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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 $q\in T$.

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 $q\in T$.

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 $Z$. 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. (R is Noetherian ring )

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 $q\in T$.

Suppose $Z'\subseteq Z\subseteq\text{Spec} R$ such that every element in $Z\backslash Z'$ is a minimal element (with respect to inclusion) in $Z$. 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. (R is Noetherian ring )

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 $Z$. 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. (R is Noetherian ring )