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let $S$ be a commutative ring and $I_1,...,I_n\unlhd S$, such that $I_i+I_j=S\ \forall i\neq j$.

Let $g_1,...,g_n\in S$.

Why are there $h_1,...,h_n,h'\in S$, such that

$h_i\big|_{V(I_i)}=g_i\big|_{V(I_i)}$ and $h_i\big|_{V(I_j)}=0\ \forall i\neq j$ and $h'-g_i\in I_i\ \forall i$ hold?

Note that $\widetilde{S}$, the étalé space belonging to $S$, is a sheaf.

Thanks for the help.

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    $\begingroup$ Are you trying to use CRT to prove your statement?. Because if it is, then the existence of $h_i$ and $h'$ follows trivially from CRT applied respectively to the $n$-tuples $(0,\ldots,0,g_i,0,\ldots,0)$ and $(g_1,\ldots,g_n)$ $\endgroup$ – Diego Nov 5 '14 at 21:04
  • $\begingroup$ No, I would like to prove the CRT by using the fact that $\widetilde{S}$ is a sheaf. $\endgroup$ – Leopold Schirmer Nov 5 '14 at 22:55
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Here there is an elemental result on sheaves from Godement's book "Topologie algebrique et theorie des faisceaux", Theorem 1.3.1:

$\bf{Theorem}:$ Let $\mathscr{F}$ be a sheaf on a topological space $X$ and let $(M_i)_{i\in I}$ be a locally finte covering of $X$ formed by closed subsets. Suppose we have sections $s_i\in\mathscr{F}(M_i)$ such that $s_i=s_j$ in $M_i\cap M_j$ for all $i,j\in I$. Then there exists a section $s\in\mathscr{F}(X)$ which concides with $s_i$ in $M_i$ for all $i\in I$.

In your case let $X=\cup_{i=1}^n V(I_i)=V(\cap_{i=1}^n I_i)$. The hypothesis on the $I_i$'s implies that $V(I_i)\cap V(I_j)=\emptyset$ for $i\neq j$ and therefore we do not need to check conditions "$s_i=s_j$ in $V(I_i)\cap V(I_j)$" if we want to apply the theorem in this setting. Then CRT follows easily: for the existence of $h'$, let $s\in \tilde{S}(X)$ such that $s=g_i$ in $V(I_i)$ for all $i$ (this follows from the Theorem). The section $s\in \tilde{S}(X)=\tilde{S}(V(\cap_{i=1}^n))$ is the restriction of a section $h'\in \tilde{S}(Spec(S))=S$. It follows that $h'=g_i$ in $V(I_i)$ for all $i$, and this is equivalent to $h'-g'\in I_i$ for all $i$.

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