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In the book "Local cohomology : An algebraic introduction with geometric application", page 289 there is a proof of the following theorem :

Assume that $R=\bigoplus_{n}R_{n}$ is positive graded and homogeneous, and let $M=\bigoplus_{n}M_{n}$ be a non-zero finitely generated graded $R$-module. Then $M$ can be generated by homogeneous elements of degrees not exceeding $\text{reg}M$.

Here is a part of the proof that I concern :

Let $N$ be the graded submodule of $M$ generated by $\bigoplus_{\le\text{reg}M}M_{n}$. It is suffices to show that $M_{\mathfrak{p}_{0}}=N_{\mathfrak{p}_{0}}$ for each $\mathfrak{p}_0$ in SpecR.....(argument).....( $M_{\mathfrak{p}_{0}}=N_{\mathfrak{p}_{0}}$ is proven). Then it is therefore enough to establish the claim in the statement of the theorem under the additional assumption that $R_0$ is local.

My question is :

1- We can't conclude that $M=N$ if we have $M_{\mathfrak{p}_{0}}$=$N_{\mathfrak{p}_{0}}$ ?

2- Why can we only consider only the case that $R_0$ is local? What is the idea behind that localization technique ?

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    $\begingroup$ For clarity's sake, the book under discussion is the one by Brodmann and Sharp. $\endgroup$
    – Fred Rohrer
    Commented Dec 4, 2012 at 13:37

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First, if $h:A\rightarrow B$ is a morphism of rings, $M$ is a $B$-module, and $N\subseteq M$ is a sub-$B$-module, then it holds $M=N$ if and only if the underlying sets of $M$ and $N$ are equal, hence if and only if the $A$-modules obtained from $M$ and $N$ by scalar restriction along of $h$ are equal.

Second, if $A$ is a ring, $M$ is an $A$-module, and $N\subseteq M$ is a sub-$A$-module, then it holds $N=M$ if and only if $N_{\mathfrak{p}}=M_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$ (cf. [Bourbaki, Algèbre commutative, II.3.3 Théorème 1]).

Third, putting the above together answers what seems to be your first question.

Finally, it is not completely clear to me, what you want as an answer to your second question. Let me just point out that in the last paragraph of the proof one applies Lemma 15.1.4 (ii) and thus needs the base ring to be local with infinite residue field, and one moreover uses the facts that $R$ is *local and thus needs the base ring to be local.

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  • $\begingroup$ @Fred Rohrer: Dear Fred, what do you mean by the $B$ modules obtain from $M$ and $N$ by scalar restriction along $h$ are equal ? can you make it more precise? As you said we can conclude that $M_{\mathfrak_{p}}=N_{\mathfrak{p}}$ on $R_0$ and therefore $M=N$ on $R_0$. So, why prooving the problem in the local case makes the proof complete ? $\endgroup$
    – Axy
    Commented Dec 4, 2012 at 14:56
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    $\begingroup$ Dear @Axy, concerning scalar restriction you should take a look at Bourbaki's Algèbre II.1.13. Concerning the proof, I explained that if $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$, then $M=N$. So, if we want to show that $M=N$, then we have to show that $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$. To do this, we take such a prime ideal and localise everything at it, obtaining a similar situation as before but this time over a local base ring... $\endgroup$
    – Fred Rohrer
    Commented Dec 4, 2012 at 15:25
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    $\begingroup$ ... If the result holds in case we have a local base ring then it holds in this situation, and thus the claim will be proven once we prove it for the case of a local base ring. $\endgroup$
    – Fred Rohrer
    Commented Dec 4, 2012 at 15:26

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