Proposition 3.7 in Eisenbud says that for a f.g. module $M$ over a Noetherian ring there is a chain $$0=M_0\subset M_1\subset\dots\subset M_n=M$$ with $M_{i+1}/M_i\simeq R/P_i$ for some prime ideal $P_i$.
I'm trying to understand the proof. Since $R$ is Noetherian, $\mathrm{Ass}(M)\ne\emptyset$. So there is an injection of $R$-modules $R/P_0\to M$ for some associated prime $P_0$ of $M$. Set $M_1=R/P_0$. Then $0=M_0\subset M_1\subset M_2=M$ where $M_1$ actually stands for the image of the previous $M_1$ under the injection above. We have $M_1/M_0\simeq R/P_0$, but why does $M_2/M_1\simeq R/P_1$?
At the next step, pick $P_1\in\mathrm{Ass}(M/M_1)$ (this set is nonempty because $M/M_1$ is again a module over a Noetherian ring). This gives an injection $\bar{M_2}\to M/M_1$ of $R$-modules. Let $M_2$ be the preimage of $\bar {M_2}$ under the quotient map $M\to M/M_1$. Then $0=M_0\subset M_1\subset M_2\subset M_3=M$. How do I see that $M_2/M_1$ and $M_3/M_2$ are of the desired form?
Setting $M_0 = 0$, we then choose $M_1$ to be the image of the injection $R/P_0 \hookrightarrow M$. Now, apply the same reasoning to $M/M_1$; there is an injection $R/P_1 \hookrightarrow M/M_1$. Let $M_2'$ be the image of this injection. If $p: M \to M/M_1$ is the canonical surjection, set $M_2 := p^{-1} (M_2')$. Now, $M_2/M_1 = M_2' \cong R/P_1$, as desired. The key point is that we can continue this process to obtain an ascending chain $$0 = M_0 \subset M_1 \subset M_2 \subset \cdots \subset M$$ which must terminate since $M$ is Noetherian. But if the chain terminates at $M_k$, then $M_k = M$, since otherwise $M/M_k$ is nonzero and we can attach another nontrivial module to the chain.