# Grading of the quotient module $M/N$

Let $S$ be a graded ring, $M$ a graded $S$-module, and $N$ a graded submodule of $M$. I'm trying to convince myself (of the well known fact) that $M/N$ is graded by $$M/N=\oplus_{i\geq0} (M_i/N\cap M_i),$$ but I can't do it.

For $x\in M/N$, I would like to see $x$ displayed as $$(m_1+N\cap M_1,m_2+N\cap M_2,...,m_r+N\cap M_r,0,...)$$ for some $r$...

I followed my nose, and begun $x=(m_1,m_2,\cdots,m_r,0,...)+N$, but fail to see the next step.

Thanks a lot.

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Don't use cosets. The essential feature about a quotient is not that it is built up out of sets, but rather that it satisfies a universal property. – Martin Brandenburg Dec 6 '11 at 10:26

We have as abelian groups $M = \oplus_i M_i$ and $N = \oplus_i (M_i \cap N)$ (since $N$ is graded). Now direct sums commute with quotients (this is an instance of a more general fact from category theory, that colimits commute with colimits - in the functorial approach, this is just trivial). Thus $M/N = \oplus_i M_i / (M_i \cap N)$. Now observe that $S_i M_j \subseteq M_{i+j}$ implies the same for the quotients.
Use the definition of $N$ being a graded submodule of $M$.
Yes, $N=\oplus N\cap M_i$. But I can't see how to use this in order to display x in the desired form. – Bart Patzer Sep 22 '11 at 8:49
choose preimages in $M$, then maybe it is easier to see $x'=m_1'+m_2'+\dots+m_r'+n$, then see how you can represent $n$. What does it mean that $N=\oplus N\cap M_i$ – Julian Kuelshammer Sep 22 '11 at 12:26