Depth of modules and regular sequences of endomorphisms

Question

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?

Answer 1

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, $r=t$.

Answer 2

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.

Answer 3

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.

Answer 4

Ok, let me answer my second question. What we discuss here says that we can generalize several concepts of sequence of elements for endomorphisms.

For my second question, it is enough to consider two endomorphisms $\varphi_1, \varphi_2$. Suppose $\varphi_1$ is not injective on $M/\varphi_2(M)$. Then we have $x \in M$ such that $x \notin \varphi_2(M)$ but $\varphi_1(x) = \varphi_2(y)$ for some $y$. Since $\varphi_1, \varphi_2$ is an $M$-regular sequence we have $y = \varphi_1(z)$ for some $x$. Thus $\varphi_1(x) = \varphi_2(y) = \varphi_2(\varphi_1(z)) = \varphi_1(\varphi_2(z))$ by commutativity. Thus $x = \varphi_2(z)$ by the injectivity of $\varphi_1$, a contradiction.

We are going to prove that $\varphi_2$ is injective on $M$. Consider the short exact sequence $$0 \to M \overset{\varphi_1}{\to} M \to M/\varphi_1(M) \to 0.$$ Since $\varphi_2$ acts injective on $M/\varphi_1(M)$ we have an isophophism $\mathrm{ker}(\varphi_2) \overset{\varphi_1}{\to} \mathrm{ker}(\varphi_2)$. Repeating Mohan's argument for the $depth = 0$ case we have $\mathrm{ker}(\varphi_2) = 0$. The proof is complete.