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Let $k$ be a perfect field, let $n$ and $m$ be two positive integers.

Consider $X=\mathbb P_k^n\times \mathbb P_k^m$. Let $x_0=(1,0,\cdots,0;1,0,\cdots,0)\in X$ be fixed.

For any pair of integers $(d_1,d_2)$, let $\mathcal O(d_1,d_2)=pr_1^\ast \mathcal O(d_1)\otimes pr_2^\ast \mathcal O(d_2)$.

Let $A$ be a affine space(vector space) over $k$ which contains all sections of $\mathcal O(d_1,d_2)$.

For any $x\in X$ and any integer $r\geq 0$, let $B(x,r)$ be the affine space whose points are $r$-jets of sections of $\mathcal O(d_1,d_2)$.

For any $x\neq x_0$ and any two positive numbers $N$ and $M$, if $d_1$ and $d_2$ are large enough, I believe that the canonical restriction map is surjective $A\longrightarrow B(x_0,N)\times B(x,M)$.

But I don't know how to show it. Any comment is very welcome!

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  • $\begingroup$ What is this $Spec$? $\ \Gamma(X,\mathcal {O}(d_1,d_2))$ is not a ring. $\endgroup$
    – abx
    Jan 17, 2014 at 14:54
  • $\begingroup$ Thank you~ A is a affine space over k, not the spectral... $\endgroup$
    – Enlin Yang
    Jan 17, 2014 at 15:06

1 Answer 1

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Let $\mathfrak{m}$ (resp. $\mathfrak{m}_0$) be the maximal ideal sheaf at $x$ (resp. $x_0$). Put $L:=\mathcal{O}(d_1,d_2)$.

If I understand correctly, you are asking whether the restriction map $$H^0(X,L)\rightarrow (L/\mathfrak{m}^rL)\times (L/\mathfrak{m}_0^rL)$$ (or maybe $\mathfrak{m}^{r+1}$, this depends on your conventions) is surjective. Because of the exact sequence $$0\rightarrow \mathfrak{m}^r\mathfrak{m}_0^rL \rightarrow L \rightarrow (L/\mathfrak{m}^rL)\times (L/\mathfrak{m_0}^rL)\rightarrow 0\ ,$$ this is equivalent to the vanishing of $H^1(X, \mathfrak{m}^r\mathfrak{m}_0^rL)$; this vanishing is guaranteed by Serre's theorem for $L$ ample enough, that is, $d_1,d_2\gg 0$.

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    $\begingroup$ the four $r$ in the exact sequence should be replaced by N and M, but the proof is still work. Thank you very much! $\endgroup$
    – Enlin Yang
    Jan 17, 2014 at 16:02

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