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Suppose $Z'\subseteq Z\subseteq\text{Spec} R$ such that every element in $Z\backslash Z'$ is a minimal element (with respect to inclusion as ideals) in $Z$. Assume further that both $Z$ and $Z'$ are table under Specialization with respect to Zariski topology on $\text{Spec} R$. If $M$ is an $R$-module, consider that $\Gamma_Z(M)=\left\{m\in M\mid\text{Supp}_R R(m)\subseteq Z\right\}$.

One defines $f\colon\Gamma_Z(M)\rightarrow\bigoplus_{p\in Z\backslash Z'}\Gamma_{pR_p}(M_p)$ such that $f(m)=(m / 1)_{p\in Z\backslash Z'}$. I need to prove that $f$ is surjective. Any guidance would be helpful for me.

Here R is Noetherian, also $T\subseteq \text{Spec R}$ is called to be stable under specialization with respect to Zariski topology if for any $q'\in T$ and any $q\in\text{Spec R}$ that $q'\subseteq q$ then $q\in T$.

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  • $\begingroup$ I cleaned up a little LaTeX but I'm not quite sure what's going on. Does $Z \subseteq \text{Spec }R$? When you say that elements are minimal in $Z$, what do you mean exactly? $\endgroup$ Commented Apr 9, 2014 at 18:38
  • $\begingroup$ Thank you for editing the text. $Z\subseteq\text{Spec } R$. From being minimal, I mean if $q\subseteq p$ for $p\in Z\backslash Z'$ and $q\in Z$ then $p=q$. $\endgroup$
    – user49402
    Commented Apr 9, 2014 at 19:26
  • $\begingroup$ Ok, one more question then. Is $Z$ closed? Is $Z'$ just some subset or is it also closed? $\endgroup$ Commented Apr 10, 2014 at 13:22
  • $\begingroup$ $Z$ and $Z'$ are not necessarily closed but both are stable under specialization with respect to Zariski topology. $\endgroup$
    – user49402
    Commented Apr 10, 2014 at 20:44

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