Let $A$ be a noetherian local ring with maximal ideal $m$. Let $p$ be a prime ideal such that if $B=A/p$, then $\mathrm{dim}\;B=1$. Take $x\not\in p$, $x\in m$, and set $C=B/xB$. Then $C$ has finite length.
Why does $C$ have finite length?
By additivity of the length, if $B$ and $C$ have finite length then $l(C)=0$ and so $C=0$, am I right?
Now take a finitely generated module $M$. Suppose to know $\mathrm{Ext}^{i+1}(C,M)\neq0$ and suppose $l(C)>1$. Then the exact sequence $0\rightarrow k\rightarrow C\rightarrow C^\prime\rightarrow0$ yields an exact sequence $\mathrm{Ext}^{i+1}(C^\prime,M)\rightarrow\mathrm{Ext}^{i+1}(C,M)\rightarrow\mathrm{Ext}^{i+1}(k,M)$. The notes where I'm studying now claim that this shows that there always exists a module $N$ with $1\leq l(N)< l(C)$ and such that $\mathrm{Ext}^{i+1}(N,M)\neq 0$. My questions now are: why such an $N$ exists and why this implies $\mathrm{Ext}^{i+1}(k,M)\neq0$?