论文标题
多项式环中理想的截断数量的分级数量
The graded Betti numbers of truncation of ideals in polynomial rings
论文作者
论文摘要
令$ r = \ mathbb {k} [x_1，\ dots，x_n] $，一个分级的代数$ s = r/i $满足$ n_ {k，p} $，如果$ k $中产生$ i $，则分级的最小值是$ k $ p $ p $ p $ p $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ s $ - $ n_ {k，p} $。 Eisenbud和goto表明，对于任何分级的环$ r/i $，然后是$ r/i _ {\ geq k} $，其中$ i _ {\ geq k} = i \ cap m^k $ and $ m^k $ and $ m =（x_1，x_1，\ dots，\ dots，x_n）$ a $ k $ $ n_ $ p $ n_ $ k \ gg0 $。对于无方形的单条件理想$ i $，我们在这里对理想的$ i_k $感兴趣，这是$ i _ {\ geq k} $的平方free部分。理想的$ i $是通过Stanley-Reisner对应关系，与Simplicial Complex $δ_i$相关联。在这种情况下，所有$ r/i_k $的$ r/i_k $对于$ k> \ min \ {\ text {deg}（u）\ in i \} $ in I \} $，当然可以比索引更不变，可以从$ r/i $ f $ f $ -vector的$ r/i $和$ f $ -vector $ -vector $ -vector $ -vector $ f $ f $ n $ $ n $ $ $ $ n $中确定。我们将结果与$ i _ {\ ge k} $的相应语句进行比较。 （这里$ i $是任意分级的理想。）在这种情况下，我们表明可以从$ r/i $的betti数字和$ r/i _ {\ ge ge k k} $的betti数字确定$ r/i _ {\ ge k} $的贝蒂数字。
Let $R=\mathbb{K}[x_1,\dots,x_n]$, a graded algebra $S=R/I$ satisfies $N_{k,p}$ if $I$ is generated in degree $k$, and the graded minimal resolution is linear the first $p$ steps, and the $k$-index of $S$ is the largest $p$ such that $S$ satisfies $N_{k,p}$. Eisenbud and Goto have shown that for any graded ring $R/I$, then $R/I_{\geq k}$, where $I_{\geq k}=I\cap M^k$ and $M=(x_1,\dots,x_n)$, has a $k$-linear resolution (satisfies $N_{k,p}$ for all $p$) if $k\gg0$. For a squarefree monomial ideal $I$, we are here interested in the ideal $I_k$ which is the squarefree part of $I_{\geq k}$. The ideal $I$ is, via Stanley-Reisner correspondence, associated to a simplicial complex $Δ_I$. In this case, all Betti numbers of $R/I_k$ for $k>\min\{\text{deg}(u)\mid u\in I\}$, which of course is a much finer invariant than the index, can be determined from the Betti diagram of $R/I$ and the $f$-vector of $Δ_I$. We compare our results with the corresponding statements for $I_{\ge k}$. (Here $I$ is an arbitrary graded ideal.) In this case we show that the Betti numbers of $R/I_{\ge k}$ can be determined from the Betti numbers of $R/I$ and the Hilbert series of $R/I_{\ge k}$.
