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Hodge level of weighted complete intersections of general type V. V. Przyjalkowski Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
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     § 1. IntroductionOne of the most basic biregular invariants of smooth varieties is Hodge numbers. In particular, if a lot of them vanish, then the variety tends to be ‘simple’ from the homological point of view. In [20], § 1, [5], § 2a, and [18], for a smooth variety X the notion of Hodge level h(X)=max{q−p∣hp,q(X)≠0} was introduced. In other words, this number measures how large the number a such that HHa(Db(CohX))≠0 can be. Let n=dim(X). By definition h(X) has the maximum possible value n if and only if h0,n(X)≠0. If X is Fano, that is, its anticanonical class is ample, then h0,n(X)=h0(ΩnX)=0. If X is Calabi-Yau, then H0,n(X)=C by definition, so it has the maximal Hodge level possible. Clearly, the Hodge level of curves of positive genus is equal to 1, so it is maximal possible again. However, even in dimension 2 there exist examples of fake projective planes, which are smooth surfaces of general type (that is, ones whose canonical class is nef and big) such that their Hodge diamond is the same as for P2; see [15] and [6], for example. To investigate the Hodge levels of varieties one needs to be able to compute Hodge numbers. However this, as well as determining merely if a variety is Fano, Calabi-Yau or of general type (or none of these), can be not easy. The most straightforward way to construct varieties is to describe them as complete intersections in something well studied, like toric varieties or homogenous spaces. There is an approach, going back to Griffiths, to compute the Hodge numbers of complete intersections in toric varieties (see [8], [11], [16] and [14]); its generalization to some complete intersections in Grassmannians can be found in [10]. Perhaps the most complicated and interesting class of varieties (from different points of view) is the ones with Picard rank 1. Let X be a quasi-smooth complete intersection in a toric variety T. By the Lefschetz-type theorem for complete intersections in toric varieties (see [14], Proposition 1.4) this means that T is either a weighted projective space or its quotient by a finite group (see [13], for example). If X is in addition a smooth Fano variety, then T is in fact a weighted projective space by Theorem 2 in [3], so X is a weighted complete intersection. This is not true if X is of general type: see [3]. However, the weighted complete intersection case is still the most approachable one. The Hodge levels of Fano weighted complete intersections were considered in [18]. More precisely, the question of how small the Hodge level of Fano weighted complete intersections can be was investigated there, and the varieties of smallest or close to smallest Hodge levels were classified. In this paper, by contrast, we consider how large the Hodge levels of weighted complete intersections of general type can be. We expect that in this case the phenomenon of fake projective planes does not occur, and for smooth weighted complete intersections of general type the Hodge levels are maximal possible. The following statement summarizes the above and shows that our expectations hold up in the case of complete intersections of Cartier divisors. Theorem 1.1. Let X⊂P(a0,…,aN) be a smooth well-formed weighted complete intersection of multidegree (d1,…,dk). Put n=N−k=dim(X)>0 and iX=∑du−∑al. Then the following holds. Corollary 1.1. In the terminology of [18] this, in particular, means that a smooth well-formed weighted Calabi-Yau complete intersection of dimension n is of n-Calabi-Yau type, while if a smooth n-dimensional well-formed weighted complete intersection of Cartier divisors is of general type, then it is never Q-homologically minimal, Hodge-Tate, of curve type (if n>1), or of m-Calabi-Yau type for m<n. In Corollary 3.1 we show that assertion (iii) of Theorem 1.1 holds for codimension 2 smooth weighted complete intersections (not necessarily of Cartier divisors). We expect that this holds in the general case, for all smooth weighted complete intersections of general type (Problem 3.1). Assertions (i) and (ii) of Theorem 1.1 hold for quasi-smooth well-formed weighted complete intersections (see Corollary 2.2). However, assertion (iii) can fail even in the hypersurface case. An example is given in Proposition 4.2. Note that h(X)=n means that h0,n(X)>0, that is, dim|OX(iX)|⩾. In fact Conjecture 4.8 in [17] claims that \dim |\mathcal{O}_X(m)|\geqslant 0 for all m\geqslant i_X. In § 2 we give some definitions and results related to the subject. In § 3 we prove Theorem 1.1 and Corollary 3.1. In § 4 we provide a counterexample to the statement of Theorem 1.1 in the quasi-smooth case (Proposition 4.2). AcknowledgementsThe author is grateful to M. Korolev for the proofs of Lemma 4.1 and Proposition 4.1, to C. Shramov for his helpful comments and the idea of the proof of Proposition 3.2, and to the referee, whose useful comments improved the paper. § 2. PreliminariesWe introduce basic definitions and results related to weighted complete intersections. For more details, see [9], [12] and [19]. Let a_0,\dots,a_N be positive integers. Consider the graded algebra \mathbb{C}[x_0,\dots,x_N], where the grading is defined by assigning the weights a_l to the variables x_l. Put
\begin{equation*}
\mathbb P=\mathbb P(a_0,\dots,a_N)=\operatorname{Proj}\mathbb C[x_0,\dots,x_N].
\end{equation*}
\notag
We use the abbreviation
\begin{equation*}
(a_0^{k_0},\dots,a_m^{k_m})= (\underbrace{a_0,\dots,a_0}_{k_0},\dots,\underbrace{a_m,\dots,a_m}_{k_m}),
\end{equation*}
\notag
where k_0,\dots,k_m are allowed to be any positive integers. If some of the k_i are equal to 1, then we drop the exponents for simplicity. The weighted projective space \mathbb{P} is said to be well formed if the greatest common divisor of any N of the weights a_l is 1. Every weighted projective space is isomorphic to a well-formed one (see [9], § 1.3.1). A subvariety X\subset \mathbb{P} is said to be well formed if \mathbb{P} is well formed and
\begin{equation*}
\operatorname{codim}_X(X\cap\operatorname{Sing}\mathbb P)\geqslant 2,
\end{equation*}
\notag
where the dimension of the empty set is defined to be -1. We say that a subvariety X\subset\mathbb{P} of codimension k is a weighted complete intersection of multidegree (d_1,\dots,d_k) if its weighted homogeneous ideal in \mathbb{C}[x_0,\dots,x_N] is generated by a regular sequence of k homogeneous elements of degrees d_1,\dots,d_k. A weighted complete intersection X\subset\mathbb{P} is said to be an intersection with linear cone if d_u=a_l for some u and l. Let p\colon \mathbb A^{N+1}\setminus \{0\}\to \mathbb{P} be the natural projection onto the weighted projective space. A subvariety X\subset \mathbb{P} is called quasi-smooth if the preimage p^{-1}(X) is smooth. Note that by Proposition 2.9 in [2], if a quasi-smooth weighted complete intersection X\subset \mathbb{P} of dimension at least 3 is general in the family of weighted complete intersections of the same multidegree in \mathbb{P}, then there exists a quasi-smooth well-formed weighted complete intersection isomorphic to X that is not an intersection with a linear cone. The smoothness of a weighted complete intersection implies some arithmetic restrictions on the weights and degrees defining it. Proposition 2.1 (cf. Proposition 4.1 in [7]). Let X\subset\mathbb{P}(a_0,\dots,a_N) be a smooth well-formed weighted complete intersection of multidegree (d_1,\dots,d_k). Then for every r and every choice of r weights a_{i_1},\dots,a_{i_r}, i_1<\dots<i_r, such that their greatest common divisor \delta is larger than 1 there exist r degrees d_{s_1},\dots,d_{s_r}, {s_1<\dots<s_r}, such that their greatest common divisor is divisible by \delta. In a similar way one can determine if a hypersurface in a weighted projective space is a Cartier divisor. Proposition 2.2 (see Proposition 8 in [22] or the proof of Theorem 3.2.4, (i), in [9]). Let \mathbb{P}=\mathbb{P}(a_0,\dots,a_N) be a well-formed weighted projective space. Then the Picard group \operatorname{Pic}(\mathbb{P}) is a free group generated by \mathcal{O}_\mathbb{P}(r), where r is the least common multiple of the weights a_l. In particular, a degree d hypersurface is Cartier if and only if a_l\mid d for l=0,\dots,N. By combinatorial reasons it is more convenient to deal just with the collections of weights and degrees defining weighted complete intersections, ignoring the geometric objects they produce. Definition 2.1 (cf. [17]). The pair (\overline{d},\overline{a}) of tuples of positive integers
\begin{equation*}
\overline{d}=(d_1,\dots,d_k),\qquad\overline{a}=(a_0,\dots,a_N)
\end{equation*}
\notag
is called a regular pair if the divisibility conditions in Proposition 2.1 hold for them. The numbers d_u are called degrees and the a_l are called weights. We call a regular pair Cartier if a_l\mid d_u for all l and u. We call a regular pair Fano if for i=\sum d_u-\sum a_l we have i<0, Calabi-Yau if i=0, and of general type if i>0. Remark 2.1. Let (\overline{d},\overline{a}) be a regular pair such that, up to a permutation, Then \overline{a}=(d_1,\dots,d_k,1^{N+1-k}). Indeed, if, for instance, a_k>1, then by regularity one of degrees, for instance, d_1, is divisible by a_k; this means that, since the two weights d_1 and a_k are divisible by a_k, some other degree is also divisible by a_k, and so on. This shows that all degrees d_1,\dots,d_k are divisible by a_k, so a_k divides at least {k+1} weights, which contradicts the regularity of the pair.The weights and degrees defining a smooth weighted complete intersection (or, in addition, an intersection of Cartier, Fano, Calabi-Yau, or general-type divisors) form a regular pair (satisfying some additional requirements). Since the proofs of statements relating to this subject are obtained by an analysis of numerical conditions on weights and degrees (and these statements are often formulated in terms of regular pairs), it is usually more convenient to use the language of regular pairs and then derive results on weighted complete intersections from ones on regular pairs. We also follow this strategy. We will refer to results formulated for smooth weighted complete intersections (which follow from results for regular pairs) by replacing weighted complete intersections in their statements by regular pairs, and then we will derive the final statements for weighted complete intersections from them. It turns out that there is a strong restriction on the minimal weight for Fano and Calabi-Yau regular pairs. Theorem 2.1 (cf. Corollary 3.4 in [2]). Consider a Fano or Calabi-Yau regular pair (\overline{d},\overline{a}), where \overline{d}=(d_1,\dots,d_k) and \overline{a}=(a_0,\dots,a_N). Set i=\sum d_u-\sum a_l. Let a_l\neq d_u for all l and u. Let a_0\leqslant\dots\leqslant a_N. Then a_{k-i-1}=1. Moreover, a_{k-i}=1 unless \overline{d}=(6^k) and \overline{a}=(1^s,2^k,3^k). Corollary 2.1. Consider a regular pair (\overline{d},\overline{a}), where \overline{d}=(d_1,\dots,d_k), \overline{a}=(a_0,\dots,a_N), and N\geqslant k. Assume that a_l>1 for all l. Then (\overline{d},\overline{a}) is of general type. Proof. By Remark 2.1 there exists a degree d_u such that d_u\neq a_l for all l. Deleting the pairs d_u, a_l such that d_u=a_l from the regular pair, one obtains a regular pair satisfying the assumptions of Theorem 2.1. It remains to apply this theorem. Now recall how we can compute the Hodge numbers of weighted complete intersections. Let X\subset \mathbb{P}=\mathbb{P}(a_0,\dots,a_N) be a quasi-smooth weighted complete intersection of hypersurfaces of degrees d_1,\dots,d_k. By [14], § 3, or [4], § 11, there is a pure Hodge structure on the cohomology of X. In particular, the Hodge numbers h^{p,q}(X) are well defined. By Theorem 10.8 and Remark 10.9 in [1], for the weighted projective space \mathbb{P} we have h^{p,q}(\mathbb{P})=1 if p=q and h^{p,q}(\mathbb{P})=0 otherwise. By Proposition 3.2 in [14] the only Hodge numbers of X that are not inherited from the ambient weighted projective space are the h^{p,q}(X) for p+q=\dim(X). Set where the weight of x_i is a_i, and Let f_1,\dots,f_k be polynomials of weighted degrees d_1,\dots,d_k in S' that generate the weighted homogeneous ideal of the weighted complete intersection X. Let Denote the ideal in S generated by
\begin{equation*}
\frac{\partial F}{\partial w_1},\dots, \frac{\partial F}{\partial w_k}, \frac{\partial F}{\partial x_0},\dots,\frac{\partial F}{\partial x_{N}}
\end{equation*}
\notag
by J=J(F). Set The algebra S is bigraded by \deg (x_l)=(0,a_l) and \deg(w_u)=(1,-d_u), so that F is a bihomogeneous polynomial of bidegree (1,0). Thus, the bigrading descends to the ring \mathcal{R}. Set n=N-k=\dim (X) and Let h_{\mathrm{pr}}^{n-q,q}(X) be primitive middle Hodge numbers of X, that is, for p\neq q and otherwise.Corollary 2.2. Denote the ideal in S' generated by f_1,\dots,f_k by J', and let be a graded polynomial ring. Then h^{0,n}=\mathcal{R}'_{i_X}. In particular, h^{0,n}(X)>0 if and only if there exists a weighted monomial of degree i_X in S'. Proof. We may assume that X does not lie in any coordinate hyperplane, that is, x_r\notin J' for any r. Let M be a monomial of degree i_X in S'. Then M\notin J', otherwise X is reducible. Thus, M descends to a nontrivial element of \mathcal{R}'. On the other hand, any monomial summand of a homogenous element in S' of degree i_X has the same degree i_X. Corollary 2.3. The inequality h^{0,n}(X)>0 holds if and only if there exist positive integers \beta_0,\dots,\beta_N such that Proof. If such integers exist, then x_0^{\beta_0-1}\dotsb x_N^{\beta_N-1} is a weighted monomial of degree -i_X in S'. On the other hand, if x_0^{\alpha_0}\dotsb x_N^{\alpha_N} is a monomial of degree -i_X, then x_0^{\alpha_0+1}\dotsb x_N^{\alpha_N+1} is a monomial of degree d_1+\dots+d_k, so that one can define \beta_l=\alpha_l+1. The corollary is proved. Definition 2.2 (see [18], Definition 1.14; cf. [20], § 1, and [5], § 2a). Given a smooth projective variety X, set The number \mathrm{h}(X) is called the Hodge level of X.§ 3. Weighted intersections of Cartier divisors of general typeIn this section we prove Theorem 1.1. The key ingredient is as follows. Proposition 3.1. Let (\overline{d},\overline{a}), where \overline{d}=(d_1,\dots,d_k) and \overline{a}=(a_0,\dots,a_N), be a Cartier regular pair of general type, and let N-k>0. Then there exist positive integers \beta_0,\dots, \beta_N such that Proof. We prove this using induction on d_1+\dots+d_k. The base (when \overline{d}=4 and \overline{a}=(1,1,1)) is trivial. Set i=\sum d_u-\sum a_l>0. Let a_l=1 for some l. Then we can choose \beta_m=1, m\neq l and \beta_l=i+1. Thus, we may assume that a_l>1 for all l. Moreover, we may assume that a_l\neq d_u for all l and u. Indeed, otherwise we can set \beta_l=1 and decrease d_1+\dots+d_k by removing a_l and d_u; clearly, after the removal we obtain a Cartier regular pair of general type with N-1>k-1, so in this case the assertion of the proposition can be established using induction.
Assume that a_l>1 for all l and a_l\neq d_u for all l and u. Choose a prime number p. Let s be a maximal p-adic valuation for a_0,\dots,a_N. We may assume that for some integer r, k\geqslant r\geqslant 1, the numbers a_0,\dots, a_{r-1} are divisible by p^s, while a_{r},\dots,a_{N} are not. Then p^s|d_m for all m=1,\dots,k by the divisibility assumption. Set
\begin{equation*}
\begin{gathered} \, d_u=pd_u', \quad u=1,\dots,k, \qquad a_l=pa_l', \quad l=0,\dots,r-1, \\ \text{and}\quad a_l=a_l', \quad l=r,\dots,N. \end{gathered}
\end{equation*}
\notag
The collections \overline{d}^{\,\prime}=(d_1',\dots,d_k') and \overline{a}^{\,\prime}=(a_0',\dots,a_N') form a Cartier regular pair. Indeed, since (\overline{d},\overline{a}) is a Cartier regular pair, for any l and u the weight a_l divides d_u. Thus, if l<r, then a_l'=a_l/p divides d_u'=d_u/p, and if l\geqslant r, then a_l'=a_l also divides d_u'=d_u/p, since the p-adic valuation of d_u is at least s. This shows that (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is a Cartier pair. It is also obviously regular since any k+1 different weights a_l' are coprime. If (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is a Calabi-Yau pair, set
\begin{equation*}
\beta_l=1, \quad l=0,\dots,r-1, \quad\text{and}\quad \beta_l=p, \quad l=r,\dots,N.
\end{equation*}
\notag
One can see that this yields the assertion of the proposition in this case. If (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is of general type, then by the induction assertion there exist positive numbers \beta_0',\dots,\beta_N' such that Set
\begin{equation*}
\beta_l=\beta_l', \quad l=0,\dots,r-1, \quad\text{and}\quad \beta_l=p\beta_l', \quad l=r,\dots,N.
\end{equation*}
\notag
Then (\bigotimes) holds, which proves the proposition.
Now assume that (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is Fano. If s>1, then a_l'>1 for all l, so by Corollary 2.1 the pair (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is of general type. Thus, we can assume that s=1 for all primes dividing a_l for some l. In other words, no square of a prime number divides any a_l. Let p,p_1,\dots, p_v be the prime divisors of all a_l. We may assume that v\geqslant 1, that is, there are at least two different prime divisors of the integers a_l, for otherwise, since a_l>1 for all l, p divides all the a_l, and the pair (\overline{d},\overline{l}) is not regular. In particular, we can choose p such that p>2. Thus, for all u=1,\dots,k we have d_u=\alpha_u\cdot p\cdot p_1\dotsb p_v for some positive integers \alpha_u. We can assume that \alpha_u=1 for all u. Indeed, if we replace d_u by d_u/\alpha_u, then we obtain a Cartier regular pair which is of general type by Corollary 2.1. If the assertion of the proposition holds for it, then it holds for the initial regular pair, since if
\begin{equation*}
\frac{d_1}{\alpha_1}+\dots+\frac{d_k}{\alpha_k}=\beta_0 a_0+\dots+\beta_N a_N,
\end{equation*}
\notag
then
\begin{equation*}
d_1+\dots+d_k=\biggl(\beta_0+\sum_{u=1}^k\frac{(\alpha_u-1)d_u}{\alpha_u a_0}\biggr) a_0 +\beta_1 a_1+\dots+\beta_N a_N.
\end{equation*}
\notag
Set e=|\{l\mid a_l'=1\}| and f=|\{l\mid a_l'=d_u'\text{ for some } u\}|. Let f=k, so that there exist k different indices r_l such that a_{r_l}'=d'_l=p_1\dotsb p_v. By Remark 2.1 applied to (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) we have
\begin{equation*}
\overline{d}=((pc)^k) \quad\text{and}\quad \overline{a}=(c^k,p^r)
\end{equation*}
\notag
for c=p_1\dotsb p_v>1 coprime with p, where k\geqslant r=N+1-k>1. If k=2, then r=2, so that \overline{d}=(pc,pc) and \overline{a}=(c,c,p,p). The assertion of the proposition in this case holds because Let k>2. If r=2, then we can set
\begin{equation*}
\beta_0=1, \quad \beta_1=p-1, \quad \beta_2=p, \quad \dots, \quad \beta_{k-1}=p, \quad \beta_{k}=1 \quad\text{and}\quad \beta_{k+1}=c-1.
\end{equation*}
\notag
If r=3 and c>2, then we can set
\begin{equation*}
\beta_0=1, \ \ \beta_1=p-1, \ \ \beta_2=p, \ \ \dots, \ \ \beta _{k-1}=p, \ \ \beta_{k}=1, \ \ \beta_{k+1}=1 \ \text{and}\ \ \beta_{k+2}=c-2.
\end{equation*}
\notag
If r=3 and c=2, then we can set
\begin{equation*}
\beta_0=1+(k-2) p-k,\ \ \beta_1=1, \ \ \dots, \ \ \beta _{k-1}=1, \ \ \beta_{k}=2, \ \ \beta_{k+1}=1 \ \ \text{and}\ \ \beta_{k+2}=1.
\end{equation*}
\notag
Let r\geqslant 4. Using (\bigoplus) we can remove the degrees pc, pc and the weights c, c, p and p from (\overline{d},\overline{a}) and obtain a Cartier regular pair. It is of general type by Corollary 2.1. Moreover, Thus, the assertion of the proposition in this case holds by induction.
Now let f<k. By Theorem 2.1 we have e>k-f>0, so that e\geqslant 2. Moreover, f>0, otherwise e>k by Theorem 2.1, while e is the number of the weights a_l equal to p, so that e\leqslant k. Let f=1. Then k\geqslant e>k-1, and therefore e=k. By Theorem 2.1 this can occur only when
\begin{equation*}
\overline{d}=((6p)^k) \quad\text{and}\quad \overline{a}=(6,2^k,3^k,p^k);
\end{equation*}
\notag
however, this pair is not regular.
Now assume that f>1. We have d_1=\dots=d_k=p\cdot p_1\dotsb p_v, two weights (say, a_0 and a_1) are equal to p, and two weights (say, a_2 and a_3) are equal to p_1\dotsb p_v. We delete two degrees and the four weights a_0, a_1, a_{2} and a_3 from the regular pair and obtain a pair (\overline{d}_0,\overline{a}_0). It is a Cartier regular pair of general type (again, by Corollary 2.1). Assume that N=k+1. Then \overline{a}=(p,p,p_1\dotsb p_v,p_1\dotsb p_v,a_4,\dots,a_{k+1}), so we can take
\begin{equation*}
\begin{gathered} \, \beta_0=1, \qquad \beta_1=(p_1\dotsb p_v-1), \qquad \beta_2=1, \qquad \beta_3=p-1 \\ \text{and}\quad \beta_l=\frac{p\cdot p_1\dotsb p_v}{a_l}\quad\text{for } l=4,\dots,k+1 \end{gathered}
\end{equation*}
\notag
to prove the proposition. Assume that N=k+2. Since a_0=a_1=p, we have {k>1}. If k=2, then
\begin{equation*}
\overline{d}=(p\cdot p_1\dotsb p_v,p\cdot p_1\dotsb p_v) \quad\text{and}\quad \overline{a}=(p,p,p_1\dotsb p_v,p_1\dotsb p_v,a_4),
\end{equation*}
\notag
so the pair (\overline{d},\overline{a}) is not regular, because a_4 divides p\cdot p_1\dotsb p_v. If k>2, then
\begin{equation*}
\overline{a}=(p,p,p_1\dotsb p_v,p_1\dotsb p_v,a_4,a_5,\dots,a_{k+2}),
\end{equation*}
\notag
so we can take
\begin{equation*}
\begin{gathered} \, \beta_0=1, \qquad \beta_1=(p_1\dotsb p_v-1), \qquad\beta_2=1, \qquad\beta_3=p-1, \qquad\beta_4=\frac{p_1\dotsb p_v}{a_4}, \\ \beta_5=\frac{(p-1)\cdot p_1\dotsb p_v}{a_5} \quad\text{and}\quad \beta_l=\frac{p\cdot p_1\dotsb p_v}{a_l} \quad\text{for } l=6,\dots,k+2 \end{gathered}
\end{equation*}
\notag
to prove the proposition.
Finally, assume that N-k>2, so that (N-4)-(k-2)=N-k-2>0. Using induction we can find the decomposition analogous to (\bigotimes) for (\overline{d}_0,\overline{a}_0). Together with
\begin{equation*}
p\cdot p_1\dotsb p_v+p\cdot p_1\dotsb p_v= 1\cdot p+(p_1\dotsb p_v-1)\cdot p +1\cdot(p_1\dotsb p_v)+(p-1)(p_1\dotsb p_v)
\end{equation*}
\notag
it gives the required decomposition for (\overline{d},\overline{a}). The proposition is proved. Remark 3.1. Note that Proposition 3.1 does not hold for N=k. Indeed, let p_0,\dots,p_{N} be different prime numbers, let P=p_0\dotsb p_N, and let P_l={P}/{p_l}, so that P_l is a product of all primes except for p_l. Consider the regular pair ((P^N),(P_0,\dots, P_{N})). It is obviously Cartier and of general type. Assume that where \beta_l>0. Then, since p_m divides P,P_0,\dots,P_{m-1},P_{m+1},\dots and P_N, it also divides \beta_m. Thus,
\begin{equation*}
\beta_0 P_0+\dots+\beta_N P_N\geqslant p_0 P_0+\dots+p_N P_N=(N+1) P>NP,
\end{equation*}
\notag
which is a contradiction. Note that a complete intersection of N hypersurfaces of degree P in \mathbb{P}(P_0,\dots,P_N) is not well formed; after wellformization it becomes a complete intersection of N hyperplanes in \mathbb{P}^N, that is, just a point. Now we are ready to prove Theorem 1.1. Proof of Theorem 1.1. Assertions (i) and (ii) obviously follow from Theorem 2.2 and Corollary 2.2. We prove (iii).
Let X be a smooth well-formed weighted complete intersection of Cartier divisors, and let it be of general type. By Proposition 3.1 in combination with Propositions 2.1 and 2.2, there exist positive integers \beta_0,\dots,\beta_N such that Now assertion (iii) follows from Corollary 2.3. Theorem 1.1 is proved.Problem 3.1. Generalize assertion (iii) of Theorem 1.1 to smooth well-formed weighted complete intersections of general type (not necessarily intersections of Cartier divisors). This can be done for codimension at most 2 by use of the following proposition, the idea of whose proof was shared with us by Shramov. To prove it we need the two-coins case of the Frobenius coin problem. Theorem 3.1 (see [23]). Let a_0 and a_1 be two coprime positive integers. Then for any integer m>a_0a_1-a_0-a_1 there exist nonnegative integers \beta_0 and \beta_1 such that m=\beta_0 a_0+\beta_1 a_1. Now we solve Problem 3.1 in the case of codimension 1 or 2. Proposition 3.2. The following holds. (i) Let (\overline{d},\overline{a}), where \overline{d}=(d_1) and \overline{a}=(a_0,\dots,a_N), N>1, be a regular pair of general type. Then there exist positive integers \beta_0,\dots, \beta_N such that (ii) Let (\overline{d},\overline{a}), where \overline{d}=(d_1,d_2) and \overline{a}=(a_0,\dots,a_N), N>2, be a regular pair of general type. Assume that there exist nonnegative integers \gamma_0,\dots,\gamma_N and \mu_0,\dots,\mu_N such that d_1=\sum \gamma_l a_l and d_2=\sum \mu_l a_l. Then there exist positive integers \beta_0,\dots, \beta_N such that Proof. In the setup of assertion (i) the regular pair (\overline{d},\overline{a}) is Cartier, so it follows from Proposition 3.1. Let us prove assertion (ii). We follow the proof of Proposition 3.1. That is, let us prove the proposition using induction on d_1+d_2. The base (when \overline{d}=(2,3) and \overline{a}=(1,1,1,1)) is trivial. Set i=d_1+d_2-\sum a_l>0. We can assume that a_l>1 for all the l, otherwise we can set \beta_m=1 for m\neq l and \beta_l=i+1. Moreover, we can assume that a_l\neq d_u for all the l and u. Indeed, if, say, d_2=a_N, then the pair ((d_1),(a_0,\dots,a_{N-1})) is regular and of general type, so by assertion (i) there exist positive integers \beta_0,\dots,\beta_{N-1} such that Then
Thus, assume that a_l\neq 1 and a_l\neq d_1,d_2 for l=0\dots,N. Let all the a_l be pairwise coprime. Assume that all the a_l divide d_1. Then by assertion (i) there exist positive integers \beta_0,\dots,\beta_N such that d_1=\sum \beta_l a_l. This means that so assertion (ii) follows. Assume that a_l does not divide d_1 for some l. Up to a permutation we can assume that for some r\leqslant N, a_0,\dots, a_{r-1} divide d_1, while a_r,\dots,a_N do not. Note that in this case the pairs ((d_1),(a_0,\dots,a_{r-1})) and ((d_2),(a_r,\dots,a_{N})) are regular. If r=1, then d_1=\alpha a_0, and by assertion (i) there exist positive integers \beta_1,\dots, \beta_N such that so that If r=2, then, since d_2 is divisible by the coprime integers a_r,\dots,a_N and N-r+1=N-1>1, one has d_2-a_r-\dots-a_N>0. Moreover, d_1-a_0-a_1\geqslant a_0a_1-a_0-a_1. By Theorem 3.1 there exist nonnegative integers \beta_0 and \beta_1 such that so that
\begin{equation*}
d_1+d_2=(\beta_0+1) a_0+(\beta_1+1) a_1+1\cdot a_2+\dots+1\cdot a_N.
\end{equation*}
\notag
Thus, we may assume that r>2 and, similarly, N+1-r>2. By assertion (i) there exist positive integers \beta_0,\dots,\beta_N such that
\begin{equation*}
d_1=\beta_0 a_0+\dots+\beta_{r-1} a_{r-1} \quad\text{and}\quad d_2=\beta_r a_r+\dots+\beta_{N} a_{N}.
\end{equation*}
\notag
This means that
\begin{equation*}
d_1+d_2=\beta_0 a_0+\dots+\beta_{r-1} a_{r-1}+\beta_r a_r+\dots+\beta_{N} a_{N}.
\end{equation*}
\notag
Finally, assume that there is a prime number p such that there exist two weights a_l and a_m divisible by p. Then d_1 and d_2 are also divisible by p. Consider the pair (\overline{d}^{\,\prime},\overline{a}^{\,\prime}), where d_1'={d_1}/{p}, d_2'={d_2}/{p}, a_l'={a_l}/{p}, a_m'={a_m}/{p} and a_w'=a_w for all w\neq l,m. Note that it is regular, because p does not divide a_s for s\neq l,m. If (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is Calabi-Yau or of general type, then similarly to the proof of Proposition 3.1 we can derive assertion (ii) directly or using induction. Assume that (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) is Fano. By Theorem 2.1 this is possible only if either d_1'=a_s' and d_2'=a_t' for some different indices s and t, or d_1'=a_s' for some s and d_2'\neq a_t' for all t\neq s, or, similarly, d_2'=a_s' for some s and d_1'\neq a_t' for all t\neq s. Let d_1'=a_s' and d_2'=a_t' for some different indices s and t. By Remark 2.1 as applied to (\overline{d}^{\,\prime},\overline{a}^{\,\prime}) and Theorem 2.1, (\overline{d},\overline{a})=((p\alpha,p\gamma),\left(\alpha,\gamma,p,p\right)). Since p and \gamma are coprime, we have p\gamma-\gamma-p>0. Since p and \alpha are coprime, Theorem 3.1 implies that there exist nonnegative integers \beta_0 and \beta_2 such that
\begin{equation*}
(p\alpha-\alpha-p)+(p\gamma-\gamma-p)=\beta_0 \alpha+\beta_2 p,
\end{equation*}
\notag
so that
\begin{equation*}
d_1+d_2=p\alpha+p\gamma=(\beta_0+1) \alpha+1\cdot \gamma+(\beta_2+1) p+1\cdot p.
\end{equation*}
\notag
Thus, we may assume without loss of generality that d_1'=a_s' for some s and d_2'\neq a_t' for all t\neq s. Then by Theorem 2.1 we have a_l=a_m=1 and (\overline{d}^{\,\prime},\overline{a}^{\,\prime})=((\alpha,6),(\alpha,2,3,1,1)). This means that (\overline{d},\overline{a})=((\alpha p,6p),(\alpha,2,3,p,p)). Then assertion (ii) is ensured by the equality
\begin{equation*}
\alpha p+6p=p\alpha+\frac{p-3}{2}\cdot 2 +1\cdot 3 +1\cdot p+ 4\cdot p.
\end{equation*}
\notag
The proposition is proved. Corollary 3.1. Let X\subset \mathbb{P}(a_0,\dots,a_N) be a smooth well-formed weighted complete intersection of two hypersurfaces, let \dim (X)=N-2>0, and let it be of general type, that is, i_X>0. Then \mathrm{h}(X)=n. The proof is similar to the proof of Theorem 1.1. § 4. Hodge level for quasi-smooth weighted complete intersections of general typeIn this section we give an example of quasi-smooth weighted complete intersection of general type whose Hodge level is not maximal possible. (In fact this weighted complete intersection is either a hypersurface or has codimension 2, depending on the parity of the dimension.) To do this we need number-theoretic results (Lemma 4.1 and Proposition 4.1), whose proofs we proposed to us by Korolev. Lemma 4.1. Let n be a positive integer, and let x\geqslant 2^n be a real number. Then the following hold. Proof. Assertion (i) obviously follows from (ii). Indeed, it can be checked directly for n=5, so we can assume that n\geqslant 6. Applying assertion (ii) to 2x\geqslant 2^{n+1} we see that there are at least n+2 primes p such that \frac{2}{3}(2x)<p<2x. In particular, there are at least n+1 primes such that x<p<2x.
We prove assertion (ii). One can check by hand that it holds for 8\leqslant n \leqslant 10. Thus we may assume that n\geqslant 11. Let \pi(z) be the number of primes not exceeding z\in \mathbb{R}. By Corollary 1 of Theorem 2 in [21] one has for \alpha=1.25506. We have
\begin{equation*}
\begin{aligned} \, \frac{\pi(x)-\pi(2x/3)}{n+1} &>\frac{{x}/{\ln x}-(2\alpha/3)(x/(\ln x-\ln(3/2)))}{n+1} \\ &=\frac{x}{\ln x\cdot (n+1)} \,\frac{(1-2\alpha/3)\ln x-\ln(3/2)}{\ln x-\ln(3/2)} \\ &\geqslant\frac{2^n}{\ln 2\cdot n(n+1)} \,\frac{(1-2\alpha/3)n\ln 2-\ln(3/2)}{\ln 2\cdot n-\ln(3/2)}, \end{aligned}
\end{equation*}
\notag
since the functions {x}/{\ln x} and ((1-2\alpha/3)\ln x-\ln(3/2))/(\ln x-\ln(3/2)) are monotonically increasing for x\geqslant e. The function
\begin{equation*}
H(t)=\frac{2^t}{\ln 2 \cdot t(t+1)} \,\frac{(1-2\alpha/3)t\ln 2-\ln(3/2)}{t\ln 2-\ln(3/2)}
\end{equation*}
\notag
is increasing for t>{2}/{\ln 2}. Since H(10)>1, we have which gives assertion (ii) of the lemma.
The lemma is proved. Let \delta(n) be the minimum nonnegative number of the form where p_i are different primes. This number is well defined, because one can see that \delta(n)>0 and \delta(n) is attained at some set of primes. Indeed, if then for P_i=p_1\dotsb p_n/p_i we have so that for any i all numbers but P_i in the last equality are divisible by p_i, which is impossible. The fact that \delta(n) is attained at a finite number of sets of primes can be proved using induction. Indeed, this is trivial for n=1. Let this hold for n but fail for n+1. This means that there exists a nonnegative number a, 0\leqslant a<\delta(n), such that for any \varepsilon>0 there exist an infinite number of sets of different primes p_1<\dots<p_{n+1} such that This means that there exist such a set for \varepsilon=(\delta(n)-a)/{2}; moreover, we may assume that p_{n+1}>{2}/(\delta(n)-a) (since for any number A there is only a finite number of sets of primes smaller than A). Thus
\begin{equation*}
1-\frac{1}{p_1}-\dots-\frac{1}{p_{n}}<a+\frac{\delta(n)-a}{2}+\frac{\delta(n)-a}{2}=\delta(n),
\end{equation*}
\notag
which is a contradiction. Proof. The cases when n<8 can be checked by hand; thus, we may assume that {n\geqslant 8}.
We claim that \delta(n)\leqslant {1}/{2^n}. Indeed, let us prove this using induction. One can easily check that this holds for n=5. Let this hold for m\geqslant 5. Let p_1,\dots,p_m be primes at which \delta(m) is attained. Then by Lemma 4.1, (i), there are at least m+1 primes between {1}/{\delta(m)} and {2}/{\delta(m)}. Let p_{m+1} be one of them that is different from p_1,\dots,p_m. Then
\begin{equation*}
\delta(m+1)\,{\leqslant}\, 1-\frac{1}{p_1}-\dots-\frac{1}{p_m}-\frac{1}{p_{m+1}}\,{=}\,\delta(m)-\frac{1}{p_{m+1}} \,{\leqslant}\, \delta(m)-\frac{\delta(m)}{2}\,{=}\,\frac{\delta(m)}{2}\,{\leqslant}\, \frac{1}{2^{m+1}}.
\end{equation*}
\notag
Now let p_1<\dots<p_n be primes such that Assume that \delta(n)\geqslant{1}/{(2p_n)}. Then, since {1}/{2^n}\geqslant\delta(n), we have p_n>2^{n-1}. By Lemma 4.1, (ii), there are at least n primes between \frac{2}{3}p_n and p_n. Let p be one of them that is different from p_1,\dots,p_{n-1}. Set
\begin{equation*}
\delta=1-\frac{1}{p_1}-\dots-\frac{1}{p_{n-1}}-\frac{1}{p}.
\end{equation*}
\notag
Since \delta=\delta(n)+{1}/{p_n}-{1}/{p}, it follows that
\begin{equation*}
\delta \geqslant \frac{1}{2p_n}+\frac{1}{p_n}-\frac{1}{p}=\frac{(3/2)p-p_n}{pp_n}>0,
\end{equation*}
\notag
because p>\frac{2}{3}p_n. However \delta<\delta(n) since p<p_n, which contradicts the minimality of \delta(n). This shows that \delta(n)<{1}/(2p_n).
Summarizing, we have obtained two inequalities, p_n\,{<}\,{1}/(2\delta(n)) and {{1}/{\delta(n)}\,{\geqslant}\, 2^n}. The latter implies by Lemma 4.1, (i), that there exists a prime number p_{n+1} such that {1}/(2\delta(n))<p_{n+1}<{1}/{\delta(n)}. We have The assertion of the proposition follows from the inequalityNow we are ready to prove the main result in this section. Proposition 4.2. For any n>2 there exists an n-dimensional quasi-smooth well-formed weighted complete intersection X of Cartier divisors such that X is of general type and h^{0,n}(X)=0. Proof. We define an integer m>1 by n=2m for n even and n=2m-1 for n odd. By Proposition 4.1 there exist m+2 primes p_0<\dots< p_{m}<p_{m+1} such that
\begin{equation*}
\frac{1}{p_0}+\dots+\frac{1}{p_{m}}<1<\frac{1}{p_0}+\dots+\frac{1}{p_{m}}+\frac{1}{p_{m+1}}.
\end{equation*}
\notag
Set s=0,\dots,m. If n is even, let X be a general hypersurface of degree 2\cdot p_0\dotsb p_{m} in \mathbb{P}(a_0,a_0,a_1,a_1,\dots,a_{m},a_{m}); otherwise let X be a general intersection of two hypersurfaces of degree p_0\dotsb p_{m} in \mathbb{P}(a_0,a_0,a_1,a_1,\dots,a_{m},a_{m}). (In contrast to Remark 3.1, it is well formed.) Note that in both cases X is quasi-smooth and well formed, and \dim (X)=n. Moreover,
\begin{equation*}
i_X=2\cdot p_0\dotsb p_{m}-2\cdot a_0-\dots-2\cdot a_{m}= 2\cdot p_0\dotsb p_{m}\cdot\biggl(1-\frac{1}{p_0}-\dots-\frac{1}{p_{m}}\biggr)>0,
\end{equation*}
\notag
so that X is of general type. On the other hand, for any s and t we have
\begin{equation*}
\begin{aligned} \, i_X-a_s-a_t &=p_0\dotsb p_{m}\cdot\biggl(2-2\frac{1}{p_0}-\dots-2\frac{1}{p_{m}}-\frac{1}{p_s}-\frac{1}{p_t}\biggr) \\ &< 2\cdot p_0\dotsb p_{m}\cdot\biggl(1-\frac{1}{p_0}-\dots-\frac{1}{p_{m}}-\frac{1}{p_{m+1}}\biggr)<0, \end{aligned}
\end{equation*}
\notag
so that there exists no monomial of weight i_X whose ordinary degree in the variables with weights a_0,\dots,a_{m} is two or greater. Finally, if for some l we have i_X=a_l, then i_X+2a_l=3a_l. However, this is impossible unless p_l=3, since the left-hand side of this equality is divisible by p_l, while the right-hand side is not. For p_l=3 we have
\begin{equation*}
\begin{aligned} \, i_X-a_l &=2\cdot p_0\dotsb p_{m}\cdot\biggl(1-\frac{1}{p_0}-\dots-\frac{1}{p_{m}}-\frac{1}{6}\biggr) \\ &< 2\cdot p_0\dotsb p_{m}\cdot\biggl(1-\frac{1}{p_0}-\dots-\frac{1}{p_{m}}-\frac{1}{p_{m+1}}\biggr)<0, \end{aligned}
\end{equation*}
\notag
since p_{m+1}>6.
Thus, there are no linear monomials of weighted degree i_X. Now Corollary 2.2 yields the assertion of the proposition. The proof is complete. 
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