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 \[X=X_0\supset X_1\supset\cdots\supset X_{n-1}=C_0.\] See (The Stacks Project Authors 2026, 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 3 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 (5). A basis with distinct least exponents, Lemma 4(i), and Lemma 5 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 \[\left(N-D\sum_{i<n}\alpha_i\right)L|_{C_0}\] 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 \[\alpha_n\leq \left(N-D\sum_{i<n}\alpha_i\right)D^{n-1}V.\] Equivalently, \(D\sum_{i<n}\alpha_i+\alpha_n/(D^{n-1}V)\leq N\), which is (4). ◻

Proof. The boundary of a convex body has measure zero, so lattice-point counting by Riemann sums gives \[\#(NP\cap\mathbb Z^n)=\mathop{\mathrm{vol}}(P)N^n+o(N^n).\] Consequently the number of missing lattice points \[b_N:=\#\bigl((NP\cap\mathbb Z^n)\setminus F_N\bigr)\] 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 \(y\in NK\cap\mathbb Z^n\), consider all lattice points \[y_1\in\mathbb Z^n, \qquad \left\|y_1-\frac{N_1}{N}y\right\|<\rho N_1, \qquad y_2:=y-y_1.\] Both \(y_1/N_1\) and \(y_2/N_2\) lie in the radius-\(\rho\) ball about \(y/N\). Thus \(y_i\in N_iP\cap\mathbb Z^n\). 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 \[\sum_{|\alpha|\le m}(x'-1)^\alpha h_{i,\alpha}(Y-c_i), \qquad \deg h_{i,\alpha}\le m-|\alpha|.\] For each \(\alpha\), the Chinese remainder theorem gives a polynomial \(h_\alpha(Y)\) satisfying \[h_\alpha(Y)\equiv h_{i,\alpha}(Y-c_i) \pmod{(Y-c_i)^{m-|\alpha|+1}} \quad(1\le i\le r), \qquad \deg h_\alpha<r(m-|\alpha|+1).\] 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 \[\sum_{|\alpha|\le m}(x'-1)^\alpha h_\alpha(Y)\] 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 8. Thus \(\prod_j a_j=V\), the support \(S_N\) lies in \(NP_0\), and \[\#S_N=\frac{V}{n!}N^n+o(N^n).\] Let \(\ell=\min_j a_j\), and choose \(r_0\) so large that \[w:=\left(\frac Vr\right)^{1/n}<\frac{\ell}{100} \qquad(r\ge r_0).\] 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 \(n\)th 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 \[w\max(3/A,2/B)\le\frac{3w}{\ell}<\frac1{16},\] as required by Proposition 9. The product of all intercepts remains \(V\). The resulting graded family \(F_N\) therefore satisfies \[ \begin{aligned} &F_N\subset NP_*\cap\mathbb Z^n, \qquad P_*:=\Delta(w,\ldots,w,rw),\\ &\#F_N=\frac{V}{n!}N^n+o(N^n) =\mathop{\mathrm{vol}}(P_*)N^n+o(N^n). \end{aligned} \tag{11}\] 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 12, we contract \(P_*\) and then translate it away from the coordinate faces. Choose \(\lambda\) with \(\theta/w<\lambda<1\), and choose a vector \(v\in\mathbb R_{>0}^n\) small enough that \[v+\lambda P_*\subset\operatorname{int}(P_*).\] For example, it suffices that \(\sum_{j<n}v_j/w+v_n/(rw)<1-\lambda\). Choose vectors \(v_N\in N^{-1}\mathbb Z^n\) 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 12 to (11) gives \[ Nv_N+\bigl(N\lambda P_*\cap\mathbb Z^n\bigr)\subset F_N \tag{12}\] 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 (10) therefore gives \[E_m\subset N\lambda P_*\cap\mathbb Z^n.\] Choose distinct \(c_1,\ldots,c_r\in k^\times\) and apply Lemma 13. Multiplication by the Laurent monomial with exponent \(Nv_N\in\mathbb Z^n\) is multiplication by a unit on the torus, hence induces an automorphism on every jet space. Thus (12) implies that \(M(F_N)\) separates jets through order \(m\) at these \(r\) points. Backward jet transfer through all reshapings and Proposition 8 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 \[ \operatorname{length}_R(R/(f))\geq m\,e(R) \qquad(0\ne f\in\mathfrak n^m), \tag{13}\] where \(e(R)\) denotes the Hilbert–Samuel multiplicity of a one-dimensional Noetherian local domain \((R,\mathfrak n)\); see (The Stacks Project Authors 2026, Definition 43.15.1, Tag 0AZV). Indeed, for positive integers \(t\), the nonzero divisor \(f\) gives \[t\operatorname{length}_R(R/(f)) =\operatorname{length}_R(R/(f^t)) \geq\operatorname{length}_R(R/\mathfrak n^{mt}) =mt\,e(R)+O(1).\] Divide by \(t\) and let \(t\) tend to infinity. Applying (13) 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: \[\pi:\widetilde Y\longrightarrow Y, \qquad E_1,\ldots,E_r\subset\widetilde Y.\] For each rational number \(0\leq s<\varepsilon\), the divisor class \[D_s=\pi^*A-s\sum_i E_i\] is nef. To check this on curves, first observe that \(D_s|_{E_i}=s\mathcal O_{\mathbb P^{d-1}}(1)\), 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 \[ E_i\cdot\widetilde C=\mathop{\mathrm{mult}}_{q_i}C. \tag{14}\] 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 (14), 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 \[\operatorname{Proj}\bigl(\operatorname{gr}_{\mathfrak n}R\bigr).\] 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 (14); 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 \[0\leq D_s^d=A^d-rs^d.\] 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. Choose \(r_0\) as in Proposition 14 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 14 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 15 yields \[\frac{L_{K_r}\cdot C}{\sum_i\mathop{\mathrm{mult}}_{p_i}C} \geq\frac{\lfloor N\theta\rfloor}{N}\] 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 16 supplies the reverse inequality. ◻