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

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?

• So, if $M$ and $N$ are graded, then $(N:_R M)$ is a homogeneous ideal. Thus, it is contained in a homogeneous maximal ideal. You can modify the standard argument in Zorn's lemma by adding in the word "homogeneous" appropriately and it should carry through. However, one must be careful; apparently if $R$ is not positively graded, $3)\Rightarrow 1)$ may not be true, see exercises 2.7 and 2.10 of Tom Marley's graded rings notes: math.unl.edu/~tmarley1/905notes.pdf – walkar Jul 13 '20 at 15:56
• Sorry, I didn't write this explicitly but every graded object here is supposed to be non negatively graded. – carciofo21 Jul 13 '20 at 16:32

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.

• I am a bit confused, also about the text of the exercise I posted: 1. How is the expression "homogeneous maximal ideal" to be understood? I thought that its significance was "a maximal ideal which is also homogeneous", while you are proving the existence of an ideal maximal among homogeneous ones. Similar confusion seems to arise in the comments here: math.stackexchange.com/questions/2338813/… – carciofo21 Jul 13 '20 at 16:30
• 2. Why is $(N:M)$ homogeneous? – carciofo21 Jul 13 '20 at 16:31
• @carciofo21 I have addressed point 2 by editing my answer. I will have to think a bit further on point 1. Is anything given about $R_0$, e.g. is $R_0$ a field in this question? – walkar Jul 13 '20 at 18:45
• No, and I think the text of the exercise, as it was proposed to me (anyway not as an homework, to be clear) was built just by glueing the two exercises of the notes you mentioned. – carciofo21 Jul 13 '20 at 18:49
• @carciofo21 I think this sort of argument works, but there might be a flaw somewhere. Let $J$ be a homogeneous ideal and let $J_0 = J \cap R_0$, which is an ideal of the ungraded ring $R_0$, and consequently $J_0$ is contained in a maximal ideal $\mathfrak{m}_0$ of $R_0$. Then $\mathfrak{m}=\mathfrak{m}_0 \oplus R_1 \oplus R_2 \oplus \cdots$ is an ideal which contains $J$, is homogeneous, and is also a maximal ideal (since $R/\mathfrak{m} = R_0/\mathfrak{m}_0$ is a field). Thus, during the Zorn's lemma argument, we would have encountered such an ideal at the top of the chain. – walkar Jul 13 '20 at 18:57