设S是域k上的多元多项式环, m是S的极大分次理想。对于有限生成分次S-模M, 当H_mⁱ(M)≠0时, 令a_i(M)=max{μ|[H_mⁱ(M)]_μ≠0};当H_mⁱ(M)=0时, 令a_i(M)=-∞.M的Castelnuovo-Mumford正则度定义为

$ {\rm{reg}}\left( M \right) = \mathop {\max }\limits_{i \ge 0} \left\{ {{a_i}\left( M \right) + i} \right\}. $

reg(M)是一类重要的衡量M的复杂程度的不变量^[1], 得到它的上界是引人关注的问题。对S的一个齐次理想I, IM的极小齐次生成元的最大次数不超过I和M的相应极小齐次生成元的最大次数之和, 所以研究reg(IM)≤reg(I)+reg(M)是否成立是一个自然的问题。当dim(S/I)≤1时, Conca和Herzog^[2]证明reg(IM)≤reg(I)+reg(M).Sturmfels^[3]给出一个单项式理想I, 满足reg(I²)＞2reg(I)。进一步限制理想I的范围, Conca和Herzog^[2]提出这样一个问题:当I₁, …, I_d都是完全交单项式理想时,

$ {\rm{reg}}\left( {{I_1}, \cdots ,{I_d}} \right) \le {\rm{reg}}\left( {{I_1}} \right) + \cdots + {\rm{reg}}\left( {{I_d}} \right) $

(1)

是否对任意的d≥1都成立?当d=2时, Chardin等^[4]证明了这一问题的正确性; 当d≥3时, 这一问题至今没有得到解决。当d=3且I₁, I₂和I₃都是由单个不定元的方幂生成的完全交理想时, Gao^[5]证明了结论的正确性。当I是一个完全交且n≥1时, Tang和Gong^[6]最近证明reg(Iⁿ)≤nreg(I)。在本文中，对4个不可约单项式理想(由不定元的方幂生成的完全交理想)I, J, K和L, 证明

$ \begin{array}{*{20}{c}} {{\rm{reg}}\left( {IJKL} \right) \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + }\\ {{\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right).} \end{array} $

1 本研究的主要工具

本研究工作所用的主要工具^[5]如下。

引理1.1  设0→N→M→P→0是一个有限生成的分次S-模的一个短正合列, 则

$ \left( {\rm{i}} \right){\rm{reg}}\left( M \right) \le \max \left\{ {{\rm{reg}}\left( N \right),{\rm{reg}}\left( P \right)} \right\}. $

$ \left( {{\rm{ii}}} \right){\rm{reg}}\left( P \right) \le \max \left\{ {{\rm{reg}}\left( M \right),{\rm{reg}}\left( N \right) - 1} \right\}. $

$ \left( {{\rm{iii}}} \right){\rm{reg}}\left( N \right) \le \max \left\{ {{\rm{reg}}\left( M \right),{\rm{reg}}\left( P \right) + 1} \right\}. $

$ \begin{array}{l} \;\;\;\;\;\;\left( {{\rm{iv}}} \right){\rm{reg}}\left( P \right) = {\rm{reg}}\left( M \right),{\rm{如果}}{\rm{reg}}\left( N \right) < \\ {\rm{reg}}\left( M \right). \end{array} $

$ \begin{array}{l} \;\;\;\;\;\;\left( {\rm{v}} \right){\rm{reg}}\left( P \right) = {\rm{reg}}\left( N \right) - 1,{\rm{如果reg}}\left( M \right) < \\ {\rm{reg}}\left( N \right). \end{array} $

引理1.2  设x是一个线性形式，I是S的一个齐次理想, 则对所有的n≥1,

$ {\rm{reg}}\left( I \right) \le \max \left\{ {{\rm{reg}}\left( {I,{x^n}} \right),{\rm{reg}}\left( {I:{x^n}} \right) + n} \right\}. $

引理1.3  设u是一个次数为d的齐次多项式，I是齐次理想且u是S/I-正则的，那么

$ {\rm{reg}}\left( {I,u} \right) = {\rm{reg}}\left( I \right) + d - 1. $

下面的引理1.4和引理1.5分别对应Gao^[5]中的引理3.1和定理3.2，为便于引用，将其列出。

引理1.4  设I, J, K是域k上多元多项式环S中的3个不可约单项式理想, 则

$ \begin{array}{*{20}{c}} {{\rm{reg}}\left( {\left( {IJ,IK,JK} \right)} \right) \le {\rm{reg}}\left( I \right) + }\\ {{\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) - 1.} \end{array} $

引理1.5  设I, J, K是域k上多元多项式环S中的3个不可约单项式理想, 则

$ {\rm{reg}}\left( {IJK} \right) \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right). $

2 主要结果

S的一个理想I称为一个完全交(complete intersection)单项式理想, 如果I可以由一些单项式生成，并且这些单项式之间没有公共的不定元。我们研究一类特殊的完全交单项式理想(不可约单项式理想), 即这些理想可以由单个不定元的方幂生成，例如I=(x₁², x₂², x₅⁶)。

引理2.1  设I, J, K, L是域k上多元多项式环S中的4个不可约单项式理想, 则

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right) $

$ \begin{array}{l} \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. \end{array} $

证明  对l₁+l₂+l₃+l₄用归纳法, 这里l₁, l₂, l₃, l₄分别是I, J, K和L的最小的单项式生成元的基数。

如果l₁=l₂=l₃=l₄=1, 设I=(x^l), J=(y^m), K=(zⁿ), L=(w^s), l≥m≥n≥s且x, y, z, w两两不相等, 则(IJ, IK, IL, JK, JL, KL)=(x^ly^m, x^lzⁿ, x^lw^s, y^mzⁿ, y^mw^s, zⁿw^s).

根据引理1.2, 引理1.3及Gao^[5]的引理3.1和Herzog^[7]的推论3.2, 有

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{x^l}{y^m},{x^l}{z^n},{y^m}{z^n},{w^s}} \right)} \right){\rm{reg}}\left( {\left( {{x^l},{y^m},} \right.} \right.} \right.\\ \left. {\;\;\;\left. {\left. {{z^n}} \right)} \right) + s} \right\}. \end{array} $

$ {\rm{reg}}\left( {\left( {{x^l}{y^m},{x^l}{z^n},{y^m}{z^n},{w^s}} \right)} \right) $

$ \le {\rm{reg}}\left( {{x^l}} \right) + {\rm{reg}}\left( {{y^m}} \right) + {\rm{reg}}\left( {{z^n}} \right) + s - 2 $

$ \begin{array}{l} = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2.\\ \end{array} $

$ \;\;\;\;{\rm{reg}}\left( {\left( {{x^l},{y^m},{z^n}} \right)} \right) + s\\ \le {\rm{reg}}\left( {{x^l}} \right) + {\rm{reg}}\left( {{y^m}} \right) + {\rm{reg}}\left( {{z^n}} \right) + s - 2 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

当x=y, x=y=z, x=y=z=w时可以用相同的方法证明有相同的结论, 因此在这种情况下结论成立。

如果I=(I₁, x^m)并且x是S/I₁, S/J, S/K, S/L的非零因子, 也就是x的任何方幂都不在I₁, J, K, L的最小单项式生成元中。则

$ \left( {IJ,IK,IL,JK,JL,KL} \right) $

$ = \left( {{I_1},{x^m}} \right)J + \left( {{I_1},{x^m}} \right)K + \left( {{I_1},{x^m}} \right)L + JK + JL + KL $

$ = {I_1}J + {I_1}K + {I_1}L + JK + JL + KL + {x^m}J + {x^m}K + {x^m}L. $

根据引理1.2, 有

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL,{x^m}} \right)} \right)} \right.,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right):{x^m}} \right) + m} \right\} \end{array} $

$ \begin{array}{l} = \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}J,{I_1}K,{I_1}L,JK,JL,KL,{x^m}} \right)} \right)} \right.,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {J,K,L} \right)} \right) + m} \right\} \end{array} $

注意到x是S/(I₁J, I₁K, I₁L, JK, JL, KL)-正则的, 根据引理1.3和归纳假设, 有

$ {\rm{reg}}\left( {\left( {{I_1}J,{I_1}K,{I_1}L,JK,JL,KL,{x^m}} \right)} \right) $

$ = {\rm{reg}}\left( {\left( {{I_1}J,{I_1}K,{I_1}L,JK,JL,KL} \right)} \right) + m - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m - 3 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

上式成立是因为reg(I)=reg(I₁)+m-1.根据Herzog^[7]的推论3.2, 有

$ \begin{array}{*{20}{c}} {{\rm{reg}}\left( {\left( {J,K,L} \right)} \right) + m \le {\rm{reg}}\left( J \right) + }\\ {{\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m - 2.} \end{array} $

因此在这种情况下结论成立。

如果I=(I₁, x^m), J=(J₁, xⁿ), m≥n≥1且x是S/K, S/L-正则的。则

$ \begin{array}{l} \left( {IJ,IK,IL,JK,JL,KL} \right) = \left( {{I_1},{x^m}} \right)\left( {{J_1},{x^n}} \right) + \\ \left( {{I_1},{x^m}} \right)K + \left( {{I_1},{x^m}} \right)L + \left( {{J_1},{x^n}} \right)K + \left( {{J_1},{x^n}} \right)L + KL\\ = {I_1}{J_1} + {I_1}K + {I_1}L + {J_1}K + {J_1}L + KL + {x^n}{I_1} + \\ \;\;\;{x^m}{J_1} + {x^n}K + {x^n}L + {x^{m + n}}. \end{array} $

根据引理1.2, 有

$ \begin{array}{l} {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right)\\ \le \max \{ {\rm{reg((}}{I_1}{J_1},{I_1}K,{I_1}L,{I_1}{K_1},{I_1}L,KL,{x^n}{I_1},\\ \;\;{x^n}K,{x^n}L,{x^m})),{\rm{reg}}(({I_1},{J_1},K,L,{x^n})) + m\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}K,{I_1}L,{J_1}K,{J_1}L,KL,{x^n}} \right)} \right)} \right.,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1},K,L,{x^{m - n}}} \right)} \right) + n,{\rm{reg}}\left( {\left( {{I_1},{J_1},K,L,{x^n}} \right)} \right) + m} \right\}. \end{array} $

根据归纳假设和Herzog^[7]的推论3.2, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}K,{I_1}L,{J_1}K,{J_1}L,KL,{x^n}} \right)} \right) $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},K,L,{x^{m - n}}} \right)} \right) + n $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},K,L,{x^n}} \right)} \right) + m $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

上面式子的成立是因为reg(I)=reg(I₁)+m-1和reg(J)=reg(J₁)+n-1.

因此在这种情况下结论是成立的。

如果I=(I₁, x^m), J=(J₁, xⁿ), K=(K₁, x^s), 并且m≥n≥s≥1.则有

$ \left( {IJ,IK,IL,JK,JL,KL} \right) $

$ \begin{array}{l} = \left( {{I_1}{J_1},{I_1}{K_1},{I_1}L,{J_1}{K_1},{J_1}L,{K_1}L,{x^s}{I_1},{x^s}{J_1},{x^n}{K_1},} \right.\\ \;\;\;\left. {{x^s}L,{x^{n + s}}} \right). \end{array} $

根据引理1.2, 有

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}L,{J_1}{K_1},{J_1}L,{K_1}L,{x^s}{I_1},} \right.} \right.} \right.\\ \;\;\;\left. {\left. {\left. {{x^s}{J_1},{x^s}L,{x^n}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},L,{x^s}} \right)} \right) + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}L,{J_1}{K_1},{J_1}L,{K_1}L,{x^s}} \right)} \right),} \right.\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1},{J_1},L,{x^{n - s}}} \right)} \right) + s,{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},L,{x^s}} \right)} \right) + n} \right\}. \end{array} $

根据归纳假设和引理1.3以及Herzog^[7]的推论3.2, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}L,{J_1}{K_1},{J_1}L,{K_1}L,{x^s}} \right)} \right) $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right) + s - 3 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},L,{x^{n - s}}} \right)} \right) + s $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},L,{x^s}} \right)} \right) + n $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

因此在这种情况下结论是成立的。

如果I=(I₁, x^m), J=(J₁, xⁿ), K=(K₁, x^s), L=(L₁, x^z)且m≥n≥s≥z≥1.则有(IJ,IK,IL,JK,JL,KL)=(I₁J₁,I₁K₁,I₁L,J₁K₁,J₁L,K₁L,x^sI₁,x^sJ₁,xⁿK₁,x^sL,x^n+s).

根据引理1.2, 有

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL,KL} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^z}{I_1},} \right.} \right.} \right.\\ \;\;\;\left. {\left. {\left. {{x^z}{J_1},{x^z}{K_1},{x^s}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + s} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^z}} \right)} \right),} \right.\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{x^{s - z}}} \right)} \right) + z,{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + s} \right\}. \end{array} $

根据归纳假设和引理1.3以及Herzog^[7]的推论3.2, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^z}} \right)} \right) $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right) + z - 3 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{x^{s - z}}} \right)} \right) + z $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) - 2. $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + s $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

因此在这种情况下结论是成立的。

综上，证明了当I, J, K, L是S中的4个不可约单项式理想时, 结论是成立的。

推论2.1  设I, J, K, L是域k上多元多项式环S中的4个不可约单项式理想, 利用证明引理2.1的方法, 可以证明

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK,JL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 2. $

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JK} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJK,IJL,IKL,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJ,IKL,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJ,IK,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJ,IK,IL,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJL,IKL,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IL,JKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

$ {\rm{reg}}\left( {\left( {IJK,IJL,IKL} \right)} \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) - 1. $

注意:类似(IJ, IK, IL, JK, JL)的其他几种情况, 即形如(IK, IL, JK, JL, KL), 也满足上面的不等式。

定理2.1  设I, J, K, L是域k上多元多项式环S中的4个不可约单项式理想, 则

$ {\rm{reg}}\left( {IJKL} \right) \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

证明  关于l₁+l₂+l₃+l₄用归纳法, 这里l₁, l₂, l₃, l₄分别是I, J, K, L的最小的单项式生成元的基数。如果l₁=l₂=l₃=l₄=1, 则定理的证明是显然的。因为

$ {\rm{reg}}\left( {IJKL} \right){\rm{ = reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

如果S的一个变量x只出现在I的最小的单项式生成元中, 而没有出现在J, K和L的最小的单项式生成元中。设I=(I₁, x^m), m≥1并且x是S/I₁-正则的。则IJKL=I₁JKL+x^mJKL并且x^m是S/I₁JKL-正则的。根据引理1.2和引理1.3。

$ {\rm{reg}}\left( {IJKL} \right) \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}JKL,{x^m}} \right)} \right),} \right. $

$ \left. {{\rm{reg}}\left( {\left( {{I_1}JKL,{x^m}JKL} \right):{x^m}} \right) + m } \right\} $

$ = \max \left\{ {{\rm{reg}}\left( {{I_1}JKL} \right) + m - 1,{\rm{reg}}\left( {JKL} \right) + m} \right\}. $

根据归纳假设，有

$ {\rm{reg}}\left( {{I_1}JKL} \right) + m - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m - 1 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

根据Gao^[5]的定理3.2, 有

$ \begin{array}{l} {\rm{reg}}\left( {JKL} \right) + m \le {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m\\ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). \end{array} $

因此定理的结论在这种情况下是成立的。

如果S的一个变量x出现在I和J的最小的单项式生成元中, 而没有出现在K和L的最小的单项式生成元中。设I=(I₁, x^m), J=(J₁, xⁿ)且m≥n。则IJKL=I₁J₁KL+xⁿI₁KL+x^mJ₁KL+x^m+nKL。

根据引理1.2

$ {\rm{reg}}\left( {IJKL} \right) $

$ \le \max \left\{ {{\rm{reg}}\left( {\left( {IJKL,{x^m}} \right)} \right),{\rm{reg}}\left( {\left( {IJKL:{x^m}} \right)} \right) + m} \right\} $

$ \begin{array}{l} = \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}KL,{x^n}{I_1}KL,{x^m}} \right)} \right)} \right.,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}KL,{J_1}KL,{x^n}KL} \right)} \right) + m} \right\}. \end{array} $

则上面最后一行的两个式子可以分写成

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}KL,{x^n}{I_1}KL,{x^m}} \right)} \right) $

$ \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}KL,{x^n}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1}KL,{x^{m - n}}} \right)} \right) + n} \right\}. $

$ {\rm{reg}}\left( {\left( {{I_1}KL,{J_1}KL,{x^n}KL} \right)} \right) + m $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}KL,{J_1}KL,{x^n}} \right)} \right) + m,{\rm{reg}}\left( {KL} \right) + } \right.\\ \;\;\;\left. {m + n} \right\}. \end{array} $

根据归纳假设, Gao^[5]的定理3.2和x的确没有出现在I₁, J₁, K, L的最小的单项式生成元中。有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}KL,{x^n}} \right)} \right) $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + n - 1 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

$ {\rm{reg}}\left( {\left( {{I_1}KL,{x^{m - n}}} \right)} \right) + n $

$ = {\rm{reg}}\left( {{I_1}KL} \right) + m - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m - 1 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

$ {\rm{reg}}\left( {KL} \right) + m + n \le {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m + n. $

注意到I₁+J₁也是一个不可约单项式理想, 根据Herzog^[7]的推论3.2和Gao^[5]的定理3.2, 有

$ {\rm{reg}}\left( {\left( {{I_1}KL,{J_1}KL,{x^n}} \right)} \right) + m $

$ = {\rm{reg}}\left( {\left( {{I_1},{J_1}} \right)KL} \right) + m + n - 1 $

$ \le {\rm{reg}}\left( {{I_1},{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m + n - 1 $

$ \begin{array}{l} \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) + m + \\ \;\;\;n - 2 \end{array} $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

因此定理的结论在这种情况下是成立的。

如果S的一个变量x出现在I, J, K的最小的单项式生成元中, 而没有出现在L的最小的单项式生成元中。设I=(I₁, x^m), J=(J₁, xⁿ), K=(K₁, x^s)且m≥n≥s≥1。则IJKL=(I₁J₁K₁L, x^sI₁J₁L, xⁿI₁K₁L, x^mJ₁K₁L, x^n+sI₁L, x^m+sJ₁L, x^m+nK₁L, x^m+n+sL)。

首先假设m≤n+s, 根据引理1.2

$ {\rm{reg}}\left( {IJKL} \right) $

$ \le \max \left\{ {{\rm{reg}}\left( {\left( {IJKL,{x^{n + s}}} \right)} \right),{\rm{reg}}\left( {\left( {IJKL:{x^{n + s}}} \right)} \right) + n + s} \right\} $

$ = \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}L,{x^s}{I_1}{J_1}L,{x^n}{I_1}{K_1}L,{x^m}{J_1}{K_1}L,{x^{n + s}}} \right)} \right)} \right., $

(2)

$ \left. {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L,{x^{m - n}}{J_1}L,{x^{m - s}}{K_1}L,{x^m}L} \right)} \right) + n + s} \right\}. $

(3)

对式(2), 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}L,{x^s}{I_1}{J_1}L,{x^n}{I_1}{K_1}L,{x^m}{J_1}{K_1}L,{x^{n + s}}} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}L,{x^s}{I_1}{J_1}L,{x^n}{I_1}{K_1}L,{x^m}} \right)} \right)} \right.,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{J_1}{K_1}L,{x^{m + s - n}}} \right)} \right) + m} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}L,{x^s}{I_1}{J_1}L,{x^n}} \right)} \right)} \right.,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{x^{m - n}}} \right)} \right) + n,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{J_1}{K_1}L,{x^{m + s - n}}} \right)} \right) + m} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}L,{x^s}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{x^{n - s}}} \right)} \right)} \right. + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{x^{m - n}}} \right)} \right) + n,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{J_1}{K_1}L,{x^{n + s - m}}} \right)} \right) + m} \right\}. \end{array} $

类似于前面几种情况, 可以证明

reg((I₁J₁K₁L, x^s)), reg((I₁J₁L, x^n-s))+s, reg((I₁J₁L, I₁K₁L, x^m-n))+n的值不会超过reg(I)+reg(J)+reg(K)+reg(L).

根据推论2.1，有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{J_1}{K_1}L,{x^{n + s - m}}} \right)} \right) + m $

$ = {\rm{reg}}\left( {\left( {{I_1}{J_1}L,{I_1}{K_1}L,{J_1}{K_1}L} \right)} \right) + n + s - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( {{L_1}} \right) + n + s - 2 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right) $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

对式(3), 有

$ {\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L,{x^{m - n}}{J_1}L,{x^{m - s}}{K_1}L,{x^m}L} \right)} \right) + n + s $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L,{x^{m - n}}} \right)} \right) + n + s,} \right.\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{x^{n - s}}{K_1}L,{x^n}L} \right)} \right) + m + s} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L,{x^{m - n}}} \right)} \right) + n + s} \right.,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{x^{n - s}}} \right)} \right) + m + s,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{K_1}L,{x^s}L} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}J,{x^{m - n}}} \right)} \right) + n + s,} \right.\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{x^{n - s}}} \right)} \right) + m + s,{\rm{reg}}\left( L \right) + m + n + s,\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{K_1}L,{x^s}} \right)} \right) + m + n} \right\}. \end{array} $

类似于前面几种情况, 易证

reg(L)+m+n+s, reg((I₁L, J₁L, x^n-s))+m+s的值不会超过reg(I)+reg(J)+reg(K)+reg(L).

根据Herzog^[7]的推论3.2, 有

$ {\rm{reg}}\left( {\left( {{I_1}L,{J_1}L,{K_1}L,{x^s}} \right)} \right) + m + n $

$ = {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1}} \right)L} \right) + m + n + s - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right) + m + n + s - 3 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

根据推论2.1, 有

$ {\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L,{x^{m - n}}} \right)} \right) + n + s $

$ = {\rm{reg}}\left( {\left( {{I_1}L,{J_1}{K_1}L} \right)} \right) + m + s - 1 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right) + m + s - 2 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

所以当m≤n+s时, 有reg(IJKL)≤reg(I)+reg(J)+reg(K)+reg(L)。当m > n+s时, 同理可证reg(IJKL)≤reg(I)+reg(J)+reg(K)+reg(L)成立。因此定理的结论在这种情况下是成立的。

如果S的一个变量x出现在I, J, K, L的最小的单项式生成元中, 设I=(I₁, x^m), J=(J₁, xⁿ), K=(K₁, x^s), L=(L₁, x^z)且m≥n≥s≥z≥1则

$ \begin{array}{l} IJKL = \left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;{x^m}{J_1}{K_1}{L_1},{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},\\ \;\;\;\;\;\;\;\;\;\;\;\;{x^{m + z}}{J_1}{K_1},{x^{m + s}}{J_1}{L_1},{x^{m + n}}{K_1}{L_1},{x^{n + s + z}}{I_1},\\ \;\;\;\;\;\;\;\;\;\;\;\;\left. {{x^{m + s + z}}{J_1},{x^{m + n + z}}{K_1},{x^{m + n + s}}{L_1},{x^{m + n + s + z}}} \right). \end{array} $

首先假设m≤s+z, 根据引理1.2

$ \begin{array}{l} {\rm{reg}}\left( {IJKL} \right) \le \max \left\{ {{\rm{reg}}\left( {\left( {IJKL,{x^{n + s + z}}} \right)} \right)} \right.,\\ \;\;\;\;\;\;\left. {{\rm{reg}}\left( {\left( {IJKL:{x^{n + s + z}}} \right)} \right) + n + s + z} \right\} \end{array} $

$ \begin{array}{l} = \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},} \right.} \right.} \right.\\ \;\;\;{x^m}{J_1}{K_1}{L_1},{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},{x^{m + z}}{J_1}{K_1},\\ \;\;\;\left. {\left. {{x^{m + s}}{J_1}{L_1},{x^{m + n}}{K_1}{L_1},{x^{n + s + z}}} \right)} \right), \end{array} $

(4)

$ \begin{array}{l} {\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}{J_1},{x^{m - s}}{K_1},{x^{m - z}}{L_1}} \right.,} \right.\\ \;\;\;\;\;\left. {\left. {\left. {{x^m}} \right)} \right) + n + s + z} \right\}. \end{array} $

(5)

对式(4)假设n+s≤m+z, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right., $

$ \left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},{x^{m + z}}{J_1}{K_1},{x^{m + s}}{J_1}{L_1},{x^{m + n}}{K_1}{L_1},{x^{n + s + z}}} \right)} \right) $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right.} \right.,\\ \;\;\;\left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},{x^{m + z}}{J_1}{K_1},{x^{m + s}}{J_1}{L_1},{x^{m + n}}} \right)} \right),\\ \;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right.} \right.,\\ \;\;\;\left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},{x^{m + z}}{J_1}{K_1},{x^{m + s}}} \right)} \right),\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right.,} \right.\\ \;\;\;\left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}{I_1}{L_1},{x^{m + z}}} \right)} \right),\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right.,} \right.\\ \;\;\;\left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}{I_1}{K_1},{x^{n + s}}} \right)} \right),\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}{J_1}{K_1}{L_1}} \right.} \right.,} \right.\\ \;\;\;\left. {\left. {{x^{s + z}}{I_1}{J_1},{x^{n + z}}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},} \right.} \right.} \right.\\ \;\;\;\left. {\left. {{x^m}{J_1}{K_1}{L_1},{x^{s + z}}} \right)} \right),{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}{I_1}{K_1}{L_1},{x^m}} \right)} \right)} \right.,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}{I_1}{J_1}{L_1},{x^n}} \right)} \right)} \right.,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{x^{m - n}}} \right)} \right) + n,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}{I_1}{J_1}{K_1},{x^s}} \right)} \right)} \right.,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{x^{n - s}}} \right)} \right) + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{x^{m - n}}} \right)} \right) + n,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1}{L_1},{x^z}} \right)} \right)} \right.,{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{x^{s - z}}} \right)} \right) + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{x^{n - s}}} \right)} \right) + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{x^{m - n}}} \right)} \right) + n,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z,\\ \;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s,\\ \left. {\;\;\;{\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n} \right\} \end{array} $

根据归纳假设和Gao^[5]的引理3.1和定理3.2易证reg((I₁J₁K₁L₁, x^z)), reg((I₁J₁K₁, x^s-z))+z, reg((I₁J₁K₁, I₁J₁L₁, x^n-s))+s的值不会超过reg(I)+reg(J)+reg(K)+reg(L).

根据引理2.1, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m + n $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( {{L_1}} \right) + n + s + z - 3 $

$ = {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

根据推论2.1, 有

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{s + z - m}}} \right)} \right) + m $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1}{L_1},{J_1}{K_1}{L_1},{x^{n - s}}} \right)} \right) + s + z $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right). $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{J_1}{K_1}{L_1},{x^{s - z}}} \right)} \right) + n + z $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( {{L_1}} \right). $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1}{L_1},{x^{\left( {m + z} \right) - \left( {n + s} \right)}}} \right)} \right) + n + s $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( L \right). $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{x^{s - z}}} \right)} \right) + m + z, $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( {{L_1}} \right). $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1},{I_1}{K_1},{I_1}{L_1},{J_1}{K_1},{J_1}{L_1},{x^{n - s}}} \right)} \right) + m + s $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( J \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( {{L_1}} \right) - 1. $

$ {\rm{reg}}\left( {\left( {{I_1}{J_1}{K_1},{I_1}{J_1}{L_1},{I_1}{K_1}{L_1},{x^{m - n}}} \right)} \right) + n $

$ \le {\rm{reg}}\left( I \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( {{L_1}} \right) - 1. $

对式(5)根据引理1.2, 有

$ \begin{array}{l} {\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}{J_1},{x^{m - s}}{K_1},{x^{m - z}}{L_1},{x^m}} \right)} \right) + \\ \;\;\;\;\;n + s + z \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}{J_1},{x^{m - s}}{K_1},{x^{m - z}}} \right)} \right) + } \right.\\ \;\;\;\;\;\;\left. {n + s + z,{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + m + n + s} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}{J_1},{x^{m - s}}} \right)} \right) + } \right.\\ \;\;\;\;\;\;n + s + z,\\ \;\;\;\;\;\;{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{x^{s - z}}} \right)} \right) + m + n + z,\\ \;\;\;\;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + m + n + s} \right\} \end{array} $

$ \begin{array}{l} \le \max \left\{ {{\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}} \right)} \right) + n + s + z,} \right.\\ \;\;\;\;\;\;{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1}{L_1},{x^{n - s}}} \right)} \right) + m + s + z,\\ \;\;\;\;\;\;{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{x^{s - z}}} \right)} \right) + m + n + z,\\ \;\;\;\;\;\;\left. {{\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{L_1},{x^z}} \right)} \right) + m + n + s} \right\}. \end{array} $

根据Herzog^[7]的推论3.2，易证

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1}{L_1},{x^{n - s}}} \right)} \right) + m + s + z, $

$ {\rm{reg}}\left( {\left( {{I_1},{J_1},{K_1},{x^{s - z}}} \right)} \right) + m + n + z, $

reg((I₁, J₁, K₁, L₁, x^z))+m+n+s的值不会超过

reg(I)+reg(J)+reg(K)+reg(L)。根据Gao^[5]的引理3.1, 有

$ {\rm{reg}}\left( {\left( {{I_1},{J_1}{K_1},{J_1}{L_1},{K_1}{L_1},{x^{m - n}}} \right)} \right) + n + s + z $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {\left( {{J_1}{K_1},{J_1}{L_1},{K_1}{L_1}} \right)} \right) + m + s + z - 2 $

$ \le {\rm{reg}}\left( {{I_1}} \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( {{K_1}} \right) + {\rm{reg}}\left( {{L_1}} \right) + m + s + z - 3 $

$ = {\rm{reg}}\left( I \right) + {\rm{reg}}\left( {{J_1}} \right) + {\rm{reg}}\left( K \right) + {\rm{reg}}\left( L \right). $

因此当n+s≤m+z时结论得证, 当n+s > m+z时可以用相同的方法证明有相同的结论; 因此当m≤s+z时定理成立, 当m > s+z时用相同的方法和完全类似的推导过程可以证明有相同的结论。

综上所述，定理被证明。