Module M/IM of finite length $\implies$ Ring A/I of finite length This question is due to a proof in an algebra book (on the topic of dimension theory) which I don't fully understand (specifically, the proof of Thm 6.9b) in Kommutative Algebra by Ischebeck). It may have a very simple answer which I currently don't see.

Let A be a semilocal, noetherian ring, $M$ a finite A-module with $\mathrm{Ann}_A(M)=0$ and $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ with $l_A(M/(a_1,\ldots,a_s)M)\lt\infty.$
  Why is $l_A(A/(a_1,\ldots,a_s))\lt\infty$ ?

(where $\mathrm{Ann}_A(M)=\{x\in A\mid xM=0\}$, $\mathrm{Jac}(A)=\bigcap\limits_{\text{m is maximal ideal in } A} m$, $l_X(Y)=$ length of $Y$ as an $X$-module)
My ideas: I know that from $l_A(M/(a_1,\ldots,a_s)M)\lt \infty$ it follows that the ring $A/\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)$ is of finite length, because it can be embedded (as an $A$-module) in $M/(a_1,\ldots,a_s)M.$ But I don't know if/why $\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)=(a_1,\ldots,a_s)$.
Also I don't know if $A$ being semilocal or $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ instead of $a_1,\ldots,a_s\in A$ is of any relevance to the question.
As further information, the final goal is to show that the degree of the polynomial p(n), which for large n gives the length $l_{A/Jac(A)}(M/Jac(A)^nM)$, is lesser or equal the minimum number s for which $l(M/(a_1,..,a_s)M)&lt∞$ as above.

$\mathrm{Supp}(M/IM)=V(I+\mathrm{Ann}M)=V(I)=\mathrm{Supp}(R/I)$. Since $M/IM$ is of finite length, $\mathrm{Supp}(M/IM)$ consists of maximal ideals, so the same is true for $\mathrm{Supp}(R/I)$. Thus $R/I$ is of finite length. The second equality above follows because $\mathrm{Ann}M=0$.
