Proving that $P, Q\subset M$ are artinian, then so is $P+Q$ Suppose that if $M$ is a left $A$-module then for submodules $P, Q\subset M$ which are Artinian, then so is $P+Q$. This is not actually the exact question. I got to know the proof but here is the difficulty.
I have the following theorem in hand.
If $M/N$ and $N$ are Artinian then so is $M$.
The lecture notes uses the fact that $\frac{P+Q}{P}$ is isomorphic to something Artinian and since $P$ is Artinian then the theorem mentioned above implies the result. But I don't know what exactly $\frac{P+Q}{P}$ is isomorphic to?
Can anyone tell me a hint, or show me how to proceed THINK from this point. I am not asking for any answer but rather how do I think from this point. Thank you.
 A: From what I understand, you are trying to prove the statement "If $P,Q\subset M$ are both artinian, then so is $P+Q$" by using the theorem "If $N\subset M$ is so that $N$ and $M/N$ are both artinian, then $M$ is artinian".
To do so, you consider the quotient module $(P+Q)/P$. Given that $P$ is artinian and in accordance with the theorem mentioned above, we are reduced to proving that $(P+Q)/P$ is artinian.
There is a classical isomorphism theorem stating that
$$(P+Q)/P \simeq Q/(P\cap Q).$$
Since $Q$ is artinian, any quotient of $Q$ is artinian as well. Thus, $(P+Q)/P$ is artinian.
Edit: I attach an explanation regarding the isomorphism theorem used above. Consider two submodules $P$ and $Q$ of an $R$-module $M$ (for some ring $R$). Consider the map
$$f:Q \rightarrow (P+Q)/P$$
defined by $f(x) = x + P$. It is well defined since $Q \subset P + Q$.
It is surjective. Indeed, any element of the quotient space $(P+Q)/P$ can be written as $(p+q) + P$ for some $p\in P, q\in Q$. Since $p\in P$, we have $(p+q) + P = q + P = f(q)$.
Its kernel is $P\cap Q$. Since $P\cap Q \subset P$ it is clear that $P\cap Q \subset \mathrm{Ker}(f)$. Now, let $x\in \mathrm{Ker}(f)$. Then $x$ is a vector in $Q$ such that $x + P = P$. It means that $x \in P$, so $x \in P\cap Q$.
Thus, $f$ defines the desired isomorphism $P/(P\cap Q) \xrightarrow{\sim} (P+Q)/P$.
