Let $R$ be a (commutative unitary) graded ring and $N \subset M$ be two graded $R$- modules. I want to prove that the follwing are equivalent:
- $M=N$;
- $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for any homogeneous prime ideal $\mathfrak p \subset R$;
- $M_{\mathfrak{m}}=N_{\mathfrak{m}}$ for any homogeneous maximal ideal $\mathfrak m \subset R$.
Now 1. implies 2. implies 3. is obvious, now my problem is how to prove 3. implies 1. Any suggestion?
EDIT: Okay, I was able to prove the implication $3 \implies 1$ for non graded-rings and generic (non homogeneous) maximal ideals by localizing both $M$ and $N$ at $\mathfrak m$ where $\mathfrak m$ is a maximal ideal containing $(N:M)$. [More in details: $M=N$ iff $(N:M)=R$; suppose the contrary, then there is $(N:M) \subsetneq \mathfrak m \subsetneq R$ and we have $M_\mathfrak{m}= N_\mathfrak{m}$, this means that every element of the form $m/s$ for $m$ in $M$ and $s \in R-\mathfrak{m}$ can be written as $n/t$ for some $n \in N$ and $t \in R-\mathfrak{m}$. Then there is $u \in R-\mathfrak{m}$ such that $utm=usn \in N$, so $ut \in (N:M)$ but this can't be the case.]
Now to generalize the argument I need to show that if $(N:M)\subsetneq R$, then there exists a maximal homogeneous ideal $\tilde {\mathfrak{m}} \subsetneq R$ containing $(N:M)$. How to do this? Does Zorn's lemma work?