$\mathfrak{o}$ is a discrete valuation ring (local P.I.D), $\mathfrak{p}$ is the unique maximal ideal. $\pi$ is the generator of $\mathfrak{p}$, i.e. $\mathfrak{p}=<\pi>$. The residue field $k=\mathfrak{o}/\mathfrak{p}$ is finite,i.e. $k\cong\mathbb{F_q}$, where $q$ is a power of a prime. Now, consider a $\mathfrak{o}$-module M, such that $$M\cong\oplus_{i=1}^s\mathfrak{o}/\mathfrak{p}^{\lambda_i}$$where $\lambda_1\geq\ldots\geq\lambda_s>0$. for any $x\in M$, $\exists r\in \mathbb{N},\ s.t. \mathfrak{p}^rx=0,\ \ \mathfrak{p}^{r-1}x\neq0$. $r$ is called the height of $x$. (height of $0$ is $0$) Define $$M_r=\{x\in M|\ \text{height of }x \leq r\}$$ Lemma (1.7): $N$ is a submodule of $M$, generated by elements of height $\geq r$. Let $x\in M$, the following are equivalento:
(1)$x$ has height $r$, $\mathfrak{o}x\bigcap N=0$
(2)$x\in M_r-(M_{r-1}+N_r)$
Clearly, (1)$\Rightarrow$(2) is right. But I cannot prove (2)$\Rightarrow$(1). Following is the proof in the book, but I cannot see why the $y$ exists.
"If satisfies (2), it is clear that height $(x)=r .$ If $\mathfrak{o}x \cap N \neq 0,$ then for some $m<r$ we shall have $\mathfrak{p}^{m} x \subset N,$ and therefore $\mathfrak{p}^{r-1} x$ is contained in the socle $N_{1}$ of $N .$ since $N$ is generated by elements of height $\geqslant r,$ it follows that $\mathfrak{p}^{r-1} x=\mathfrak{p}^{r-1} y$ for some $y \in N ;$ hence $x-y \in M_{r-1}$ and therefore $x \in\left(M_{r-1}+N\right)\cap M_{r}=M_{r-1}+N_{r}$"