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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 $\mathrm{h}(X)=\max\{q-p\mid h^{p,q}(X)\neq 0\}$ was introduced. In other words, this number measures how large the number $a$ such that ${HH_a(D^b(\operatorname{Coh} X))\neq 0}$ can be. Let $n=\dim (X)$. By definition $\mathrm{h}(X)$ has the maximum possible value $n$ if and only if $h^{0,n}(X)\neq 0$. If $X$ is Fano, that is, its anticanonical class is ample, then $h^{0,n}(X)=h^0(\Omega_X^n)=0$. If $X$ is Calabi-Yau, then $H^{0,n}(X)=\mathbb{C}$ 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 $\mathbb{P}^2$; 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 $X\subset\mathbb{P}(a_0,\dots,a_N)$ be a smooth well-formed weighted complete intersection of multidegree $(d_1,\dots,d_k)$. Put $n=N-k=\dim (X)>0$ and $i_X=\sum d_u-\sum a_l$. 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 $\mathbb{Q}$-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 $\mathrm{h}(X)=n$ means that $h^{0,n}(X)>0$, that is, $\dim |\mathcal{O}_X(i_X)|\geqslant 0$. 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 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 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, 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
$$
\begin{equation*}
d_1+d_2=(\beta_0+\mu_0) a_0+\dots+(\beta_N+\mu_N) a_N,
\end{equation*}
\notag
$$
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
$$
\begin{equation*}
(d_1-a_0-a_1)+(d_2-a_2-\dots-a_N)=\beta_0 a_0+\beta_1 a_1,
\end{equation*}
\notag
$$
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
$$
\begin{equation*}
a<1-\frac{1}{p_1}-\dots-\frac{1}{p_{n+1}}<a+\varepsilon.
\end{equation*}
\notag
$$
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 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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