Depth of modules and regular sequences of endomorphisms

To prove the full statement, the key point is:

Lemma: Let $\phi: M\to M$ be a map such that $\phi(M)\subset mM$. Then the induced map $\phi_i: H^i_m(M) \to H^i_m(M)$ on each local cohomology module satisfies: $\ker(\phi_i)$ is nonzero if $H_m^i(M)\neq 0$.

Proof: Let $N=H^i_m(M)$. Let $z\in N$ be non-zero. By definition of local cohomology, one can check that $\phi_i^r(z)=0$ for $r\gg0$ (write $z=(\frac{a_1}{x_1},...\frac{a_n}{x_n})$ with $a_j\in M$ and $x_j\in m$, then $\phi_i^r(z)=(\frac{\phi_i^r(a_1)}{x_1},...\frac{\phi_i^r(a_n)}{x_n})$ . Let $L$ be the $N$-submodule $N\cap(\oplus \frac{M}{x_j})$, namely the collection of elements $(\frac{b_1}{x_1},...\frac{b_n}{x_n})$ in $N$ with $b_j\in M$. As the $x_j$ are fixed, $L$ is a finitely generated submodule of $H^i_m(M)$ and therefore has finite length. Then $\phi_i^r(z)\in m^rL=0$ for $r\gg0$). Choose $r$ smallest, then $\phi_i^{r-1}(z)\in \ker(\phi_i)$.

Now, if you have an injective map $\phi$, then the Lemma implies that $M$ has positive depth since we can apply it with $i=0$ to get that $H^0_m(M)=0$ (this is similar to that part in Mohan's proof). The long exact sequence of local cohomology coming from $0 \to M \to M \to M/\phi(M)\to 0$ and the Lemma again tells us that $depth(M/\phi(M))= depth(M)-1$. Induction finishes the statement.

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, $r=t$.

Here I write a sketch of proof which answers both the questions. Consider the $R$ sub-algebra $S$ of $\operatorname{End} M$ generated by the $\phi_i$s. Then, by our assumption, $S$ is commutative, it is a finite type $R$-module and using the assumption $\phi_i(M)\subset\mathfrak{m}M$, it is also local. $M$ is naturally an $S$-module. Under these hypotheses, it is easy to check that $\operatorname{depth}_R M=\operatorname{depth}_S M$ and that you can easily see answers both the questions.