Given a filtration of a finitely generated module over a noetherian ring that "looks" split, is it split?

Question

The following question came up while trying to determine whether the extension problems in a spectral sequence are trivial.

Given a noetherian ring $R$ and a finitely generated $R$-module $M$ with a filtration $M=F_0 \supset F_1 \supset \ldots \supset F_n \supset F_{n+1}=0$ such that $M \cong \bigoplus_{i=0}^n F_i/F_{i+1}$ (abstractly, but I do not want to assume that this isomorphism is induced by the maps in the filtration), is the filtration split in the sense that all the short exact sequences $0 \to F_{i+1} \to F_i \to F_i/F_{i+1} \to 0$ are split?

I can prove this for $R=\mathbb{Z}$, but my proof does not seem to generalize well. In that case, one can work $p$-locally and then show that a short exact sequence $0 \to A \to B \to C \to 0$ of $p$-groups,

$|\{x \in B|ord(x)\text{ divides }p^k\}|\geq |\{x \in A|ord(x)\text{ divides }p^k\}|\cdot|\{x \in C|ord(x)\text{ divides }p^k\}|$

, with equality iff the sequence is split, and then use that by the assumed isomorphisms, the number of these elements can never strictly increase.

Steven Landsburgs answer in Do all exact sequences $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split for finitely generated abelian groups? got me hoping that something like this might hold for finitely generated modules over noetherian rings, but a simple induction argument does not seem to work.

Does anyone know if this generalisation is true, and knows a proof?

Answer 1

Take the direct sum of the short exact sequences $$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ for $0\leq i\leq n$.

This has the form $$0\to \bigoplus_{i=1}^n F_i\to \bigoplus_{i=0}^n F_i\to F_0\to 0$$ and so splits by the linked answer of Steven Landsburg.

A direct summand of a split short exact sequence is split, so $$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ is split for each $i$.