Notations 2.2.

1. (1) Let

   $$\begin{eqnarray}P_{(R,\mathbf{m})}(m)=\tilde{e}_{0}\binom{m+d-1}{d}-\tilde{e}_{1}\binom{m+d-2}{d-1}+\cdots +(-1)^{d}\tilde{e}_{d}\end{eqnarray}$$
   be the Hilbert–Samuel polynomial of $R$ with respect to the graded maximal ideal  $\mathbf{m}$ . Therefore,
   $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D712}(X,{\mathcal{O}}_{X}(m)) & = & \displaystyle \tilde{e}_{0}\binom{m+d-1}{d-1}-\tilde{e}_{1}\binom{m+d-2}{d-2}\nonumber\\ \displaystyle & & \displaystyle +\cdots +(-1)^{d-1}\tilde{e}_{d-1}.\nonumber\end{eqnarray}$$

2. (2) Let $\bar{m}$ be a positive integer such that

   1. (a) $\bar{m}$ is a regularity number for $(X,{\mathcal{O}}_{X}(1))$ , and

   2. (b) $R_{m}=h^{0}(X,{\mathcal{O}}_{X}(m))$ , for all $m\geqslant \bar{m}$ . In particular, $\ell (R/\mathbf{m}^{m})=P_{(R,\mathbf{m})}(m)$ , for all $m\geqslant \bar{m}$ .

3. (3) Let $l_{1}=h^{0}(X,{\mathcal{O}}_{X}(1))$ .

4. (4) Let $n_{0}\geqslant 1$ be an integer such that $R_{n_{0}}\subseteq I$ , where $R=\oplus _{n\geqslant 0}R_{n}$ .

5. (5) We denote $\dim _{k}\text{Coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{Q}})$ by $\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{Q}})$ .

Proof. Let $\tilde{m}$ be a regularity number for ${\mathcal{Q}}$ ; then, by Definition 2.4, we have

$$\begin{eqnarray}h^{0}(X,{\mathcal{Q}}(m+q))\leqslant D_{0}({\mathcal{Q}})(m+q)^{\bar{d}}\quad \text{for all}~n,m\geqslant 0.\end{eqnarray}$$

Let ${\mathcal{Q}}_{\bar{d}}={\mathcal{Q}}$ . Let $a_{\bar{d}},\ldots ,a_{1}\in H^{0}(X,{\mathcal{O}}_{X}(1))$ , with the exact sequence of ${\mathcal{O}}_{X}$ -modules

$$\begin{eqnarray}0\longrightarrow {\mathcal{Q}}_{i}(-1)\stackrel{a_{i}}{\longrightarrow }{\mathcal{Q}}_{i}\longrightarrow {\mathcal{Q}}_{i-1}\longrightarrow 0,\end{eqnarray}$$

where ${\mathcal{Q}}_{i}={\mathcal{Q}}_{\bar{d}}/(a_{\bar{d}},\ldots ,a_{i+1}){\mathcal{Q}}_{\bar{d}}$ , for $0\leqslant i\leqslant \bar{d}$ , and realizing the minimal value $C_{0}({\mathcal{Q}})$ . Now, by the exact sequence (5), we have the following short exact sequence of ${\mathcal{O}}_{X}$ -sheaves:

$$\begin{eqnarray}0\longrightarrow F^{n\ast }P\otimes {{\mathcal{Q}}_{i}\longrightarrow \oplus }_{j}{\mathcal{Q}}_{i}(-qb_{j})\longrightarrow F^{n\ast }P^{\prime \prime }\otimes {\mathcal{Q}}_{i}\longrightarrow 0.\end{eqnarray}$$

This implies $H^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i}){\hookrightarrow}\oplus _{j}H^{0}(X,{\mathcal{Q}}_{i}(-qb_{j}))$ . Therefore,

(6) $$\begin{eqnarray}h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i})\leqslant \mathop{\sum }_{j}h^{0}(X,{\mathcal{Q}}_{i}(-qb_{j}))\leqslant (\tilde{\unicode[STIX]{x1D707}})h^{0}(X,{\mathcal{Q}}_{i}),\end{eqnarray}$$

as $-b_{j}\leqslant 0$ . Since $F^{n\ast }P$ is a locally free sheaf of ${\mathcal{O}}_{X}$ -modules, we have

$$\begin{eqnarray}0\longrightarrow F^{n\ast }P\otimes {\mathcal{Q}}_{i}(m-1)\stackrel{a_{i}}{\longrightarrow }F^{n\ast }P\otimes {\mathcal{Q}}_{i}(m)\longrightarrow F^{n\ast }P\otimes {\mathcal{Q}}_{i-1}(m)\longrightarrow 0,\end{eqnarray}$$

which is a short exact sequence of ${\mathcal{O}}_{X}$ -sheaves. Now, by induction on  $i$ , we prove that, for $m\geqslant 1$ ,

$$\begin{eqnarray}h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i}(m))\leqslant (\tilde{\unicode[STIX]{x1D707}})\left[h^{0}(X,{\mathcal{Q}}_{i})+\cdots +h^{0}(X,{\mathcal{Q}}_{0})\right](m^{i}).\end{eqnarray}$$

For $i=0$ , the inequality holds as $h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{0}(m))\leqslant (\tilde{\unicode[STIX]{x1D707}})h^{0}(X,{\mathcal{Q}}_{0})$ (as $\dim {\mathcal{Q}}_{0}=0$ ).

Now, for $m\geqslant 1$ , by the inequality (6) and by induction on  $i$ , we have

$$\begin{eqnarray}\displaystyle h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i}(m)) & {\leqslant} & \displaystyle h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i})+h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i-1}(1))\nonumber\\ \displaystyle & & \displaystyle +\cdots +h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{i-1}(m))\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (\tilde{\unicode[STIX]{x1D707}})h^{0}(X,{\mathcal{Q}}_{i})+\tilde{\unicode[STIX]{x1D707}}[h^{0}(X,{\mathcal{Q}}_{i-1})+\cdots +h^{0}(X,{\mathcal{Q}}_{0})]\nonumber\\ \displaystyle & & \displaystyle \times \,(1+2^{i-1}+\cdots +m^{i-1})\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (\tilde{\unicode[STIX]{x1D707}})[h^{0}(X,{\mathcal{Q}}_{i})+\cdots +h^{0}(X,{\mathcal{Q}}_{0})]m^{i}.\nonumber\end{eqnarray}$$

This implies

$$\begin{eqnarray}h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}(m))=h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}_{\bar{d}}(m))\leqslant \tilde{\unicode[STIX]{x1D707}}C_{0}({\mathcal{Q}})m^{\bar{d}},\end{eqnarray}$$

for all $m\geqslant 1$ . Therefore, for all $0\leqslant i\leqslant \bar{d}$ and for all $m\geqslant 0$ , we have $h^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}(m))\leqslant \tilde{\unicode[STIX]{x1D707}}C_{0}({\mathcal{Q}})(m^{\bar{d}}+1)$ . This proves the lemma.◻

Proof. Note that $X=\text{Proj}~R$ , where $R=\oplus _{n\geqslant 0}R_{n}$ , is a standard graded domain such that $R_{0}$ is an algebraically closed field. Therefore, there exists a Noether normalization

$$\begin{eqnarray}k[X_{0},\ldots ,X_{d-1}]\longrightarrow R,\end{eqnarray}$$

which is an injective, finite separable graded map of degree 0 (as $k$ is an algebraically closed field). This induces a finite separable affine map $\unicode[STIX]{x1D70B}:X\longrightarrow \mathbb{P}_{k}^{d-1}=S$ .

Note that there is also an isomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D702}:{\mathcal{O}}_{S}^{\oplus n_{0}}\oplus {\mathcal{O}}_{S}(-1)^{\oplus n_{1}}\oplus \cdots \oplus {\mathcal{O}}_{S}(-d+1)^{\oplus n_{d-1}}\longrightarrow F_{\ast }{\mathcal{O}}_{S}\end{eqnarray}$$

of ${\mathcal{O}}_{S}$ -modules, where $\sum \,n_{i}=p^{d-1}$ .

Now, the isomorphism of $\unicode[STIX]{x1D702}$ implies that the map

$$\begin{eqnarray}\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D702}):\oplus _{i=0}^{d-1}{\mathcal{O}}_{X}(-i)^{\oplus n_{i}}\longrightarrow \unicode[STIX]{x1D70B}^{\ast }F_{\ast }{\mathcal{O}}_{s}\end{eqnarray}$$

is an isomorphism of ${\mathcal{O}}_{X}$ -sheaves. Since $0\leqslant i\leqslant d-1$ , we also have an injective and generically isomorphic map of ${\mathcal{O}}_{X}$ -sheaves

$$\begin{eqnarray}\oplus ^{p^{d-1}}{\mathcal{O}}_{X}(-d)\longrightarrow \oplus _{i=0}^{d-1}{\mathcal{O}}_{X}(-i)^{\oplus n_{i}}.\end{eqnarray}$$

Composing this map with $\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D702})$ gives an injective and generically isomorphic map of ${\mathcal{O}}_{X}$ -sheaves

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}:\oplus ^{p^{d-1}}{\mathcal{O}}_{X}(-d)\longrightarrow \unicode[STIX]{x1D70B}^{\ast }F_{\ast }{\mathcal{O}}_{S}.\end{eqnarray}$$

Since $\unicode[STIX]{x1D70B}$ is separable, there is a canonical map $\unicode[STIX]{x1D6FD}:\unicode[STIX]{x1D70B}^{\ast }F_{\ast }{\mathcal{O}}_{S}\longrightarrow F_{\ast }{\mathcal{O}}_{X}$ , of sheaves of ${\mathcal{O}}_{X}$ -modules, which is generically isomorphic.

Now, we have the composite map

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}:\oplus ^{p^{d-1}}{\mathcal{O}}_{X}(-d)\longrightarrow \unicode[STIX]{x1D70B}^{\ast }F_{\ast }{\mathcal{O}}_{S}\rightarrow F_{\ast }{\mathcal{O}}_{X},\end{eqnarray}$$

which is generically an isomorphism. Hence, $\dim \text{Coker}(\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC})<\dim X=d-1$ , and the map $\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}:\oplus ^{p^{d-1}}{\mathcal{O}}_{X}(-d)\longrightarrow F_{\ast }{\mathcal{O}}_{X}$ is injective, as $X$ is an integral scheme. This proves the lemma.◻

Proof. (1) The above short exact sequence of ${\mathcal{O}}_{X}$ -sheaves gives a long exact sequence of cohomologies

$$\begin{eqnarray}\displaystyle \cdots & \longrightarrow & \displaystyle \oplus ^{p^{d-1}}H^{i}(X,{\mathcal{O}}_{X}(m-d))\longrightarrow H^{i}(X,{\mathcal{O}}_{X}(mp))\longrightarrow H^{i}(X,{\mathcal{Q}}(m))\nonumber\\ \displaystyle & \longrightarrow & \displaystyle \oplus ^{p^{d-1}}H^{i+1}(X,{\mathcal{O}}_{X}(m-d))\longrightarrow \cdots \,.\nonumber\end{eqnarray}$$

However, $h^{i}(X,{\mathcal{O}}_{X}(m-d-i))=0$ , for all $m\geqslant \bar{m}+d$ and $i\geqslant 1$ , which implies that if $m\geqslant \bar{m}+d$ , then $h^{i}(X,{\mathcal{Q}}(m-i))=0$ , for $i\geqslant 1$ , and the canonical map

$$\begin{eqnarray}f_{1,m}:H^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m))\longrightarrow H^{0}(X,{\mathcal{Q}}(m))\end{eqnarray}$$

is surjective. Moreover, the canonical map

$$\begin{eqnarray}H^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m))\otimes H^{0}(X,{\mathcal{O}}_{X}(1))\longrightarrow H^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m+1))\end{eqnarray}$$

is the same as the canonical map

$$\begin{eqnarray}f_{2,m}:H^{0}(X,{\mathcal{O}}_{X}(mp))\otimes H^{0}(X,{\mathcal{O}}_{X}(1))^{[p]}\longrightarrow H^{0}(X,{\mathcal{O}}_{X}(mp+p)),\end{eqnarray}$$

which is surjective for $m\geqslant \tilde{m}$ . The map $f_{2m}$ fits into the following canonical diagram:

$$\begin{eqnarray}\begin{array}{@{}ccc@{}}R_{mp}\otimes R_{1}^{[p]} & \longrightarrow & R_{mp+p}\\ \downarrow & & \downarrow \\ H^{0}(X,{\mathcal{O}}_{X}(mp))\otimes H^{0}(X,{\mathcal{O}}_{X}(1))^{[p]} & \stackrel{f_{2,m}}{\longrightarrow } & H^{0}(X,{\mathcal{O}}_{X}(mp+p)),\end{array}\end{eqnarray}$$

where $H^{0}(X,{\mathcal{O}}_{X}(1))^{[p]}$ , as a set, is $H^{0}(X,{\mathcal{O}}_{X}(1))$ , and the upper symbol $^{[p]}$ indicates that the $k$ -space structure is through the $p$ th power map of  $k$ , where the top horizontal map is surjective for $m\geqslant l_{1}-1$ , and the right vertical map is identity as $mp+p\geqslant \bar{m}$ . Now, the commutative diagram of canonical maps

$$\begin{eqnarray}\begin{array}{@{}ccc@{}}H^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m))\otimes H^{0}(X,{\mathcal{O}}_{X}(1)) & \longrightarrow & H^{0}(X,{\mathcal{Q}}(m))\otimes H^{0}(X,{\mathcal{O}}_{X}(1))\\ \downarrow f_{2,m} & & \downarrow \\ H^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m+1)) & \stackrel{f_{1,m+1}}{\longrightarrow } & H^{0}(X,{\mathcal{Q}}(m+1))\end{array}\end{eqnarray}$$

implies that the second vertical map is surjective, for $m\geqslant \tilde{m}$ , as the maps $f_{2,m}$ and $f_{1,m+1}$ are surjective. This proves that ${\mathcal{Q}}$ is $\tilde{m}$ -regular. Hence, the assertion (1).

(2) If

(7) $$\begin{eqnarray}\unicode[STIX]{x1D712}(X,{\mathcal{Q}}(m))=q_{0}\binom{m+d-2}{d-2}-q_{1}\binom{m+d-3}{d-3}+\cdots +(-1)^{d-2}q_{d-2},\end{eqnarray}$$

then, by Lemma A.1(1) (in the appendix, below),

$$\begin{eqnarray}|q_{i}|\leqslant p^{d-1}P_{i}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{i+1}),\end{eqnarray}$$

where $P_{i}^{d}(X_{0},\ldots ,X_{i+1})$ is a universal polynomial function with rational coefficients.

Now, ${\mathcal{Q}}$ is $\tilde{m}$ -regular implies that, for $0\leqslant i<d$ , ${\mathcal{Q}}_{i}:={\mathcal{Q}}/(a_{\bar{d}},\ldots ,a_{i+1}){\mathcal{Q}}$ is $\tilde{m}$ -regular, for any ${\mathcal{Q}}$ -sequence $a_{1},\ldots ,a_{\bar{d}}\in H^{0}(X,{\mathcal{O}}_{X}(1))$ . Therefore, the short exact sequence of ${\mathcal{O}}_{X}$ -modules

$$\begin{eqnarray}0\longrightarrow {\mathcal{Q}}_{i+1}(-1)\longrightarrow {\mathcal{Q}}_{i+1}\longrightarrow {\mathcal{Q}}_{i}\longrightarrow 0\end{eqnarray}$$

gives the short exact sequence of $k$ -vector spaces

$$\begin{eqnarray}0\longrightarrow H^{0}(X,{\mathcal{Q}}_{i+1}(\tilde{m}-1))\longrightarrow H^{0}(X,{\mathcal{Q}}_{i+1}(\tilde{m}))\longrightarrow H^{0}(X,{\mathcal{Q}}_{i}(\tilde{m}))\longrightarrow 0,\end{eqnarray}$$

for $m\geqslant \tilde{m}$ . Hence,

$$\begin{eqnarray}\displaystyle h^{0}(X,{\mathcal{Q}}_{i}) & {\leqslant} & \displaystyle h^{0}(X,{\mathcal{Q}}_{i}(\tilde{m}))\leqslant \cdots \leqslant h^{0}(X,{\mathcal{Q}}(\tilde{m}))=\unicode[STIX]{x1D712}(X,{\mathcal{Q}}(\tilde{m}))\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \displaystyle |q_{0}|\binom{\tilde{m}+d-2}{d-2}+|q_{1}|\binom{\tilde{m}+d-3}{d-3}+\cdots +|q_{d-2}|.\nonumber\end{eqnarray}$$

This implies that $h^{0}(X,{\mathcal{Q}}_{i})\leqslant p^{d-1}P^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d-1},\tilde{m})$ , where $P^{d}(X_{0},\ldots ,X_{d-1},Y)$ is a universal polynomial function with rational coefficients. Therefore,

$$\begin{eqnarray}C_{0}({\mathcal{Q}})\leqslant (d-1)p^{d-1}P^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d-1},\tilde{m}).\end{eqnarray}$$

The inequality for $D_{0}({\mathcal{Q}})$ follows similarly. This proves the assertion (2) and hence the lemma.◻

Proof. Without loss of generality, one can assume that $m_{0}\geqslant 1$ and $n_{2}\geqslant 1$ . Since ${\mathcal{O}}_{X}$ is $\bar{m}$ -regular, the sheaf ${\mathcal{M}}_{1}$ is an $\bar{m}+m_{0}$ -regular sheaf of ${\mathcal{O}}_{X}$ -modules of dimension $d-2$ . Therefore, for any ${\mathcal{M}}_{1}$ -sequence $a_{1},\ldots ,a_{d-2}$ , the sheaf of ${\mathcal{O}}_{X}$ -modules ${\mathcal{M}}_{1i}:={\mathcal{M}}_{1}/(a_{d-2},\ldots ,a_{i+1}){\mathcal{M}}_{1}$ is $\bar{m}+m_{0}$ -regular. This implies

$$\begin{eqnarray}\displaystyle h^{0}(X,{\mathcal{M}}_{1i}) & {\leqslant} & \displaystyle h^{0}(X,{\mathcal{M}}_{1i}(\bar{m}+m_{0}))\nonumber\\ \displaystyle & {\leqslant} & \displaystyle h^{0}(X,{\mathcal{M}}_{1}(\bar{m}+m_{0}))\leqslant h^{0}(X,{\mathcal{O}}_{X}(\bar{m}+m_{0}))\nonumber\\ \displaystyle & = & \displaystyle \tilde{e}_{0}\binom{\bar{m}+m_{0}+d-1}{d-1}-\tilde{e}_{1}\binom{\bar{m}+m_{0}+d-2}{d-2}\nonumber\\ \displaystyle & & \displaystyle +\cdots +(-1)^{d-1}\tilde{e}_{d-1}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle {\tilde{e}_{0}}^{2}\binom{\bar{m}+d+d-1}{d-1}m_{0}^{d-1}+{\tilde{e}_{1}}^{2}\binom{\bar{m}+d+d-2}{d-2}m_{0}^{d-2}\nonumber\\ \displaystyle & & \displaystyle +\cdots +{\tilde{e}_{d-1}}^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle {m_{0}}^{d-1}\tilde{P}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d-1},\bar{m}),\nonumber\end{eqnarray}$$

where the second last inequality follows as $\tilde{e}_{i}\leqslant {\tilde{e}_{i}}^{2}$ and $\bar{m}+m_{0}+k\leqslant (\bar{m}+d+k)m_{0}$ , for any $k\geqslant 0$ , and $\tilde{P}^{d}(X_{0},\ldots ,X_{d-1},Y)$ is a universal polynomial function with rational coefficients.

Let $e_{i}({\mathcal{M}}_{1})$ denote the $i$ th coefficient of the Hilbert polynomial of ${\mathcal{M}}_{1}$ with respect to the line bundle ${\mathcal{O}}_{X}(1)$ . Then, by Lemma A.1, we have $e_{i}({\mathcal{M}}_{1})\leqslant m_{0}^{i+1}P_{i}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{i})$ , where $P_{i}^{d}(X_{0},\ldots ,X_{i})$ is a universal polynomial with rational coefficients.

Now, the bound for $2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1})$ follows. The identical proof follows for ${\mathcal{M}}_{2}$ .◻

Notations 2.9. For a pair $(R,I)$ , where $R$ is a standard graded ring of $\text{char}~p>0$ and of dimension $d\geqslant 2$ , and $I\subset R$ is a graded ideal of finite colength, we define (similarly to the sequence of functions that had been defined in [Reference TrivediT4]) a sequence of functions $\{f_{n}:[1,\infty )\rightarrow [0,\infty )\}_{n}$ as follows.

Fix $n\in \mathbb{N}$ and denote $q=p^{n}$ . Let $x\in [1,\infty )$ ; then, there exists a unique nonnegative integer $m$ such that $(m+q)/q\leqslant x<(m+q+1)/q$ . Then,

$$\begin{eqnarray}f_{n}(x):=f_{n}(R,I)(x)=\frac{\ell (R/I^{[q]})_{m+q}}{q^{d-1}}.\end{eqnarray}$$

Proof. Since $R$ is a standard graded ring, for $m\geqslant n_{0}\unicode[STIX]{x1D707}q$ , we have $R_{m}\subseteq (R_{n_{0}})^{\unicode[STIX]{x1D707}q}\subseteq I^{\unicode[STIX]{x1D707}q}\subseteq I^{[q]}$ . This implies that $\ell (R/I^{[q]})_{m}=0$ , if $m\geqslant n_{0}\unicode[STIX]{x1D707}q$ . Therefore, support $f_{n}\subseteq [1,n_{0}\unicode[STIX]{x1D707}]$ , for every $n\geqslant 0$ . This proves the lemma.◻

Proof. Fix $x\in [1,\infty )$ . Therefore, for given $q=p^{n}$ , there exists a unique integer $m\geqslant 0$ , such that $(m+q)/q\leqslant x<(m+q+1)/q$ and

$$\begin{eqnarray}\frac{(m+q)p+n_{2}}{qp}\leqslant x<\frac{(m+q)p+n_{2}+1}{qp},\quad \text{for some}~0\leqslant n_{2}<p.\end{eqnarray}$$

Hence,

$$\begin{eqnarray}f_{n}(x)=\frac{1}{q^{d-1}}\ell \left(\frac{R}{I^{[q]}}\right)_{m+q}\end{eqnarray}$$

and

$$\begin{eqnarray}f_{n+1}(x)=\frac{1}{(qp)^{d-1}}\ell \left(\frac{R}{I^{[qp]}}\right)_{mp+qp+n_{2}}.\end{eqnarray}$$

Now, by Equation (4) in Remark 2.3, we have

(9) $$\begin{eqnarray}\left|f_{n}(x)-\frac{\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})}{q^{d-1}}\right|<\frac{C_{R}}{q^{d-1}}\quad \text{and}\end{eqnarray}$$

$$\begin{eqnarray}\left|f_{n+1}(x)-\frac{\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})}{(qp)^{d-1}}\right|<\frac{C_{R}}{(qp)^{d-1}}.\end{eqnarray}$$

Consider the short exact sequence of ${\mathcal{O}}_{X}$ -modules

$$\begin{eqnarray}0\longrightarrow {\mathcal{O}}_{X}(-d)\longrightarrow {\mathcal{O}}_{X}\longrightarrow {\mathcal{M}}_{1}\longrightarrow 0.\end{eqnarray}$$

Then, for any locally free sheaf $P$ of ${\mathcal{O}}_{X}$ -modules and for $m\geqslant 0$ , we have the following short exact sequence of ${\mathcal{O}}_{X}$ -modules:

$$\begin{eqnarray}0\longrightarrow F^{n\ast }P\otimes {\mathcal{O}}_{X}(-d+m)\longrightarrow F^{n\ast }P\otimes {\mathcal{O}}_{X}(m)\longrightarrow F^{n\ast }P\otimes {\mathcal{M}}_{1}(m)\longrightarrow 0.\end{eqnarray}$$

Since

$$\begin{eqnarray}\displaystyle \text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X}) & = & \displaystyle h^{0}(X,{\mathcal{O}}_{X}(m+q))-\mathop{\sum }_{i}h^{0}(X,{\mathcal{O}}_{X}(m+q-qd_{i}))\nonumber\\ \displaystyle & & \displaystyle +\,h^{0}(X,(F^{n\ast }V)(m)),\nonumber\end{eqnarray}$$

we have (by taking $P=V~\text{and}=\sum \,{\mathcal{O}}_{X}(1-d_{i})$ , respectively)

$$\begin{eqnarray}\displaystyle & & \displaystyle |\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{m-d,q}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant |h^{0}(X,{\mathcal{O}}_{X}(m+q))-h^{0}(X,{\mathcal{O}}_{X}(m-d+q))|\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\mathop{\sum }_{i}|h^{0}(X,{\mathcal{O}}_{X}(m+q-qd_{i}))\nonumber\\ \displaystyle & & \displaystyle \qquad -\,h^{0}(X,{\mathcal{O}}_{X}(m-d+q-qd_{i}))|+|h^{0}(X,(F^{n\ast }V)(m))\nonumber\\ \displaystyle & & \displaystyle \qquad -\,h^{0}(X,(F^{n\ast }V)(m-d))|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant h^{0}(X,{\mathcal{M}}_{1}(m+q))+h^{0}\left(X,\mathop{\sum }_{i}{\mathcal{O}}_{X}(q-qd_{i})\otimes {\mathcal{M}}_{1}(m)\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,h^{0}(X,F^{n\ast }V\otimes {\mathcal{M}}_{1}(m)).\nonumber\end{eqnarray}$$

Let $d-2=0$ . Then, ${\mathcal{M}}_{1}$ is a zero-dimensional sheaf, which implies that $h^{0}(X,{\mathcal{M}}_{1}(m))=h^{0}(X,{\mathcal{M}}_{1})$ , for every $m\in \mathbb{Z}$ . Moreover, if $P$ is a locally free sheaf of ${\mathcal{O}}_{X}$ -modules, then $h^{0}(X,P\otimes {\mathcal{M}}_{1})=(\text{rank}~P)h^{0}(X,{\mathcal{M}}_{1})$ . Therefore, we get

$$\begin{eqnarray}\displaystyle |\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{m-d,q}({\mathcal{O}}_{X})| & {\leqslant} & \displaystyle [1+\unicode[STIX]{x1D707}+(\unicode[STIX]{x1D707}-1)]h^{0}(X,{\mathcal{M}}_{1})\nonumber\\ \displaystyle & = & \displaystyle 2\unicode[STIX]{x1D707}C_{0}({\mathcal{M}}_{1}).\nonumber\end{eqnarray}$$

If $d-2>0$ then, by Lemma 2.5,

$$\begin{eqnarray}\displaystyle & & \displaystyle |\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{m-d,q}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant D_{0}({\mathcal{M}}_{1})(m+q)^{d-2}+2(\unicode[STIX]{x1D707})C_{0}({\mathcal{M}}_{1})(m^{d-2}+1)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (\unicode[STIX]{x1D707})[2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1})](m+q)^{d-2}.\nonumber\end{eqnarray}$$

Therefore, for $d-1\geqslant 1$ , we have

(10) $$\begin{eqnarray}\displaystyle & & \displaystyle |p^{d-1}\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})-p^{d-1}\text{coker}~\unicode[STIX]{x1D719}_{m-d,q}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (\unicode[STIX]{x1D707})[2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1})](m+q)^{d-2}p^{d-1}.\end{eqnarray}$$

Since, for a locally free sheaf $P$ , we have

$$\begin{eqnarray}h^{0}(X,F^{n\ast }P\otimes (F_{\ast }{\mathcal{O}}_{X})(m))=h^{0}(X,F^{(n+1)\ast }P\otimes {\mathcal{O}}_{X}(mp)),\end{eqnarray}$$

the short exact sequence

$$\begin{eqnarray}0\longrightarrow \oplus ^{p^{d-1}}{\mathcal{O}}_{X}(-d)\longrightarrow F_{\ast }{\mathcal{O}}_{X}\longrightarrow {\mathcal{Q}}\longrightarrow 0,\end{eqnarray}$$

as given in the statement of Lemma 2.7, gives a canonical long exact sequence

$$\begin{eqnarray}\displaystyle 0 & \longrightarrow & \displaystyle \oplus H^{0}(X,(F^{n\ast }P)(m-d))\longrightarrow H^{0}(X,(F^{(n+1)\ast }P)(mp))\nonumber\\ \displaystyle & \longrightarrow & \displaystyle H^{0}(X,F^{n\ast }P\otimes {\mathcal{Q}}(m)),\nonumber\end{eqnarray}$$

which implies (by taking $P=V$ and $V=\sum \,{\mathcal{O}}_{X}(1-d_{i})$ , respectively)

$$\begin{eqnarray}\displaystyle \text{coker}~\unicode[STIX]{x1D719}_{mp,qp}({\mathcal{O}}_{X}) & = & \displaystyle h^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(m+q))\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{i}h^{0}(X,(F_{\ast }{\mathcal{O}}_{X})(q-qd_{i}+m))\nonumber\\ \displaystyle & & \displaystyle +\,h^{0}(X,F^{n\ast }V\otimes (F_{\ast }{\mathcal{O}}_{X})(m+q)).\nonumber\end{eqnarray}$$

Therefore, we have

(11) $$\begin{eqnarray}\displaystyle & & \displaystyle |p^{d-1}\text{coker}~\unicode[STIX]{x1D719}_{(m-d),q}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{mp,qp}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (\unicode[STIX]{x1D707})[2C_{0}({\mathcal{Q}})+D_{0}({\mathcal{Q}})](m+q)^{d-2}.\end{eqnarray}$$

The short exact sequence of ${\mathcal{O}}_{X}$ -modules

$$\begin{eqnarray}0\longrightarrow {\mathcal{O}}_{X}\longrightarrow {\mathcal{O}}_{X}(n_{2})\longrightarrow {\mathcal{M}}_{2}\longrightarrow 0\end{eqnarray}$$

gives

$$\begin{eqnarray}\displaystyle 0 & \longrightarrow & \displaystyle H^{0}(X,(F^{(n+1)\ast }P)(mp))\longrightarrow H^{0}(X,(F^{(n+1)\ast }P)(mp+n_{2}))\nonumber\\ \displaystyle & \longrightarrow & \displaystyle H^{0}(X,(F^{(n+1)\ast }P)\otimes {\mathcal{M}}_{2}(mp))\longrightarrow \cdots \,,\nonumber\end{eqnarray}$$

which gives

$$\begin{eqnarray}\displaystyle & & \displaystyle \!\!|\text{coker}~\unicode[STIX]{x1D719}_{mp,qp}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \!\!\quad \leqslant \left[h^{0}(X,F^{(n+1)\ast }V\otimes {\mathcal{M}}_{2}(mp))+h^{0}\!\left(\!X,\mathop{\sum }_{i}{\mathcal{O}}_{X}(qp-qpd_{i})\otimes {\mathcal{M}}_{2}(mp)\!\right)\right.\nonumber\\ \displaystyle & & \displaystyle \!\!\qquad +\left.h^{0}(X,{\mathcal{M}}_{2}(mp+qp))\vphantom{\left(\mathop{\sum }_{i}\right)}\right]\leqslant 2(\unicode[STIX]{x1D707})C_{0}({\mathcal{M}}_{2})((mp)^{d-2}+1)\nonumber\\ \displaystyle & & \displaystyle \!\!\qquad +\,(\unicode[STIX]{x1D707})D_{0}({\mathcal{M}}_{2})(mp+qp)^{d-2}.\nonumber\end{eqnarray}$$

Therefore,

(12) $$\begin{eqnarray}\displaystyle & & \displaystyle |\text{coker}~\unicode[STIX]{x1D719}_{mp,qp}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (\unicode[STIX]{x1D707})[2C_{0}({\mathcal{M}}_{2})+D_{0}({\mathcal{M}}_{2})](mp+qp)^{d-2}.\end{eqnarray}$$

Combining Equations (10)–(12), we get

$$\begin{eqnarray}\displaystyle (A) & := & \displaystyle |p^{d-1}\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})-\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (\unicode[STIX]{x1D707})[2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1})](m+q)^{d-2}p^{d-1}\nonumber\\ \displaystyle & & \displaystyle +\,(\unicode[STIX]{x1D707})[2C_{0}({\mathcal{Q}})+D_{0}({\mathcal{Q}})](m+q)^{d-2}\nonumber\\ \displaystyle & & \displaystyle +\,(\unicode[STIX]{x1D707})[2C_{0}({\mathcal{M}}_{2})+D_{0}({\mathcal{M}}_{2})](mp+qp)^{d-2}.\nonumber\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}\displaystyle (A) & {\leqslant} & \displaystyle (\unicode[STIX]{x1D707})(m+q)^{d-2} [\,(2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1}))p^{d-1}+(2C_{0}({\mathcal{Q}})+D_{0}({\mathcal{Q}}))\nonumber\\ \displaystyle & & \displaystyle +\,(2C_{0}({\mathcal{M}}_{2})+D_{0}({\mathcal{M}}_{2}))p^{d-2} ]\!.\nonumber\end{eqnarray}$$

Now, if we denote $\bar{P}_{i}^{d}=\bar{P}_{i}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d-1},\tilde{m})$ , for $i=1,2$ , and 3, then, by Lemma 2.7,

$$\begin{eqnarray}2C_{0}({\mathcal{Q}})+D_{0}({\mathcal{Q}})\leqslant p^{d-1}\tilde{P}_{1}^{d},\end{eqnarray}$$

and, by Lemma 2.8, we have

$$\begin{eqnarray}2C_{0}({\mathcal{M}}_{1})+D_{0}({\mathcal{M}}_{1})\leqslant d^{d-1}\tilde{P}_{2}^{d}\end{eqnarray}$$

and

$$\begin{eqnarray}2C_{0}({\mathcal{M}}_{2})+D_{0}({\mathcal{M}}_{2})\leqslant n_{2}^{d-1}\tilde{P}_{3}^{d}\leqslant p^{d-1}\tilde{P}_{3}^{d},\end{eqnarray}$$

where the last inequality follows as $n_{2}<p$ . Therefore, we have

$$\begin{eqnarray}(A)\leqslant (\unicode[STIX]{x1D707})(m+q)^{d-2}[p^{d-1}d^{d-1}\bar{P}_{2}^{d}+p^{d-1}\bar{P}_{1}^{d}+p^{d-1}p^{d-2}\bar{P}_{3}^{d}].\end{eqnarray}$$

Now, multiplying the above inequality by $1/(qp)^{d-1}$ , we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\frac{\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})}{q^{d-1}}-\frac{\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})}{(qp)^{d-1}}\right|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (\unicode[STIX]{x1D707})\frac{(m+q)^{d-2}}{q^{d-1}}[d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{1}^{d}+p^{d-2}\bar{P}_{3}^{d}].\nonumber\end{eqnarray}$$

Moreover, by Remark 2.3,

$$\begin{eqnarray}\displaystyle m+q & {\geqslant} & \displaystyle \bar{m}+n_{0}\left(\mathop{\sum }_{i}d_{i}\right)q+q\Rightarrow \text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})\nonumber\\ \displaystyle & = & \displaystyle \text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})=0.\nonumber\end{eqnarray}$$

Furthermore, $m+q\leqslant \bar{m}+n_{0}(\sum _{i}d_{i})q+q,\Rightarrow (m+q)^{d-2}\leqslant L_{0}q^{d-2}$ , where

$$\begin{eqnarray}L_{0}=\left(\bar{m}+n_{0}\left(\mathop{\sum }_{i}d_{i}\right)+1\right)^{d-2}.\end{eqnarray}$$

Therefore, for every $m\geqslant 0$ and $n\geqslant 1$ , where $q=p^{n}$ , we have

$$\begin{eqnarray}\displaystyle & & \displaystyle |\text{coker}~\unicode[STIX]{x1D719}_{m,q}({\mathcal{O}}_{X})/q^{d-1}-\text{coker}~\unicode[STIX]{x1D719}_{mp+n_{2},qp}({\mathcal{O}}_{X})/(qp)^{d-1}|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant ((\unicode[STIX]{x1D707})L_{0}q^{d-2}[d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{1}^{d}+p^{d-2}\bar{P}_{3}^{d}])/q^{d-1}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant ((\unicode[STIX]{x1D707})L_{0}[d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{1}^{d}+\bar{P}_{3}^{d}]p^{d-2}q^{d-2})/q^{d-1}.\nonumber\end{eqnarray}$$

Now, by Equation (9), we have

$$\begin{eqnarray}\displaystyle |f_{n}(x)-f_{n+1}(x)| & {\leqslant} & \displaystyle \frac{C_{R}}{q^{d-1}}+\frac{C_{R}}{(qp)^{d-1}}+(\unicode[STIX]{x1D707})L_{0}[d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{1}^{d}+\bar{P}_{3}^{d}]\frac{p^{d-2}}{q}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle C\frac{p^{d-2}}{q},\nonumber\end{eqnarray}$$

where $C=2C_{R}+(\unicode[STIX]{x1D707})L_{0}(d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{1}^{d}+\bar{P}_{3}^{d})$ , which proves the proposition. ◻

Proof. Note that

$$\begin{eqnarray}\mathop{\sum }_{m=0}^{\infty }\frac{1}{p^{nd}}\ell \left(\frac{R}{I^{[p^{n}]}}\right)_{m+q}=\int _{1}^{\infty }f_{n}(x)\,dx=\int _{1}^{n_{0}\unicode[STIX]{x1D707}}f_{n}(x)\,dx,\end{eqnarray}$$

where the last equality follows from Lemma 2.10. Hence,

$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\frac{1}{(p^{n})^{d}}\ell \left(\frac{R}{I^{[p^{n}]}}\right)-\frac{1}{(p^{n+1})^{d}}\ell \left(\frac{R}{I^{[p^{n+1}]}}\right)\right|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \left|\frac{1}{p^{nd}}\ell \left(\frac{R}{\mathbf{m}^{p^{n}}}\right)-\frac{1}{p^{(n+1)d}}\ell \left(\frac{R}{\mathbf{m}^{p^{n+1}}}\right)\right|\nonumber\\ \displaystyle & & \displaystyle \qquad +\left|\int _{1}^{n_{0}\unicode[STIX]{x1D707}}f_{n}(x)\,dx-\int _{1}^{n_{0}\unicode[STIX]{x1D707}}f_{n+1}(x)\,dx\right|.\nonumber\end{eqnarray}$$

If $p^{n}\leqslant \bar{m}$ , then

$$\begin{eqnarray}\displaystyle \left|\frac{1}{p^{nd}}\ell \left(\frac{R}{\mathbf{m}^{p^{n}}}\right)-\frac{1}{p^{(n+1)d}}\ell \left(\frac{R}{\mathbf{m}^{p^{n+1}}}\right)\right| & {\leqslant} & \displaystyle \left|\frac{P_{(R,\mathbf{m})}(\bar{m})}{p^{nd}}+\frac{P_{(R,\mathbf{m})}(\bar{m}^{2})}{p^{(n+1)d}}\right|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{P_{(R,\mathbf{m})}(\bar{m}^{2})}{p^{n}},\nonumber\end{eqnarray}$$

where $P_{(R,\mathbf{m})}$ is the Hilbert polynomial of $R$ with respect to $\mathbf{m}$ . If $p^{n}>\bar{m}$ , then there exists a universal polynomial function $P_{6}^{d}(X_{0},\ldots ,X_{d})$ with rational coefficients such that

$$\begin{eqnarray}\text{L.H.S.}\leqslant \left|\frac{P_{(R,\mathbf{m})}(p^{n})}{p^{nd}}-\frac{P_{(R,\mathbf{m})}(p^{n+1})}{p^{(n+1)d}}\right|\leqslant \frac{P_{6}^{d}(\tilde{e}_{0},\tilde{e}_{1},\ldots ,\tilde{e}_{d})}{p^{n}}.\end{eqnarray}$$

Therefore, combining this with Proposition 2.11, part (1), we get a universal polynomial function $P_{4}^{d}(X_{0},\ldots ,X_{d},Y)$ with rational coefficients such that

$$\begin{eqnarray}\!\left|\frac{1}{(p^{n})^{d}}\ell \left(\frac{R}{I^{[p^{n}]}}\right)\!-\frac{1}{(p^{n+1})^{d}}\ell \left(\frac{R}{I^{[p^{n+1}]}}\right)\right|\leqslant \frac{P_{4}^{d}(\tilde{e}_{0},\tilde{e}_{1},\ldots ,\tilde{e}_{d})}{p^{n}}+\frac{(n_{0}\unicode[STIX]{x1D707}-1)C}{p^{n-d+2}}.\end{eqnarray}$$

Since $d\geqslant 2$ , the corollary follows.◻

Notations 3.3. Given a spread $(A,R_{A},I_{A})$ as above, for a closed point $s\in \text{Spec}(A)$ , we define $R_{s}=R_{A}\otimes _{A}\bar{k}(s)$ and the ideal $I_{s}=\text{Im}(I_{A}\otimes _{A}\bar{k}(s))\subset R_{s}$ . Similarly, for $X_{A}:=\text{Proj}~R_{A}$ , we define $X_{s}:=X\otimes \bar{k}(s)=\text{Proj}~R_{s}$ , and, for a coherent sheaf $V_{A}$ on $X_{A}$ , we define $V_{s}=V_{A}\otimes \bar{k}(s)$ .

Proof of Theorem 1.1.

For given $(R,I)$ , and a given spread $(A,R_{A},I_{A})$ , we can choose a spread $(A^{\prime },R_{A^{\prime }},I_{A^{\prime }})$ , where $A\subset A^{\prime }$ , such that the induced projective morphism of Noetherian schemes $\tilde{\unicode[STIX]{x1D70B}}:X_{A^{\prime }}\longrightarrow A^{\prime }$ is flat, and, for every $s\in \text{Spec}(A^{\prime })$ , the scheme $X_{s}$ is an integral scheme over $\bar{k}(s)$ of dimension $=d-1$ . Let $R_{A^{\prime }}=\oplus _{i\geqslant 0}(R_{A^{\prime }})_{i}$ , and let $(R_{A^{\prime }})_{1}$ be generated by $N$ elements as an $A^{\prime }$ -module. Then, the canonical graded surjective map

$$\begin{eqnarray}A^{\prime }[X_{0},\ldots ,X_{N}]\longrightarrow R_{A^{\prime }}\end{eqnarray}$$

gives a closed immersion $X_{A^{\prime }}=\text{Proj}~R_{A^{\prime }}\longrightarrow \mathbb{P}_{A^{\prime }}^{N}$ such that ${\mathcal{O}}_{X_{A^{\prime }}}(1)={\mathcal{O}}_{\mathbb{P}_{A^{\prime }}^{N}}(1)\mid _{X_{A^{\prime }}}$ . Let $X_{s}=X_{A^{\prime }}\otimes \bar{k}(s)$ . Then, $X_{s}=\text{Proj}~R_{s}$ , and ${\mathcal{O}}_{X_{s}}(1)$ is the canonical line bundle induced by ${\mathcal{O}}_{X_{A^{\prime }}}(1)$ . Let $s_{0}=\text{Spec}Q(A)=\text{Spec}Q(A^{\prime })$ be the generic point of $\text{Spec}(A^{\prime })$ . We now have the following.

1. (1) The Hilbert polynomial for the pair $(X_{s},{\mathcal{O}}_{X_{s}}(1))$ is

   $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D712}(X_{s},{\mathcal{O}}_{X_{s}}(m)) & = & \displaystyle \tilde{e}_{0}\binom{m+d-1}{d-1}-\tilde{e}_{1}\binom{m+d-2}{d-2}\nonumber\\ \displaystyle & & \displaystyle +\cdots +(-1)^{d-1}\tilde{e}_{d-1},\nonumber\end{eqnarray}$$
   where the coefficients $\tilde{e}_{i}$ are as above for $(X,{\mathcal{O}}_{X}(1))$ (from $\text{char}~0$ ).

   In particular, $\dim ~X_{s}=d-1$ and we have the following.

2. (2) By Remark 3.1, there exists $\bar{m}=P_{5}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d-1})$ such that $(R_{s})_{m}=H^{0}(X_{s},{\mathcal{O}}_{X_{s}}(m))$ for all $m\geqslant \bar{m}$ , and $(X_{s},{\mathcal{O}}_{X_{s}}(1))$ is $\bar{m}$ -regular.

3. (3) Moreover, by the semicontinuity theorem (Chapter III, Theorem 12.8 in [H]), by shrinking $\text{Spec}(A^{\prime })$ further, we have $h^{i}(X_{s},{\mathcal{O}}_{X_{s}}(\bar{m}))$ , and $h^{0}(X_{s},{\mathcal{O}}_{X_{s}})$ is independent of  $s$ , for all $i\geqslant 0$ .

4. (4) Again, by shrinking $\text{Spec}(A^{\prime })$ (if necessary), can choose $n_{0}\in \mathbb{N}$ such that $(R_{A^{\prime }})_{1}^{n_{0}}\subseteq I_{A^{\prime }}$ . This implies that $(R_{s})_{1}^{n_{0}}\subseteq I_{s}$ , for all $s\in \text{Spec}(A^{\prime })$ .

Let $s\in \text{Spec}(A^{\prime })$ , and let $p_{s}=\text{char}~k(s)$ . We sketch the proof of the existence of the map $f_{R_{s},I_{s}}:[1,\infty )\rightarrow \mathbb{R}$ and its relation to $e_{HK}(R,I)$ (note that we have proved this in a more general setting in [Reference TrivediT4]). By Proposition 2.11, for any given $s$ , the sequence ${\{f_{n}^{s}\}}_{n}$ of functions is uniformly convergent. Let $f_{R_{s},I_{s}}(x)=\lim _{n\rightarrow \infty }f_{n}(R_{s},I_{s})(x)$ . This implies that $\lim _{n\rightarrow \infty }\int _{1}^{\infty }f_{n}(R_{s},I_{s})(x)=\int _{1}^{\infty }f_{R_{s},I_{s}}(x)$ , as, by Lemma 2.10, there is a compact set containing $\cup _{n}\text{supp}~f_{n}(R_{s},I_{s})$ . On the other hand,

$$\begin{eqnarray}\displaystyle e_{HK}(R_{s},I_{s}) & = & \displaystyle \lim _{n\rightarrow \infty }\frac{1}{p_{s}^{nd}}\ell \left(\frac{R_{s}}{I_{s}^{[p_{s}^{n}]}}\right)\nonumber\\ \displaystyle & = & \displaystyle \lim _{n\rightarrow \infty }\frac{1}{p_{s}^{nd}}\ell \left(\frac{R_{s}}{\mathbf{m}_{s}^{p_{s}^{n}}}\right)+\lim _{n\rightarrow \infty }\frac{1}{p_{s}^{nd}}\mathop{\sum }_{m\geqslant 0}\ell \left(\frac{R}{I_{s}^{[p_{s}^{n}]}}\right)_{m+p_{s}^{n}}\nonumber\\ \displaystyle & = & \displaystyle \frac{e_{0}(R_{s})}{d!}+\lim _{n\rightarrow \infty }\int _{1}^{\infty }f_{n}(R_{s},I_{s})(x)\,dx\nonumber\\ \displaystyle & = & \displaystyle \frac{e_{0}(R_{s})}{d!}+\int _{1}^{\infty }f_{R_{s},I_{s}}(x)\,dx,\nonumber\end{eqnarray}$$

where $\mathbf{m}_{s}$ is the graded maximal ideal of $R_{s}$ and $e_{0}(R_{s})$ denotes the Hilbert–Samuel multiplicity of $R_{s}$ with respect to $\mathbf{m}_{s}$ .

Now, by Proposition 2.11, there exists a constant

$$\begin{eqnarray}C=2C_{R_{s}}+\unicode[STIX]{x1D707}\left(\bar{m}+n_{0}\left(\mathop{\sum }_{i=1}^{\unicode[STIX]{x1D707}}d_{i}\right)+1\right)^{d-2}(\bar{P}_{1}^{d}+d^{d-1}\bar{P}_{2}^{d}+\bar{P}_{3}^{d}),\end{eqnarray}$$

which is independent of the choice of $s$ in $\text{Spec}(A^{\prime })$ (as $C_{R_{s}}=\unicode[STIX]{x1D707}h^{0}(X_{s},{\mathcal{O}}_{X_{s}}(1))$ ), such that

(13) $$\begin{eqnarray}\Vert f_{n}(R_{s},I_{s})-f_{n+1}(R_{s},I_{s})\Vert \leqslant C/p_{s}^{n-d+2},\quad \text{for all}~n\geqslant 1.\end{eqnarray}$$

In particular, for given $m\geqslant d-1$ ,

$$\begin{eqnarray}\Vert f_{m}(R_{s},I_{s})-f_{R_{s},I_{s}}\Vert \leqslant \left[\frac{C}{p_{s}}+\frac{C}{p_{s}^{2}}+\frac{C}{p_{s}^{3}}+\cdots \,\right]\frac{1}{p_{s}^{m-(d-1)}}\leqslant \frac{2C}{p_{s}^{m-d+2}}.\end{eqnarray}$$

As $s\rightarrow s_{0}$ , we have $p_{s}=\text{char}~k(s)\rightarrow \infty$ , which implies

(14) $$\begin{eqnarray}\text{for any}\quad m\geqslant d-1,\quad \text{we have}~\lim _{p_{s}\rightarrow \infty }\Vert f_{m}(R_{s},I_{s})-f_{R_{s},I_{s}}\Vert =0.\end{eqnarray}$$

This completes the proof of Assertion (1) of the theorem.

It is easy to check that there exists a universal polynomial function $\bar{P}_{9}^{d}(X_{0},\ldots ,X_{d})$ with rational coefficients such that for $C_{2}:=\bar{P}_{9}^{d}(\tilde{e}_{0},\ldots ,\tilde{e}_{d})$ we have

$$\begin{eqnarray}\left|\frac{1}{p_{s}^{nd}}\ell \left(\frac{R_{s}}{\mathbf{m}_{s}^{p_{s}^{n}}}\right)-\frac{e_{0}(R_{s})}{d!}\right|\leqslant \frac{C_{2}}{p_{s}^{n}}.\end{eqnarray}$$

Moreover, for every $n\geqslant 1$ , the function $f_{n}(R_{s},I_{s})$ has support in the compact interval $[1,n_{0}\unicode[STIX]{x1D707}]$ . Therefore,

$$\begin{eqnarray}\displaystyle (A_{1}) & := & \displaystyle \left|\frac{1}{p_{s}^{nd}}\ell \left(\frac{R_{s}}{I_{s}^{[p_{s}^{n}]}}\right)-\frac{1}{p_{s}^{(n+1)d}}\ell \left(\frac{R_{s}}{I_{s}^{[p_{s}^{n+1}]}}\right)\right|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \left|\frac{1}{p_{s}^{nd}}\ell \left(\frac{R_{s}}{\mathbf{m}_{s}^{p_{s}^{n}}}\right)-\frac{e_{0}(R_{s})}{d!}\right|\nonumber\\ \displaystyle & & \displaystyle +\,\int _{1}^{n_{0}\unicode[STIX]{x1D707}}|f_{n}(R_{s},I_{s})(x)-f_{n+1}(R_{s},I_{s})(x)|\,dx.\nonumber\end{eqnarray}$$

By (13), we have

$$\begin{eqnarray}(A_{1})\leqslant \frac{C_{2}}{p_{s}^{n}}+\frac{(2C)(n_{0}\unicode[STIX]{x1D707}-1)}{p_{s}^{n-d+2}}.\end{eqnarray}$$

This proves Assertion (2). Now, similarly, Assertion (3) easily follows from (14).◻

Proof of Corollary 1.2.

This is easy to deduce from Theorem 1.1. ◻

Proof. Let us denote $f^{\infty }=f_{R,I}^{\infty }$ and $g^{\infty }=f_{S,J}^{\infty }$ . For $q=p_{s}^{n}$ , where $p_{s}=\text{char}~k(s)$ , we denote $f_{n}^{s}=f_{n}(R_{s},I_{s})$ and $g_{n}^{s}=f_{n}(S_{s},J_{s})$ , where $s\in \text{Spec}(A)$ denotes a closed point and $(A,R_{A},I_{A})$ and $(A,S_{A},J_{A})$ are spreads.

For any $n\geqslant 1$ , we have

$$\begin{eqnarray}f_{n}(R_{s}\#S_{s},I_{s}\#J_{s})(x)=F_{n}(R_{s})(x)g_{n}^{s}(x)+F_{n}(S_{s})(x)f_{n}^{s}(x)-f_{n}^{s}(x)g_{n}^{s}(x).\end{eqnarray}$$

For a spread $(A,R_{A},I_{A})$ , let $n_{0}$ and $\unicode[STIX]{x1D707}$ be positive integers such that ${(R_{A})_{1}}^{n_{0}}\subseteq I_{A}$ and ${(S_{A})_{1}}^{n_{0}}\subseteq J_{A}$ , and also $\unicode[STIX]{x1D707}(I_{A})$ , $\unicode[STIX]{x1D707}(J_{A})\leqslant \unicode[STIX]{x1D707}$ . Then, by Lemma 2.10,

(15) $$\begin{eqnarray}\mathop{\bigcup }_{n\geqslant 0,s\in \text{Spec}(A)}\text{Support}~(f_{n}^{s})\bigcup \mathop{\bigcup }_{n\geqslant 0,s\in \text{Spec}(A)}\text{Support}~(\mathop{g}_{n}^{s})\subseteq [1,n_{0}\unicode[STIX]{x1D707}].\end{eqnarray}$$

Moreover, there is a constant $C_{1}$ such that, for any $n\geqslant 1$ and every closed point $s\in \text{Spec}(A)$ , we have

$$\begin{eqnarray}f_{n}^{s}(x)\leqslant F_{n}(R_{s})(x)\leqslant C_{1}\quad \text{and}\quad g_{n}^{s}(x)\leqslant F_{n}(S_{s})(x)\leqslant C_{1},\end{eqnarray}$$

for all $x\in [1,n_{0}\unicode[STIX]{x1D707}]$ .

Since $f^{\infty }$ and $g^{\infty }$ exist, by Theorem 1.1(1), for given $n\geqslant d_{1}+d_{1}-2$ , we have

$$\begin{eqnarray}\lim _{p_{s}\rightarrow \infty }f_{n}^{s}=f^{\infty }\quad \text{and}\quad \lim _{p_{s}\rightarrow \infty }g_{n}^{s}=g^{\infty }.\end{eqnarray}$$

Therefore, by (15), for given $n\geqslant d_{1}+d_{2}-2$ , we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{p_{s}\rightarrow \infty }F_{n}(R_{s})(x)g_{n}^{s}(x)+F_{n}(S_{s})(x)f_{n}^{s}(x)-f_{n}^{s}(x)g_{n}^{s}(x)\nonumber\\ \displaystyle & & \displaystyle \quad =F_{R}(x)g^{\infty }(x)+F_{S}(x)f^{\infty }(x)-f^{\infty }(x)g^{\infty }(x).\nonumber\end{eqnarray}$$

Hence, for any $n\geqslant d_{1}+d_{2}-2$ ,

$$\begin{eqnarray}\lim _{p_{s}\rightarrow \infty }f_{n}(R_{s}\#S_{s},I_{s}\#J_{s})(x)=F_{R}(x)g^{\infty }(x)+F_{S}(x)f^{\infty }(x)-f^{\infty }(x)g^{\infty }(x).\end{eqnarray}$$

Now, by Theorem 1.1(1), the proposition follows. ◻

Proof. (1) Let us denote $f^{\infty }=f_{R,I}^{\infty }$ and $g^{\infty }=g^{\infty }(S,J)$ , and denote $f^{s}=f_{R_{s},I_{s}}$ and $g^{s}=f_{S_{s},J_{s}}$ .

We know, by the multiplicative property of the HK density functions (see [Reference TrivediT4, Proposition 2.18]), that

$$\begin{eqnarray}\displaystyle f_{R_{s}\#S_{s},I_{s}\#J_{s}}(x) & = & \displaystyle F_{R}(x)g^{s}(x)+F_{S}(x)f^{s}(x)-f^{s}(x)g^{s}(x)\nonumber\\ \displaystyle & = & \displaystyle (F_{R}(x)-f^{s}(x))g^{s}(x)+F_{S}(x)f^{s}(x)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle (F_{R}(x)-f^{s}(x))g^{\infty }(x)+F_{S}(x)f^{s}(x)\nonumber\\ \displaystyle & = & \displaystyle F_{R}(x)g^{\infty }(x)+f^{s}(x)[F_{S}(x)-g^{\infty }(x)]\nonumber\\ \displaystyle & {\geqslant} & \displaystyle F_{R}(x)g^{\infty }(x)+f^{\infty }(x)[F_{S}(x)-g^{\infty }(x)]\nonumber\\ \displaystyle & = & \displaystyle f_{R\#S,I\#J}^{\infty }(x),\nonumber\end{eqnarray}$$

where the third and fifth inequalities hold as $F_{R}(x)\geqslant f^{s}(x)$ and $F_{S}(x)\geqslant g^{s}(x)$ , for every $s\in \text{Spec}\,A$ , and the last equality follows from Proposition 4.3.

(2) Suppose that $I$ and $J$ are the ideals of $R$ and $S$ , respectively, and $s\in \text{Spec}(A)$ is a closed point such that $I_{s}\cap (R_{s})_{1}\neq 0$ and $J_{s}\cap (S_{s})_{1}\neq 0$ . Then, we make the following claim.

Claim. $F_{R}(x)>f^{s}(x)$ and $G_{S}(x)>g^{s}(x)$ , for all $x\geqslant 1$ .

Proof of the Claim. It is sufficient to prove that $F_{R}(x+1)>f^{s}(x+1)$ , for $x>0$ . Choose an integer $n_{0}$ such that $x\geqslant 1/p_{s}^{n_{0}}$ , where $p_{s}=\text{char}~k(s)$ . Let $q=p_{s}^{n}$ for some $n$ . For a given nonzero $y\in I_{s}\cap (R_{s})_{1}$ , we have an injective map of the $R_{s}$ -linear map ( $R$ is a domain) $\oplus _{m\geqslant 0}{(R_{s})_{m}\longrightarrow \oplus }_{m\geqslant 0}(I_{s}^{[q]})_{m+q},$ of degree $q$ , given by the multiplication by element $y^{q}$ . Therefore, $\ell (I_{s}^{[q]})_{m+q}\geqslant \ell (R_{s})_{m}$ , for all $m\geqslant 0$ . Since $\lfloor xq\rfloor =m$ if and only if $\lfloor (x+1)q\rfloor =m+q$ , we have $\ell (I_{s}^{[q]})_{\lfloor (x+1)q\rfloor }\geqslant \ell (R_{s})_{\lfloor xq\rfloor }$ . Hence, $\ell (R_{s}/I_{s}^{[q]})_{\lfloor (x+1)q\rfloor }\leqslant \ell (R_{s})_{\lfloor (x+1)q\rfloor }-\ell (R_{s})_{\lfloor xq\rfloor }$ .

Therefore, $f_{n}(R_{s},I_{s})(x+1)\leqslant F_{n}(R_{s})(x+1)-F_{n}(R_{s})(x)$ .

However,

$$\begin{eqnarray}\lim _{n\rightarrow \infty }F_{n}(R_{s})(x)=F_{R_{s}}(x)=\frac{e_{0}(R)x^{d-1}}{(d-1)!}\geqslant \frac{1}{(d-1)!}\frac{e_{0}(R)}{(p_{s}^{n_{0}})^{d-1}}>0,\end{eqnarray}$$

where $d=\dim R$ . This implies that $f^{s}(x+1)=f_{R_{s},I_{s}}(x+1)<F_{R}(x+1)$ . This proves the claim.

Now, retracing the above argument, we note that $f_{R_{s}\#S_{s},I_{s}\#J_{s}}=f_{R\#S,I\#J}^{\infty }$ if and only if

$$\begin{eqnarray}[F_{R}(x)-f^{s}(x)]g^{s}(x)=[F_{R}(x)-f^{s}(x)]g^{\infty }(x)\end{eqnarray}$$

and

$$\begin{eqnarray}[F_{S}(x)-g^{\infty }(x)]f^{s}(x)=[F_{S}(x)-g^{\infty }(x)]f^{\infty }(x).\end{eqnarray}$$

Hence, by the above claim, we have $f^{s}(x)=f^{\infty }(x)$ and $g^{s}(x)=g^{\infty }(x)$ for all $x>1$ . For $x=1$ , we have $F_{R}(x)=f^{s}(x)=f^{\infty }(x)$ and $F_{S}(x)=g^{s}(x)=g^{\infty }(x)$ . This proves the proposition.◻

Proof. We fix such an $s\in \text{Spec}~A$ and let $d=\deg X_{s}$ ( $d$ independent of  $s$ ), and let the HN filtration of $V_{s}$ be

$$\begin{eqnarray}0=E_{0}\subset E_{1}\subset \cdots \subset E_{l}\subset E_{l+1}=V_{s}.\end{eqnarray}$$

By [Reference LangerL, Theorem 2.7], there is $n_{1}\geqslant 1$ such that $F^{n_{1}\ast }V_{s}$ has the strong HN filtration. (Note that $n_{1}$ may depend on  $s$ .)

Then, by [Reference TrivediT2, Lemma 1.8], for $p_{s}=\text{char}~k(s)>4(\text{genus}(X_{s}))\text{rank}(V_{s})^{3}$ , the HN filtration of $F^{n_{1}\ast }V_{s}$ is

$$\begin{eqnarray}\displaystyle & 0=E_{00}\subset E_{01}\subset \cdots \subset E_{0t_{0}}\subset E_{0,(t_{0}+1)}=F^{n_{1}\ast }E_{1}=E_{10}\subset \cdots \subset & \displaystyle \nonumber\\ \displaystyle & E_{i-1(t_{i-1}+1)}=F^{n_{1}\ast }E_{i}=E_{i0}\subset E_{i1}\subset \cdots \subset E_{it_{i}}\subset E_{i(t_{i}+1)} & \displaystyle \nonumber\\ \displaystyle & E_{i(t_{i}+1)}=F^{n_{1}\ast }E_{i+1}=E_{i+1,0}\subset \cdots \subset F^{n_{1}\ast }V_{s}. & \displaystyle \nonumber\end{eqnarray}$$

Let, for $i\geqslant 0$ and $j\geqslant 1$ ,

$$\begin{eqnarray}a_{ij}=\frac{1}{p_{s}^{n_{1}}}\unicode[STIX]{x1D707}(E_{ij}/E_{i,j-1})\quad \text{and}\quad r_{ij}=\text{rank}(E_{ij}/E_{i,(j-1)}).\end{eqnarray}$$

Let

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{0}=1,~\text{and, for}~i\geqslant 1,~\text{let}~\unicode[STIX]{x1D707}_{i}=\unicode[STIX]{x1D707}(E_{i}/E_{i-1})\quad \text{and}\quad r_{i}=\text{rank}(E_{i}/E_{i-1}).\end{eqnarray}$$

Note that, for any $i\geqslant 1$ , the only possible inequalities are

$$\begin{eqnarray}a_{01}\geqslant \unicode[STIX]{x1D707}_{1}\geqslant a_{0,(t_{0}+1)}>\cdots >a_{i0}\geqslant \unicode[STIX]{x1D707}_{i+1}\geqslant a_{i,(t_{i}+1)},\end{eqnarray}$$

and also note that $a_{ij}\leqslant 0$ . By [Reference TrivediT2, Lemma 1.14], for a given  $i$ ,

(18) $$\begin{eqnarray}a_{ij}=\unicode[STIX]{x1D707}_{i+1}+O(1/p_{s}),\end{eqnarray}$$

where, by $O(1/p_{s})$ , we mean $O(1/p_{s})=C/p_{s}$ , where $|C|$ is bounded by a constant depending only on the degree of $X$ and rank of $V$ (and hence is independent of  $p_{s}$ ).

Claim. If $1-a_{ij_{0}}/d\leqslant x<1-a_{i(j_{0}+1)}/d$ , for some $i\geqslant 0$ and $j_{0}\geqslant 1$ , then we have the following.

1. (1) $-[a_{ij}r_{ij}+d(x-1)r_{ij}]=-[\unicode[STIX]{x1D707}_{i+1}r_{ij}+d(x-1)r_{ij}]+O(1/p_{s})$ , for any $1\leqslant k\leqslant t_{i}+1$ , and $-[a_{ij}r_{ij}+d(x-1)r_{ij}]\leqslant 0$ if $k\leqslant j_{0}$ .

2. (2) $-\sum _{k\geqslant j_{0}+1}[a_{ik}r_{ik}+d(x-1)r_{ik}]\geqslant -[\unicode[STIX]{x1D707}_{i+1}r_{i+1}+d(x-1)r_{i+1}]$ .

We skip the proof of the claim.

We also recall that, for $x$ , as in the above claim, by [Reference TrivediT4, Example 3.3], we have

$$\begin{eqnarray}f_{R_{s},I_{s}}(x)=-\mathop{\sum }_{j\geqslant j_{0}+1}[a_{ij}r_{ij}+d(x-1)r_{ij}]-\mathop{\sum }_{k\geqslant i+1,j\geqslant 1}[a_{kj}r_{kj}+d(x-1)r_{kj}].\end{eqnarray}$$

Let $x\geqslant 1$ ; then, $1-\unicode[STIX]{x1D707}_{i}/d\leqslant x<1-\unicode[STIX]{x1D707}_{i+1}/d$ , for some $i\geqslant 0$ . (Note that $-\unicode[STIX]{x1D707}_{i}\geqslant 0$ .) Now, there are three possibilities.

1. (1) $1-\unicode[STIX]{x1D707}_{i}/d\leqslant x<1-a_{i-1,(t_{i-1}+1)}/d$ . Then,

   $$\begin{eqnarray}1-\frac{a_{i-1,j_{0}}}{d}\leqslant x<1-\frac{a_{i-1,(j_{0}+1)}}{d},\quad \text{for some}~j_{0}\geqslant 1,\end{eqnarray}$$
   and
   $$\begin{eqnarray}\displaystyle f_{R_{s},I_{s}}(x) & = & \displaystyle -\mathop{\sum }_{j\geqslant j_{0}+1}[a_{i-1,j}r_{i-1,j}+d(x-1)r_{i-1,j}]\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{k\geqslant i+2}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}].\nonumber\end{eqnarray}$$
   Therefore, by the above claim part (2),
   $$\begin{eqnarray}f_{R_{s},I_{s}}(x)\geqslant -\mathop{\sum }_{k\geqslant i+1}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}].\end{eqnarray}$$

2. (2) $1-a_{i-1,(t_{i-1}+1)}/d\leqslant x<1-a_{i1}/d$ . Then,

   $$\begin{eqnarray}f_{R_{s},I_{s}}(x)=-\mathop{\sum }_{k\geqslant i+1}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}].\end{eqnarray}$$

3. (3) $1-a_{i1}/d\leqslant x<1-\unicode[STIX]{x1D707}_{i+1}/d$ . Then

   $$\begin{eqnarray}1-\frac{a_{ij_{0}}}{d}\leqslant x<1-\frac{a_{i,(j_{0}+1)}}{d},\quad \text{for some}~j_{0}\geqslant 1,\end{eqnarray}$$
   and
   $$\begin{eqnarray}f_{R_{s},I_{s}}(x)=-\mathop{\sum }_{j\geqslant j_{0}+1}[a_{ij}r_{ij}+d(x-1)r_{ij}]-\mathop{\sum }_{k\geqslant i+2}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}].\end{eqnarray}$$

   Therefore, again by the above claim part (2),

   $$\begin{eqnarray}f_{R_{s},I_{s}}(x)\geqslant -\mathop{\sum }_{k\geqslant i+1}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}].\end{eqnarray}$$

By (18), we have that $f_{R,I}^{\infty }=\lim _{p_{s}\rightarrow \infty }f_{R_{s},I_{s}}$ exists, and

$$\begin{eqnarray}\begin{array}{@{}lcl@{}}1\leqslant x<1-\unicode[STIX]{x1D707}_{1}/d & \Rightarrow & f_{R,I}^{\infty }(x)=-\left[\displaystyle \mathop{\sum }_{i\geqslant 1}\unicode[STIX]{x1D707}_{i}r_{i}+d(x-1)r_{i}\right],\\ 1-\unicode[STIX]{x1D707}_{i}/d\leqslant x<1-\unicode[STIX]{x1D707}_{i+1}/d & \Rightarrow & f_{R,I}^{\infty }(x)=-\left[\displaystyle \mathop{\sum }_{k\geqslant i+1}\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}\right]\!.\end{array}\end{eqnarray}$$

This implies that $f_{R_{s},I_{s}}\geqslant f_{R,I}^{\infty }$ for $1\leqslant x<1-a_{l,(t_{l}+1)}/d$ , and $f_{R_{s},I_{s}}=f_{R,I}^{\infty }=0$ otherwise. This proves part (1) of the theorem.

(2) If $V_{s}$ has strongly semistable HN filtration, then it is obvious that $f_{R_{s},I_{s}}=f_{R,I}^{\infty }$ . Let, as before, $n_{1}$ be such that $F^{n_{1}\ast }V$ has a strongly semistable HN filtration in the sense of [Reference LangerL, Theorem 2.7].

If the HN filtration of $V_{s}$ is not strongly semistable, then

$$\begin{eqnarray}0=F^{n_{1}\ast }E_{0}\subset F^{n_{1}\ast }E_{1}\subset \cdots \subset F^{n_{1}\ast }E_{l}\subset F^{n_{1}\ast }E_{l+1}=F^{n_{1}\ast }V\end{eqnarray}$$

is not the HN filtration of $F^{n_{1}\ast }V$ . Therefore, there exists $i\geqslant 0$ such that

$$\begin{eqnarray}F^{n_{1}\ast }E_{i}=E_{i0}\subset E_{i1}\subset \cdots \subset F^{n_{1}\ast }E_{i+1},\end{eqnarray}$$

where $E_{i1}\subset F^{n_{1}\ast }E_{i+1}$ . Since $a_{i1}>\unicode[STIX]{x1D707}_{i+1}$ , one can choose $1-a_{i1}/d<x_{0}\leqslant 1-\unicode[STIX]{x1D707}_{i+1}/d\leqslant 1-a_{i2}/d$ . Now,

$$\begin{eqnarray}\displaystyle f_{R_{s},I_{s}}(x) & = & \displaystyle -\mathop{\sum }_{j\geqslant 2}\left[a_{ij}r_{ij}+d(x-1)r_{ij}\right]-\mathop{\sum }_{k\geqslant i+2}\left[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}\right].\nonumber\\ \displaystyle & = & \displaystyle [a_{i1}r_{i1}+d(x-1)r_{i1}]-\mathop{\sum }_{k\geqslant i+1}[\unicode[STIX]{x1D707}_{k}r_{k}+d(x-1)r_{k}]>f_{R,I}^{\infty }.\nonumber\end{eqnarray}$$

This proves the theorem. ◻

Proof. The proof follows by Proposition 4.4 and Theorem 4.6. ◻

Notations 4.9. Let $R=k[x,y,z]/(h)$ be a plane trinomial curve of degree $d$ . That is, $h=M_{1}+M_{2}+M_{3}$ , where the $M_{i}$ are monomials of degree  $d$ . As given in [Reference MonskyMo2, Lemma 2.2], one can divide such an $h$ into two types.

1. (1) $h$ is irregular; that is, one of the points $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ of $\mathbb{P}^{2}$ has multiplicity ${\geqslant}d/2$ on the plane curve $h$ . Here, we define $\unicode[STIX]{x1D706}_{R}=1$ .

2. (2) $h$ is regular and hence is one of the following types (up to a change of variables).

   1. (a) $h=x^{a_{1}}y^{a_{2}}+y^{b_{1}}z^{b_{2}}+z^{c_{1}}x^{c_{2}}$ , where $a_{1},b_{1},c_{1}>d/2$ . Here, we define $\unicode[STIX]{x1D6FC}=a_{1}+b_{1}-d$ , $\unicode[STIX]{x1D6FD}=a_{1}+c_{1}-d$ , $\unicode[STIX]{x1D708}=b_{1}+c_{1}-d$ , and $\unicode[STIX]{x1D706}=a_{1}b_{1}+a_{2}c_{2}-b_{1}c_{2}$ .

   2. (b) $h=x^{d}+x^{a_{1}}y^{a_{2}}z^{a_{3}}+y^{b}z^{c}$ , where $a_{2},c>d/2$ . Here, we define $\unicode[STIX]{x1D6FC}=a_{2}$ , $\unicode[STIX]{x1D6FD}=c$ , $\unicode[STIX]{x1D708}=a_{2}+c-d$ , and $\unicode[STIX]{x1D706}=a_{2}c-a_{3}b$ .

   We denote $\unicode[STIX]{x1D706}_{h}=\unicode[STIX]{x1D706}/a$ , where $a=\text{g.c.d.}(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D708},\unicode[STIX]{x1D706})$ .

Proof. We can choose spreads $(A,S_{iA})$ with the property that $\text{char}~k(s)>\max {\{d_{1},\ldots ,d_{n}\}}^{2}$ , for every closed point $s\in \text{Spec}(A)$ . Now, for any irreducible plane curve given by $S=k[x,y,z]/(h)$ , let $S\longrightarrow \tilde{S}$ be the normalization of $S$ . Then, it is a finite graded map of degree 0 and $Q(S)=Q(\tilde{S})$ such that $\tilde{S}$ is a finitely generated $\mathbb{N}$ -graded two-dimensional domain over  $k$ . Now, for pairs $(S,\mathbf{m})$ and $(\tilde{S},\mathbf{m}\tilde{S})$ , we can choose a spread $(A,S_{A},\mathbf{m}_{A})$ and $(A,\tilde{S}_{A},\mathbf{m}\tilde{S}_{A})$ such that for every closed point $s\in \text{Spec}(A)$ , the natural map $S_{s}=S_{A}\otimes k(s)\longrightarrow \tilde{S}_{A}\otimes k(s)$ is a finite graded map of degree 0. This implies, for every $x\geqslant 0$ ,

$$\begin{eqnarray}\lim _{q\rightarrow \infty }\frac{1}{q}\ell \left(\frac{S_{s}}{\mathbf{m}^{[q]}}\right)_{\lfloor xq\rfloor }=\lim _{q\rightarrow \infty }\frac{1}{q}\ell \left(\frac{\tilde{S}_{s}}{\mathbf{m}{\tilde{S}_{s}}^{[q]}}\right)_{\lfloor xq\rfloor },\end{eqnarray}$$

as kernel and cokernel of the map $S_{s}\longrightarrow \tilde{S}_{s}$ are zero-dimensional. Therefore, $f_{S_{s},\mathbf{m}_{s}}=f_{\tilde{S}_{s},\mathbf{m}\tilde{S}_{s}}$ and $f_{S,\mathbf{m}}^{\infty }=f_{\tilde{S},\mathbf{m}\tilde{S}}^{\infty }$ . This also implies that $e_{HK}(S_{s},\mathbf{m}_{s})=e_{HK}(\tilde{S}_{s},\mathbf{m}\tilde{S}_{s})$ . (This inequality about $e_{HK}$ can also be found in [Reference MonskyMo1, Lemma 1.3], [Reference Watanabe and YoshidaWY1, Theorem 2.7], and [Reference Buchweitz, Chen and PardueBCP].) Let $\unicode[STIX]{x1D70B}:\tilde{X}_{s}=\text{Proj}~\tilde{S}_{s}\longrightarrow X_{s}=\text{Proj}~S_{s}$ be the induced map. We also choose a spread $(A,X_{A},V_{A})$ , where $V_{A}$ is given by

$$\begin{eqnarray}0\longrightarrow V_{A}\longrightarrow {\mathcal{O}}_{X_{A}}\oplus {\mathcal{O}}_{X_{A}}\oplus {\mathcal{O}}_{X_{A}}\longrightarrow {\mathcal{O}}_{X_{A}}(1)\longrightarrow 0\end{eqnarray}$$

and gives the syzygy bundle $V_{s}$ with its HN filtration as given in Example 4.5.

This gives a short exact sequence of sheaves of ${\mathcal{O}}_{\tilde{X}_{s}}$ -modules

$$\begin{eqnarray}0\longrightarrow \unicode[STIX]{x1D70B}^{\ast }V_{s}\longrightarrow {\mathcal{O}}_{\tilde{X}_{s}}\oplus {\mathcal{O}}_{\tilde{X}_{s}}\oplus {\mathcal{O}}_{\tilde{X}_{s}}\longrightarrow {\mathcal{O}}_{\tilde{X}_{s}}(1)\longrightarrow 0.\end{eqnarray}$$

Moreover, $\tilde{X}_{s}$ is a nonsingular curve. If $S$ is regular trinomial given by $h$ , then, by [Reference TrivediT5, Theorem 5.6], the bundle $\unicode[STIX]{x1D70B}^{\ast }(V_{s})$ is strongly semistable, provided that $\text{char}~k(s)\equiv \pm 1~(\text{mod}~2\unicode[STIX]{x1D706}_{h_{s}})$ . Therefore, by Theorem 4.6, we have $f_{\tilde{S}_{s},\mathbf{m}\tilde{S}_{s}}=f_{\tilde{S},\mathbf{m}\tilde{S}}^{\infty }$ . This implies that $f_{S_{s},\mathbf{m}_{s}}=f_{S,\mathbf{m}}^{\infty }$ , for $\text{char}~k(s)\equiv \pm 1~(\text{mod}~2\unicode[STIX]{x1D706}_{S_{s}})$ .

If $S$ is an irregular trinomial, then, by [Reference TrivediT5, Theorem 1.1], $\unicode[STIX]{x1D70B}^{\ast }V$ has an HN filtration $0\subset {\mathcal{L}}\subset \unicode[STIX]{x1D70B}^{\ast }V$ . Therefore, $0\subset {\mathcal{L}}_{s}\subset \unicode[STIX]{x1D70B}^{\ast }V_{s}$ is the HN filtration and hence the strong HN filtration (as $\text{rank}~V=2$ ), for $\unicode[STIX]{x1D70B}^{\ast }V_{s}$ , for every closed point $s\in \text{Spec}~A$ . In particular, by Theorem 4.6, $f_{S_{s},\mathbf{m}_{s}}=f_{S,\mathbf{m}}^{\infty }$ , for all such $s$ . Now, assertion (1) follows by Proposition 4.4(2).

If $S_{1}=k[x,y,z]/(h_{1})$ , where $h_{1}$ is as in statement (2) of the corollary, then $\unicode[STIX]{x1D70B}^{\ast }V_{s}$ is semistable, but not strongly semistable, if $\text{char}~k(s)\equiv \pm 2~(\text{mod}~\unicode[STIX]{x1D706}_{{h_{1}}_{s}})$ . In particular, by Corollary 4.7, $f_{{S_{1}}_{s},\mathbf{m}{S_{1}}_{s}}>f_{{S_{1}}_{s},\mathbf{m}{S_{1}}_{s}}^{\infty }$ , for such  $s$ . Therefore, the statement (2) follows from Proposition 4.4(2).◻