Does a finite coresolution with Specht-filtered modules imply a Specht filtration?

Question

$\newcommand{\sym}{\mathfrak{S}}$ $\newcommand{\rarr}{\rightarrow}$ Given a partition $\lambda$ of $n$ and a commutative ring $R$, writing $\sym_n$ for the symmetric group, there is a Specht module $S^\lambda$ well-defined as an $R\sym_n$-module. For example James does this in his book using polytabloids.

By a Specht filtration on an $R\sym_n$-module $I$, I mean a finite filtration of $I$ where each cofactor is isomorphic to a Specht module.

Suppose there is an exact sequence $$0 \rarr M \rarr I^0 \rarr I^1 \rarr \cdots \rarr I^r \rarr 0$$ of $R\sym_n$-modules where each $I^j$ for $0 \leq j \leq r$ has a Specht filtration. Does this imply $M$ should also have a Specht filtration in general? What if $R$ is a field whose characteristic is at least 5? (Hemmer and Nakano have papers showing Specht filtrations are better behaved in this situation)

Answer 1

No. Working with modules for $\mathbb{F}_5S_5$, take the Specht module $S^{(2,1,1,1)}$. Using $S^{(2,1,1,1)} \cong (S^{(4,1)} \otimes \mathrm{sgn})^\star$ and that $S^{(4,1)}$ has top $D^{(4,1)}$ and socle $\mathbb{F}_5$, we get a short exact sequence

$$0 \rightarrow D^{(4,1)} \otimes \mathrm{sgn} \rightarrow S^{(2,1,1,1)} \rightarrow \mathrm{sgn}\rightarrow 0. $$

Here $S^{2,1,1,1)}$ and $\mathrm{sgn} \cong S^{(1^5)}$ are Specht modules, but the simple module $D^{(4,1)} \otimes \mathrm{sgn}$ has dimension $3$. Since the Specht modules for $\mathbb{F}_5S_5$ (other than $S^{(5)} \cong \mathbb{F}_5$ and $S^{(1^5)} \cong \mathrm{sgn}$) all have dimension at least $4$, $D^{(4,1)} \otimes \mathrm{sgn}$ does not have a Specht filtration.