Maximal multipoint Seshadri constants in positive characteristic

Proof. Choose the smallest exponent occurring as \nu(f) for a nonzero member of W, and choose a member whose coefficient there is 1. The kernel of that coefficient functional has codimension one. Repeating in this kernel gives a basis with distinct least exponents. Every nonzero linear combination has as its least exponent the smallest least exponent among the basis vectors with nonzero coefficients. This proves the first assertion; the second follows label by label and from (2.1).□

Proof. In the lexicographic case set \lambda_h=1, and successively choose

If an exponent is lexicographically greater than e_j, its gain at the first differing coordinate outweighs every possible loss in the later coordinates.

For (ii), the indicated order is a monomial order because every bounded weight range contains only finitely many nonnegative integral exponents. Choose a real bound larger than the weights of all the e_j. Among the finitely many exponents below that bound, replacing t by a sufficiently close rational t'>t preserves all strict comparisons with the e_j and resolves equal-weight comparisons by increasing \beta. Exponents above the bound cannot interfere: their weights only increase, while the new weights of the e_j stay below the bound. Clearing the denominator of (1,t') gives the required positive integral weights.□

Proof. Let (Q_i,U_i), 1\leq i\leq r, be a tuple at which the monomials separate the stated jets. Shrink the neighborhood of y to an affine one, denoted Y^\circ, on which all the functions are defined. Put

and write \delta=(\delta',\delta'') according to the two factors. For each i, substitute

The first substitution is defined on the completed local ring, since B is s-adically complete and every \lambda_a is positive. The second is defined on Laurent polynomials, since each U_{i,b} is nonzero. In particular, these substitutions define ring maps from the coordinate ring of Y^\circ\times\T^{d-h} to B.

Use the classes of \delta^\alpha, |\alpha|\leq m, as an R-basis of B. The images of f_ju^{\gamma_j} under all r substitutions give the jth column of a matrix with r\binom{d+m}{d} rows. Divide this column by c_js^{\lambda\cdot e_j}. These normalized columns still belong to B^{\oplus r}, and their reductions modulo s are precisely the jets of Z^{e_j}u^{\gamma_j} at (Q_i,U_i) in translated coordinates. The assumed surjectivity supplies a maximal minor nonzero modulo s. The original columns therefore span B[1/s]^{\oplus r} over F.

We now interpret this rank statement as jet separation on a variety. Setting \delta=0 in (2.2) gives actual F-points P_i of Y^\circ\times\T^{d-h}. After inverting s, the corresponding ring map factors through the order-m jet algebra of (Y^\circ\times\T^{d-h})_F at P_i: elements nonvanishing at P_i map to units, and its maximal ideal maps into (\delta). Both this jet algebra and B[1/s] have dimension \binom{d+m}{d} over F. Surjectivity onto the direct sum of the latter spaces forces each induced map from a jet algebra to be an isomorphism. It follows that the original functions separate the order-m jets at the P_i themselves. These points are distinct, since two repeated centers would make two target blocks factor through one jet algebra, contradicting surjectivity.

Thus the open jet-separation locus has an F-point and is nonempty. The configuration space is irreducible, so this locus meets the configuration space of any prescribed dense open subset. A nonempty open subset of a variety over the algebraically closed field k has a k-point. This proves the assertion. A local frame reduces the line-bundle version to the one just proved.□

Proof. An invertible integral linear map preserves cardinalities and addition. Its action on characters is induced by a torus automorphism, which identifies the corresponding jet evaluations.□

Proof. Expand the row spaces in the parameter z-1. Lemma 2.2 proves the cardinality and gradedness assertions, because multiplication adds both degrees and labels. Lemma 2.4, with Y=\mathbb G_m, y=1, and weight 1, proves backward jet transfer.

For a nonempty row let a_- and a_+ be its smallest and largest exponents. Multiplication by z^{-a_-} preserves orders at 1 and turns each member into a polynomial of degree at most a_+-a_-\leq b. Its order of vanishing is at most its degree, proving the bound.□

Proof. Bertini’s theorem for hyperplane sections, valid in every characteristic for a projective embedding over an algebraically closed field, gives sections \sigma_1,\ldots,\sigma_{n-1}\in H^0(X,DL) whose successive zero loci form a smooth integral flag

See [?, Tags 0FD6 and 0G4G] for the smoothness and integrality assertions. At each step the variety being cut has dimension at least two. Choose x\in C_0(k) and a local frame e of L near x. The functions z_i=\sigma_i/e^D, i<n, form part of a regular system of parameters at x; complete them by z_n. These parameters identify \widehat{\mathcal O}_{X,x} with k[[z_1,\ldots,z_n]].

Expand every section of NL using the frame e^N, and let S_N be the set of lexicographically least exponents of the nonzero expansions. The expansion map is injective: a section with zero germ is zero by integrality, and the local ring injects into its completion. Lemma 2.2 therefore gives \#S_N=h^0(X,NL) and the graded property. The Hilbert polynomial of the ample line bundle gives the last equality in (3.2). A basis with distinct least exponents, Lemma 2.3(i), and Lemma 2.4 with h=n prove the jet-transfer assertion.

It remains to prove the simplex bound. Let \alpha=(\alpha_1,\ldots,\alpha_n)\in S_N, coming from a nonzero section s. We successively divide by the equations of the flag and restrict to the next member. The details matter because the lexicographic order first measures vanishing along an entire divisor, not just at the chosen point.

Suppose we have reached a nonzero section on X_{i-1} with least exponent (\alpha_i,\ldots,\alpha_n) in its formal expansion at x. If \alpha_i>0, restriction of this section to X_i has zero formal expansion. Injectivity into the completed local ring, followed by integrality of X_i, shows that this restriction is identically zero. The exact sequence for the effective Cartier divisor X_i\subset X_{i-1} therefore allows division by \sigma_i|_{X_{i-1}}. Repeating this argument \alpha_i times, and then restricting to X_i, gives a nonzero section whose least exponent is (\alpha_{i+1},\ldots,\alpha_n). Nonvanishing follows from the nonzero coefficient of the original least term. Each division subtracts DL from the line bundle.

After these steps, the resulting nonzero section on C_0 belongs to

and has order exactly \alpha_n at x. Since L\cdot C_0=D^{n-1}V, comparison with the degree of its zero divisor gives

Equivalently, D\sum_{i<n}\alpha_i+\alpha_n/(D^{n-1}V)\leq N, which is (3.1).□

Proof. Write g=y^2+xy-x^3. The parametrization

identifies the complement of x=0 in g=0 with \operatorname{Spec}k[v,1/(v(v+1))]: the inverse is v=y/x. Consequently, a polynomial vanishes under (4.2) if and only if it is divisible by g. Indeed, vanishing implies membership in (g) after inverting x, and g is relatively prime to x in the unique factorization domain k[x,y].

The polynomial Z^2+Z-x has simple roots 0 and -1 modulo x. Hensel lifting gives roots r_1,r_2\in k[[x]] with these respective constant terms. Set

Their linear terms are y and y+x, so they are formal coordinates, and \xi\eta=g. These assertions also hold in characteristic two: the roots 0 and -1=1 remain distinct and the linear terms remain independent. Along (4.2) near v=0, uniqueness of the lifted root gives r_1(x(v))=v, so \eta=0 and \xi has order one in v. Near v=-1, one similarly has \xi=0 and \eta has order one in v+1.

Factor f=g^e f_0 with e\ge0 and g\nmid f_0. Give x,y the positive weights 1/A,1/B, and write \deg_{A,B} for the maximum weight of a polynomial’s monomials. This degree is additive on products, because their highest weight forms multiply nontrivially. Moreover,

here 1/A+1/B\le\max\{3/A,2/B\}=d. Thus ed\le\kappa. The nonzero polynomial

has degree at most W(\kappa-ed), since

for every monomial x^iy^j of f_0. Its expansions at v=0 and v=-1 are both nonzero, since k[v] injects into each completed local ring. These expansions are exactly the restrictions of f_0 to the two formal branches described above.

Let \nu_t(f_0)=(\alpha,\beta) and \mu=\alpha+t\beta. Every term of its formal expansion has weight at least \mu. Restricting to \eta=0 therefore gives order at least \mu on the branch at v=0; restricting to \xi=0 gives order at least \mu/t on the branch at v=-1. The nonzero polynomial F_0 consequently satisfies

Equivalently, (\alpha,\beta)\in(\kappa-ed)\Delta(a,b). Because g=\xi\eta, multiplicativity of least exponents gives

This also covers \kappa=0, when f is constant.□

Proof. Put R=H/w and a_0=WT/(1+T). As t ranges over (T,\infty), the value a=Wt/(1+t) ranges over (a_0,W). For u=(i,j), the heights of (-b,0),(a,0),(0,c) are jb,-ja,i/d. We seek j>0 and i such that

The first equality turns the second inequality into R-jW>0. Thus we first choose jW\le R/2, and then use the freedom in a to obtain the equality with i coprime to j.

Take j to be the largest power of two at most

Then j>1/(4dw)>4. The open interval

has length

so it contains an odd integer i. Since j is a power of two, (i,j) is primitive. There is a unique t>T such that

The resulting heights satisfy

This gives the required height range and distinctness.□

Proof of Proposition 4.1. Fix t and u from Lemma 4.3; these choices will be used in every degree. Keep coordinates 3,\ldots,n unchanged. For each degree N and each label \gamma=(\gamma_3,\ldots,\gamma_n) occurring in S_N, put

The pair exponents with this label lie in \kappa\Delta(A,B). Take their polynomial span in k[x,y] and replace it by its least exponents in the coordinates \xi,\eta and order \nu_t. Lemma 2.2 preserves cardinality and gradedness: the row spaces multiply into the row with degree and label added, and the same coordinates and order are used throughout. For fixed N, choose a basis with distinct least exponents in each row and apply Lemma 2.3 to the finite union of these bases. Its common weight vector allows Lemma 2.4 to apply to all labels at once, giving backward jet transfer. The old points can be chosen in the dense open torus of \A^2\times\T^{n-2}.

Lemma 4.2 places the new pair exponents in \kappa Q. Apply the shear above and compress in the second coordinate. Lemmas 2.5 and 2.6, together with (4.3), give the bound \kappa P. It remains to turn this triangle into \Delta(w,H/w).

Complete the primitive vector u to an integral basis (u,v) with \det(u,v)=1, and use the coordinates in that basis. This is a unimodular change, and its second coordinate is precisely z\mapsto\det(u,z). Write q_0<q_1<q_2 for the three vertex heights of the transformed triangle. Its horizontal slice length is linear from zero at q_0 to a maximum at q_1, and linear back to zero at q_2. The maximum is w, since area is preserved and

Compressing in the first coordinate therefore gives the bound

Its vertical slice at first coordinate h\in[0,w] has length R(1-h/w). Compression in the second coordinate gives \Delta(w,R), as illustrated in Figure 2. These slice statements commute with multiplication by \kappa, so the resulting pair bound is \kappa\Delta(w,R) for every label. If \kappa=0, its only possible pair exponent is (0,0), which remains unchanged at every step.

Restoring the labels gives S'_N\subset N\Delta(w,R,a_3,\ldots,a_n). Every step preserves cardinality and gradedness and has backward jet transfer, so their composition has all the asserted properties.□

Proof. The boundary of a convex body has measure zero, so lattice-point counting by Riemann sums gives

Consequently the number of missing lattice points

is o(N^n). We show that a point of NK has too many decompositions into two points in nearly equal degrees for all of them to involve a missing point.

Assume K is nonempty, and choose \rho>0 so that the closed ball of radius \rho about every point of K lies in P. Write N=N_1+N_2 with N_1=\lfloor N/2\rfloor and N_2=\lceil N/2\rceil. For y\in NK\cap\Z^n, consider all lattice points

Both y_1/N_1 and y_2/N_2 lie in the radius-\rho ball about y/N. Thus y_i\in N_iP\cap\Z^n. The ball from which y_1 is chosen contains at least cN^n lattice points for a constant c>0 independent of y and all sufficiently large N: an inscribed cube gives this bound uniformly in the center. Among these choices, at most b_{N_1} have y_1\notin F_{N_1}, and at most b_{N_2} have y_2\notin F_{N_2}, since y_1\mapsto y-y_1 is injective. As b_{N_1}+b_{N_2}=o(N^n), at least one choice has y_i\in F_{N_i}. Gradedness then gives y=y_1+y_2\in F_N. All bounds are uniform in y, which proves the assertion.□

Proof. Write the coordinates as (x_1,\ldots,x_{n-1},Y) and put x'-1=(x_1-1,\ldots,x_{n-1}-1). An arbitrary jet through order m at q_i has a unique expression

For each \alpha, the Chinese remainder theorem gives a polynomial h_\alpha(Y) satisfying

Indeed, the ideals in this display are pairwise comaximal, and their product is generated by a monic polynomial of degree r(m-|\alpha|+1). The polynomial

realizes the prescribed jets. Each monomial x'^\beta Y^b in its expansion satisfies \beta_j\le\alpha_j for every j and b<r(m-|\alpha|+1) for some \alpha. Hence |\beta|+b/r<m+1, so the polynomial belongs to M(E_m). This uses powers of local parameters and the Chinese remainder theorem, and therefore works in every characteristic.□

Proof. Take the graded family S_N and simplex P_0=\Delta(a_1,\ldots,a_n) from Proposition 3.1. Thus \prod_j a_j=V, the support S_N lies in NP_0, and

Let \ell=\min_j a_j, and choose r_0 so large that

Fix one such r. Reshape the first two intercepts to (w,a_1a_2/w), then reshape the second and third intercepts, and continue through the (n-1)st and nth intercepts. At every step, both intercepts A,B being reshaped are at least \ell: untouched intercepts have this property, and the carried intercept AB/w is larger than A because B>w. Consequently

as required by Proposition 4.1. The product of all intercepts remains V. The resulting graded family F_N therefore satisfies

Here the last intercept is V/w^{n-1}=rw. The transformations preserve backward jet transfer, so jet separation by M(F_N) will imply jet separation by H^0(X,NL).

Fix 0<\theta<w. To apply Lemma 5.1, we contract P_* and then translate it away from the coordinate faces. Choose \lambda with \theta/w<\lambda<1, and choose a vector v\in\R_{>0}^n small enough that

For example, it suffices that \sum_{j<n}v_j/w+v_n/(rw)<1-\lambda. Choose vectors v_N\in N^{-1}\Z^n tending to v. For all sufficiently large N, the sets v_N+\lambda P_* lie in a common compact subset K of \operatorname{int}(P_*). Applying Lemma 5.1 to (5.2) gives

for all sufficiently large N.

Put m=\lfloor N\theta\rfloor. Since \theta<\lambda w, we have m+1\le N\lambda w for all sufficiently large N. The defining inequality (5.1) therefore gives

Choose distinct c_1,\ldots,c_r\in k^\times and apply Lemma 5.2. Multiplication by the Laurent monomial with exponent Nv_N\in\Z^n is multiplication by a unit on the torus, hence induces an automorphism on every jet space. Thus (5.3) implies that M(F_N) separates jets through order m at these r points. Backward jet transfer through all reshapings and Proposition 3.1 gives the asserted jet separation at an ordered tuple on X.

For each fixed r, \theta, and sufficiently large N, the locus in U_r where this evaluation map is surjective is a nonempty open set, by the openness of jet separation established in Section 2. The variety U_r is integral, so its generic point belongs to that open set. Base change of the jet evaluation map to K_r preserves surjectivity and identifies its targets with the jets at p_1,\ldots,p_r. The same geometric generic tuple therefore satisfies every required jet-separation statement.□

Proof. Choose such a point q_j. Write R=\mathcal O_{C,q_j} and let \mathfrak n be its maximal ideal. The vector space \mathfrak n^m/\mathfrak n^{m+1} is nonzero: otherwise Nakayama’s lemma would give \mathfrak n^m=0, contrary to R being a one-dimensional local domain. In a local frame of A, lift a nonzero class in this vector space to an ambient jet of order m at q_j. This lift can be chosen in the image of \mathfrak m_{Y,q_j}^m. Prescribe that jet at q_j and the zero jet at all other marked points. Jet separation produces a section \sigma\in H^0(Y,A) which vanishes to order at least m at every q_i and whose restriction to C is nonzero.

For completeness, the local multiplicity estimate used here is

where e(R) denotes the Hilbert–Samuel multiplicity of a one-dimensional Noetherian local domain (R,\mathfrak n); see [?, Definition 43.15.1, Tag 0AZV]. Indeed, for positive integers t, the nonzero divisor f gives

Divide by t and let t tend to infinity. Applying (6.1) to the local equations of \sigma|_C at the marked points proves the inequality: the sum of all local zero lengths of a nonzero section of A|_C is \deg(A|_C)=A\cdot C.□

Proof. Denote the infimum by \varepsilon. If \varepsilon=0, the conclusion is immediate. Otherwise, blow up the marked points:

For each rational number 0\leq s<\varepsilon, the divisor class

is nef. To check this on curves, first observe that D_s|_{E_i}=s\mathcal O_{\PP^{d-1}}(1), which has nonnegative degree on every curve in E_i. Every other integral curve in \widetilde Y is the strict transform \widetilde C of an integral curve C\subset Y, and

Thus D_s\cdot\widetilde C\geq0 by the definition of \varepsilon; for curves missing all marked points this follows from the ampleness of A.

Here is a local justification of (6.2), which also covers singular curves. If q_i\in C, the Rees algebra construction identifies \widetilde C with the blow-up of C at the marked points. Indeed, the ambient Rees algebra surjects onto the Rees algebra of the restricted ideal. The latter is integral, and its Proj agrees with C away from the centers, so its closed image is exactly the strict transform. For R=\mathcal O_{C,q_i} with maximal ideal \mathfrak n, its exceptional fiber is

It is the effective Cartier divisor cut out on \widetilde C by E_i. Its length is e(R): the Hilbert function \dim_{k(q_i)}(\mathfrak n^t/\mathfrak n^{t+1}) equals e(R) for all sufficiently large t, and its eventual value is the length of this zero-dimensional projective scheme. This proves (6.2); if q_i\notin C, both sides vanish.

A nef class on a projective variety is a limit of ample classes and therefore has nonnegative top self-intersection. The exceptional divisors are disjoint, \pi^*A restricts trivially to each of them, and E_i^d=(-1)^{d-1}. Consequently

If \varepsilon>(A^d/r)^{1/d}, even if \varepsilon=+\infty, there is a rational number s strictly between (A^d/r)^{1/d} and \varepsilon. The displayed inequality excludes such an s, proving the bound.□

Proof of Theorem 1.1. Choose r_0 as in Proposition 5.3 and fix r\geq r_0. Put V=L^n and w=(V/r)^{1/n}. Work over K_r at the geometric generic tuple p_1,\ldots,p_r. For every fixed real number 0<\theta<w and all sufficiently large integers N, Proposition 5.3 gives separation of jets through order m=\lfloor N\theta\rfloor by H^0(X_{K_r},L_{K_r}^{\otimes N}) at this fixed tuple.

For large N we have m\geq1, so Lemma 6.1 yields

for every integral curve in the defining infimum. Letting N tend to infinity gives \varepsilon_r(X,L)\geq\theta. Since \theta<w was arbitrary, \varepsilon_r(X,L)\geq w. Intersection numbers are unchanged by extension of the algebraically closed ground field, so (L_{K_r})^n=V; Lemma 6.2 supplies the reverse inequality.□