Gradation of the quotient module

Question

Let $A$ be a graded ring (of course, commutative with identity), $M$ a graded $A$-module, and $N$ a homogeneous submodule of $M$. I'm trying to prove that $$\frac MN=\bigoplus_{n\geq0} \frac{M_n}{N_n}$$ is again a graded $A$-module.

So consider some homogeneous components $\dfrac{M_n}{N_n}$ and $A_m$ of $M/N$ and $A$ respectively. I need to show that $A_m\left(\dfrac{M_n}{N_n}\right)\subseteq \dfrac{M_{m+n}}{N_{m+n}}.$

Let $x_n\in M_n$ and $r_m\in A_m$. Then $r_mx_n\in M_{m+n}$ as $M$ is a graded $A$-module. So $r_mx_n+{N_n}\in \frac{M_{m+n}}{N_n}$. But how to show $$\frac{M_{m+n}}{N_n}\subseteq \dfrac{M_{m+n}}{N_{m+n}}\,?$$ Since $N_n=N\cap M_n$, I thought I could use the second isomorphism theorem, but it turned out to be a useless approach. Correspondence theorem also wasn't helpful.

How should I proceed? Thank you.

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Edit: I almost found a solution, but I believe there are some subtle issues.

Here is my tentative solution:

By the Second Isomorphism theorem, we have $$\dfrac{M_n}{N_n}=\dfrac{M_n}{N\cap M_n}\cong \dfrac{M_n+N}{N}.$$ Now taken $\overline x_n=x_n+ N$, where $x_n\in M_n$. As $A_mM_n\subseteq M_{m+n}$, for $r_m\in A_m$, we see that $$r_m(x_n+ N)=r_mx_n+ N\in \dfrac{M_{m+n}+N}{N}\cong \dfrac{M_{m+n}}{N_{m+n}}.$$

So $A_m\left(\dfrac{M_n}{N_n}\right)\subseteq \dfrac{M_{m+n}}{N_{m+n}}.$

The problem is that $M_n$ need not be an $A$-module, it's only an $A_0$-module. So when I'm applying the Second Isomorphism theorem, I'm actually defining an $A_0$-module isomorphism. Is that a problem? If it is not, is my proposed solution correct?

Answer 1

I only discovered this question now, so I apologise if the topic is already outdated.

What you essentially need is the following result, for which I will first set up the terminology. Given an inclusion $S \subseteq T$ between arbitrary subsets, $\mathrm{i}^S_T$ denotes the inclusion map of $S$ into $T$. If $f \colon A \to B$ is an arbitrary map and $M \subseteq A$ I will use the notation $f[M]$ for the direct image on $M$ (preference for square brackets).

Rings always have units in my vocabulary. Given arbitrary left $A$-module $M$ I will write $\mathscr{S}_{\operatorname{A-\mathrm{Mod}}}(M)$ for the set of all its $A$-submodules. $T \leqslant_A M$ means that $T$ is an $A$-submodule of $M$. Given a family $N \in \mathscr{S}_{\operatorname{A-\mathrm{Mod}}}(M)^I$ of submodules of $M$ indexed by set $I$, I use the syntagm "$N$ yields a direct sum decomposition of $M$" if the canonical $A$-linear map $\displaystyle\bigoplus_{i \in I}N_i \to M$ induced via the universal property of direct sums by the family $\left(\mathrm{i}^{N_i}_M\right)_{i \in I}$ of inclusion morphisms is actually an isomorphism.

The result is as follows:

Consider an arbitrary ring $A$, a left $A$-module $M$, families $P, Q \in \mathscr{S}_{\operatorname{A-\mathrm{Mod}}}(M)^I$ of submodules such that $Q_i \leqslant_A P_i$ for every $i \in I$ and such that $P$ yields a direct sum decomposition of $M$. Also consider $N=\displaystyle\sum_{i \in I}Q_i \leqslant_A M$ -- the submodule generated by $\displaystyle\bigcup_{i \in I}Q_i$ -- and write $\sigma \colon M \to M/N$ for the canonical surjection of the quotient in question. Then the family $R=\left(\sigma[P_i]\right)_{i \in I} \in \mathscr{S}_{\operatorname{A-\mathrm{Mod}}}(M/N)^I$ of submodules yields a direct sum decomposition of the quotient.

Let us see how this applies to gradings on quotients. Consider a commutative monoid of degrees, $(\Delta, +)$ -- given in additive notation -- and arbitrary (need not be commutative) ring $A$ with $\Delta$-grading $\mathbf{A}$. To be fully committed to my syntax rules, I should denote the term of index $\delta \in \Delta$ in this grading (which is in essence a family of additive subgroups of $A$ indexed by $\Delta$) by $\mathbf{A}_{\delta}$, but I shall resort to the simplified notation $\mathbf{A}_{\delta}=A_{\delta}$. The fact that $\mathbf{A}$ is a $\Delta$-grading means explicitly that $A_{\gamma}A_{\delta} \subseteq A_{\gamma+\delta}$ for any two degrees $\gamma, \delta \in \Delta$.

Let us further consider a left $A$-module $M$ equipped with a $\Delta$-grading $\mathbf{M}$ which is compatible with the grading $\mathbf{A}$ on $A$, rendering $(M, +, \mathbf{M})$ into a graded module over the graded ring $(A, +, \cdot, \mathbf{A})$. This compatibility between the gradings means explicitly that $A_{\gamma}M_{\delta} \subseteq M_{\gamma+\delta}$. Let us finally assume that the submodule $N \leqslant_A M$ is graded (compatible with $\mathbf{M}$). This property can be introduced through a number (3, to be precise) of equivalent formulations, one of which being that $N=\displaystyle\sum_{\delta \in \Delta}(N \cap M_{\delta})$ (the remaining two are "$N$ is generated by homogenous elements" respectively "for every $x \in N$ all the homogenous components of $x$ also belong to $N$").

By virtue of the result stated above, the family $\mathbf{M'}=\left(\sigma[M_{\delta}]\right)_{\delta \in \Delta}$ will yield a direct sum decomposition of the quotient $M/N$ (whose canonical surjection $\sigma$ denotes). For any degrees $\gamma, \delta \in \Delta$ the relation: $${\vphantom{\brac{A}}A}_{\gamma}M'_{\delta}=A_{\gamma}\sigma[M_{\delta}]=\sigma[A_{\gamma}M_{\delta}] \subseteq \sigma[M_{\gamma+\delta}]=M'_{\gamma+\delta}$$ is easily ascertained and signifies that the grading $\mathbf{M'}$ is indeed compatible with the one on $A$.

This is the canonical structure of the graded quotient $(M/N, +, \mathbf{M'})$.