# If M is a Artinian Module and A submodule of M, show that M/A is Artinian.

In the proof we assume that $$A_{1}/A\supseteq A_{2}/A\supseteq A_{3}/A ......$$

is a descending chain of submodules of $M/A$ and so we get that $$A_{1}\supseteq A_{2}\supseteq A_{3} ......$$ is a descending chain of submodules of $M$, but I do not know why, could anyone clarify this form me please?

Let $x\in A_{n+1}$. As $x+A\in A_{n+1}/A\subseteq A_{n}/A$, then $x+A\in A_{n}/A$, say, $x+A=y+A$ for some $y\in A_{n}$, then $x-y\in A\subseteq A_{n}$, so $x-y\in A_{n}$. As $A_{n}$ is a module, then $x=y+(x-y)\in A_{n}$.
• Is it true that $A_{i} \subset A_{i}/A$? if so why? for Noetherian – Intuition Jan 9 '18 at 12:28
• That is not true because the elements of $A_{i}/A$ are cosets which have totally different structure than elements of $A_{i}$. – user284331 Jan 10 '18 at 2:58
The notation seems a bit unfortunate. Let $\pi: M\to M/A$ be the canonical projection. Then, let us rather assume that $B_1\supseteq B_2 \supseteq B_2\supseteq \cdots$ is a descending chain of submodules of $M/A$ - after all, this is how we should check that it is Artinian. Let then $A_i:=\pi^{-1}(B_i)$. It follows immediately that $A_i\supseteq A_{i+1}$ for all $i$.
In your case, someone simply wrote $A_i/A$ instead of $B_i$. This is a little unprecise with respect to the fact that you might have a choice regarding $A_i$, i.e. there might be several submodules of $M$ that project to $B_i$. Instead of taking just any of these, you can always pick one that completely contains $A$. This way, the inclusion is no longer confusing.