(N:M) is contained in a homogeneous maximal ideal in a graded ring

Question

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:

1. $M=N$;
2. $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for any homogeneous prime ideal $\mathfrak p \subset R$;
3. $M_{\mathfrak{m}}=N_{\mathfrak{m}}$ 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?

Answer 1

First, note $(N:_R M)$ is homogeneous. To see this, it suffices to replace $M$ with $M/N$ (still a graded module) and prove $(0:_R M)$ is homogeneous. Now, let $r \in (0:_R M)$ and let $m \in M_t$ be any homogeneous element of $M$. Since $R$ is graded, $r$ can be written as $r=\sum_{i=0}^n r_i$, each $r_i$ homogeneous of degree $i$, explicitly some of the $r_i$ may be zero.

Then, $0=rm=\sum_{i=0}^n r_i m$, but each $r_i m$ is of a distinct degree $t+i$, so the sum is zero if and only if each summand is zero. Thus, $r_i m = 0$ for each $i$, and since $m$ was an arbitrary homogeneous element of $M$ and $r_i$ annihilates all such, each $r_i$ is in $(0:_R M)$ since $M$ can be generated by homogeneous elements. Thus $(0:_R M)$ is also homogeneous.

Now, as noted, one can modify the standard Zorn's lemma argument for the existence of maximal ideals to show that $(N:_R M)$ is contained in an ideal maximal among homogeneous ideals. For brevity, call such an ideal "maximally homogeneous".

Claim: Let $R$ be a graded ring (commutative with unity) and $I$ be a homogeneous ideal. Then, there exists a maximally homogeneous proper ideal containing $I$.

Proof: Partially order the set of homogeneous proper ideals containing $I$ by inclusion. The set is nonempty, as $I$ itself is in the set. Now, let $\{J_\lambda\}_{\lambda \in \Lambda}$ be any linear chain in the set, in particular, each of the $J_\lambda$ are homogeneous. Then, $J=\cup_{\lambda} J_\lambda$ is an ideal, as it is an increasing union.

Now, this is the only place where we need a bit of an extra argument. Can you show why $J$ must be homogeneous? I've hidden the proof of that below.

$J$ is homogeneous, as any $x \in J$ is inside some $J_\lambda$, wherein it can be expressed as a sum of homogeneous generators of $J_\lambda$. As every element in $J$ can be expressed as a sum of homogeneous elements, $J$ is then homogeneous.

Thus, every linear chain in the set has an upper bound, implying that it has a maximal element by Zorn's lemma -- a maximally homogeneous ideal containing $I$, as required.