Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ is a finitely generated $R$-module of depth $t$. It is well known that every maximal regular sequence of $M$ has length $t$. Recall that $x\_1, \dotsc, x\_t \in \mathfrak{m}$ is an $M$-regular sequence if $x\_i$ is a non-zero divisor of $M/(x\_1, \dotsc, x\_{i-1})M$ for all $i = 1, \dotsc, t$, i.e. the multiplicative map $x\_i: M/(x\_1, \dotsc, x\_{i-1})M \to M/(x\_1, \dotsc, x\_{i-1})M$ is injective.


Now we consider a sequence of endomorphisms instead of multiplications.


**Definition.** A sequence of endomorphism $\varphi\_1, \dotsc, \varphi\_t \in \operatorname{End}(M)$ is called a $M$-regular sequence if


(1) For all $i = 1, \dotsc, t$, $\operatorname{Im}(\varphi\_i) \subseteq \mathfrak{m}M$.


(2) For all $i =1, \dotsc, t$, $\varphi\_i$ induces an injective endomorphism on $M/(\operatorname{Im}(\varphi\_1), \ldots, \operatorname{Im}(\varphi\_{i-1}))$.


**Question 1.** Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ be a finitely generated $R$-module of depth $t$. Does every maximal $M$-regular sequence of endomorphims of $M$ have length $t$?


**Update:** Based on Mohan's answer we will assume the our endomorphisms commute. It is natural to ask the following question.


**Question 2.** Suppose $\varphi\_1, \ldots, \varphi\_t$ is an $M$-regular sequence of endomorphisms. Is every permutation of $\varphi\_1, \ldots, \varphi\_t$ an $M$-regular sequence of endomorphisms?



Depth of modules and regular sequences of endomorphisms
Let me give one proof of what I said in the comment. Proof is by induction on the depth. Endomorphism of a module, to avoid repetition, will mean an injective map with image
contained in maximal ideal times the module.


First we deal with depth zero. Then I claim there are no such endomorphisms. If $\phi:M\to M$ is any endomorphism, and $N\subset M$ is the maximal finite length submodule with $M/N$ having positive depth, it is immediate that $\phi(N)\subset N$. But $\phi$ is injective implies, by length considerations, $\phi(N)=N$. So, we get $\phi^r(N)=N$ for all $r$. Since $\phi(M)\subset\mathfrak{m}M$, we see that $N\subset \mathfrak{m}^rM$ for all $r$. But this implies $N=0$ and contradicting our assumption on $M$.


So, assume by induction we have proved this for all smaller depth and now let $M$ have depth $t>0$. Let $\phi\_i, 1\leq i\leq r$ be a maximal sequence as you have. If $r<t$, you can easily check that $M/(\phi\_1(M),\ldots,\phi\_r(M))$ has depth $t-r>0$ since $\phi\_i\phi\_j=\phi\_j\phi\_i$. Then picking a non-zero divisor $x\in \mathfrak{m}$ for this module, we see that we can take $\phi\_{r+1}$ to be multiplication by $x$. So, we may assume $r\geq t$. But, same argument says, $M/(\phi\_1(M),\ldots, \phi\_t(M))$ has depth zero and so by the first argument, $r=t$.


