the form $$e'=e'_1\otimes\cdots\otimes e'_n$$ for standard idempotents $e'_i\in H^0({\mathbb{P}}^d, \Lambda_i)$. We prove $e'X=0$ by showing $H^i({\mathbb{P}}^d, e'X(-\ell))=0$ for sufficiently many $\ell$ (depending on the support of $e'X$) and then invoke Theorem \[beilinson\]. We proceed by using the induction with respect to $N:=|\{i\mid 1\le i\le n,\ e'_i\neq e_i\}|$. First we show when case $N=0$ (i.e. $e'=e$):

\[eX=0\] We have $eX=0$.

For each $\ell$ with $0\le \ell\le d$, we have $\ell{\vec c}\in[0,d{\vec c}]$. By Lemma \[cohomology\] and our assumption we have $$H^i({\mathbb{P}}^d, eX(-\ell))={\operatorname{Ext}\nolimits}^i_\Lambda(P(\ell{\vec c}),X)\subset{\operatorname{Ext}\nolimits}^i_\Lambda(T,X)=0$$ for all $i\ge0$. By Theorem \[beilinson\], we have $eX=0$.

The case $N=1$ follows from the following two lemmas:

Let $Y$ be a $\Lambda_i$-module such that $e_iY=0$. Then for any standard idempotent $e_i'\in H^0({\mathbb{P}}^d, \Lambda_i)$, the ${{\mathcal O}}$-module $e_i'Y$ is annihilated by ${{\mathcal O}}(-L_i)$. \[killebyideal\]

Under the natural identification $e'_i\Lambda_ie'_i={{\mathcal O}}$, we have $e_i'\Lambda e_i\Lambda e'_i={{\mathcal O}}(-L_i)$. We thus have ${{\mathcal O}}(-L_i)e'_iY=(e_i'\Lambda e_i\Lambda e'_i)(e'_iY)
=(e_i'\Lambda e_i)(\Lambda e'_iY)\subset e_i'\Lambda e_iY=0$.

Let $e':=e'_1\otimes e_2\otimes\cdots\otimes e_n\in H^0({\mathbb{P}}^d, \Lambda)$ where $e_1'$ is standard. Then $e'X=0$.

By Lemma \[eX=0\], we have $eX=0$. Furthermore the previous lemma implies ${{\mathcal O}}(-L_1)(e'X)={{\mathcal O}}(-L_1)(e_1'\otimes
    1\otimes\dots\otimes 1)(1\otimes e_2\otimes\dots\otimes e_n)(e'X)=0$. Thus we can regard $e'X$ as a sheaf on $L_1\cong{\mathbb{P}}^{d-1}$.

For each $\ell$ with $0\le \ell<d$, and $0<m<p_1$ we have 