form a regular sequence with quotient of length $\prod_i a_i$. If some $a_i>1$, the resulting zero scheme is nonreduced, yet the Koszul resolution remains exact and its coordinate ring still maps onto the residue field at the origin. The saved finite algebra check computes these quotient lengths in small examples; smoothness of the intersection is not part of the generation argument.

What the argument does require is the availability of sections of $L$ through the selected point and their ability to cut to a finite scheme. A general ample $L$ can have a base locus or even have no sections. Ampleness says that a sufficiently large power has enough sections; it gives no permission to apply $3$ with $L$ itself.

5. What replacing $L$ by a power really gives.

Choose $a>0$ with $L^a$ globally generated. Applying paragraph 3 to $A=L^a$ proves \[ K_X+(n+1)aL\quad\text{is globally generated}. \tag{5} \] This is an actual conclusion, but its coefficient depends on the power chosen for $L$. There is no operation that divides a generated line bundle by $a$ and preserves generation. On an elliptic curve the degree-one example of paragraph 2 has a generated square and a base point itself. Even numerically, $5$ is not the $a$th tensor power of the desired bundle: \[ a\left(K_X+(n+1)L\right) =aK_X+(n+1)aL, \] which has a different canonical coefficient. Neither treating $5$ as a root nor ignoring this coefficient is legitimate.

The same issue explains why choosing a very ample multiple of $L$ does not settle Fujita's bound. One would need a mechanism to return to the fixed exponent $n+1$ uniformly over all ample polarizations, not just a qualitative eventual-generation theorem.

6. A local positivity criterion from the blowup of a point.

Let $n\ge2$, and blow up $p$: \[ \pi:Y\longrightarrow X,\qquad E\simeq\mathbb P^{n-1}. \] The canonical bundle formula is $K_Y=\pi^*K_X+(n-1)E$. Consequently \[ \pi^*M-E =K_Y+\left((n+1)\pi^*L-nE\right). \tag{6} \] Define $A_p=(n+1)\pi^*L-nE$. If $A_p$ is ample, Kodaira vanishing on $Y$ yields $H^1(Y,\pi^*M-E)=0$. The standard blowup identities \[ \pi_*\mathcal O_Y(-E)=\mathcal I_p, \qquad R^j\pi_*\mathcal O_Y(-E)=0\ (j>0) \] then identify this group with $H^1(X,M\otimes\mathcal I_p)$. Projection formula and Leray justify the identification. One way to check the higher-direct-image assertion is formal functions: successive restrictions near the exceptional divisor involve the sheaves $\mathcal O_{\mathbb P^{n-1}}(k)$ with $k\ge1$, whose higher cohomology vanishes. Thus this route does establish generation at $p$ whenever $A_p$ is ample.

In terms of the Seshadri constant \[ \varepsilon(L;p)=\sup\{t\ge0:\pi^*L-tE\text{ is nef}\}, \] a sufficient condition is \[ \varepsilon(L;p)>{n\over n+1}. \tag{7} \] For an ample $L$, the classes $\pi^*L-tE$ are ample for $0<t<\varepsilon(L;p)$. Indeed they are ample for sufficiently small positive $t$; convex combination of one such ample class and a nef class at a larger parameter proves ampleness at each smaller interior parameter. Taking $t=n/(n+1)$ gives the claimed implication.

The inequality in $7$ is strict in this simple Kodaira-vanishing proof. At an endpoint one needs additional nef-and-big analysis or another method; ampleness of $L$ alone does not say that $7$ holds at every point. Establishing the needed uniform local positivity would itself be a substantive theorem, not a consequence of the definition of an ample line bundle.

7. Checking the exceptional coefficient and the sharp model.

The coefficient of $E$ in $6$ is $n$, rather than $1$ or $n-1$, because the ideal of the point contributes one copy of $-E$ and the canonical blowup discrepancy contributes another $n-1$. This is the same positivity cost that disappears incorrectly in a naive direct use of Kodaira vanishing.

For a line $\ell\subset E$, one has $\pi^*L\cdot\ell=0$ and $E\cdot\ell=-1$, so $A_p\cdot\ell=n>0$. On the strict transform of a curve $C$ through $p$, \[ A_p\cdot\widetilde C=(n+1)L\cdot C -n\operatorname{mult}_pC. \tag{8} \] Thus the difficult curves are those through the center, not the exceptional lines themselves. Formula $8$ also explains the threshold in $7$.

On $X=\mathbb P^n$ with $L=\mathcal O(1)$, the canonical bundle is $\mathcal O(-n-1)$, so \[ K_X+nL=\mathcal O(-1),\qquad K_X+(n+1)L=\mathcal O. \tag{9} \] The first is not generated and the second is generated. Hence the universal coefficient cannot be lowered below $n+1$. For $n\ge1$, the generated bundle in $9$ defines a constant morphism and is not very ample. This verifies that the current target and the separate very-ampleness target must not be conflated. The same model has $\varepsilon(L;p)=1$: a line through $p$ gives the upper bound, and hyperplanes through $p$ show that $\pi^*L-E$ is nef, giving the lower bound. Therefore $7$ holds in the sharp model.

8. Another explicit family: products with product polarizations.

Let $X=C_1\times\cdots\times C_n$ be a product of smooth projective curves, and let $L=\bigotimes_i\operatorname{pr}_i^*L_i$ with $\deg L_i=d_i\ge1$. This is ample. We have \[ K_X+(n+1)L\simeq \bigotimes_i\operatorname{pr}_i^*(K_{C_i}+(n+1)L_i). \] Each factor on the right is generated. Indeed for every $p_i\in C_i$, Serre duality reduces the required $H^1$ vanishing to absence of sections of a bundle of degree $1-(n+1)d_i<0$. Given a point of $X$, choose in each factor a section nonzero at its coordinate; their tensor product is nonzero at the chosen point of $X$. This proves global generation directly.

Here $L_i$ need not be generated: a degree-one polarization on an elliptic factor is permitted. Thus this family goes beyond the hypothesis of paragraph 3 in a genuine way. It does not include arbitrary ample line bundles on products, which can contain mixed factors in the Picard group. The factorization of the polarization is an explicit assumption, not a conclusion from the geometry of the underlying product.

9. The exact remaining obstacle.

For an arbitrary ample $L$, the first route lacks a finite complete intersection cut out by sections of $L$ through each point. Passing to a power loses the required coefficient. The blowup route instead needs enough positivity of $A_p$, which is not supplied uniformly by the hypotheses. More refined methods using singular metrics or multiplier ideals can address weaker bounds and many subcases, but one would have to prove their quantitative estimates at exactly $n+1$; naming the method does not provide those estimates.

The exact checks saved with this attempt concern the Koszul local models, the exponent range in $4$, the discrepancy identity $6$, and the sharp projective-space calculation $9$. These are checks on the reductions, not numerical evidence for every smooth variety. The known dimension-five result and the new reported linear bound remain attributed external results, not consequences of these checks.

Outcome: UNRESOLVED. This attempt proves the curve case, gives a full Koszul proof for ample globally generated polarizations, proves a family with nongenerated product polarizations, and isolates a sufficient local positivity criterion. It neither establishes the universal local estimates nor constructs the sections required for arbitrary ample $L$ in all dimensions at the sharp coefficient $n+1$.

Sources

• S1 Takumi Murayama, Moving Seshadri constants and effective Fujita-type conjectures, arXiv:2408.06530v2, Conjecture 1.1.1(i). https://arxiv.org/abs/2408.06530v2. Formulation source.
• E1 Jingjun Han, A linear bound for Fujita's freeness conjecture, arXiv:2609.01574v2, 11 September 2026. https://arxiv.org/abs/2609.01574v2. Primary abstract and version record inspected for the reported linear bound; full proof not audited here.