## $Ass(M)$ is finite

Let $$M$$ be a f.g. module over a Noetherian ring. I’m trying to fill in the details in the proof of the fact that $$Ass(M)$$ is finite. It should follow from Proposition 3.7 and Lemma 3.6 in Eisenbud.

First, consider a filtration $$0=M_0\subset\dots\subset M_n=M$$ where $$M_{i+1}/M_{i}\simeq R/P_i$$ for some prime ideals $$P_i$$ (Proposition 3.7 in Eisenbud). Let us prove by induction on $$n$$ that $$Ass(M_n)\subset \{P_0,\dots, P_{n-1}\}$$.

First of all, I have problems with the base case: if $$n=1$$, the chain looks like $$0=M_0\subset M_1=M$$, and $$M_1\simeq M_1/M_0\simeq R/P_0$$ for some prime ideal $$P_0$$. So $$P_0$$ is an associated prime of $$M$$, so $$\{P_0\}\subset Ass(M)$$. Why does the reverse inclusion hold?

Assume the result holds for all $$k. Consider the exact sequence $$0\to M_{n-1}\to M_n\to M_n/M_{n-1}\to 0$$. By lemma 3.6, $$Ass(M_n)\subset Ass(M_{n-1})\cup Ass(M_n/M_{n-1})$$

By hypothesis $$Ass(M_{n-1})\subset \{P_0,\dots,P_{n-2}\}$$. But what is $$Ass$$ of the quotient, and why?