= Solution
The <standard polytabloid> <basis> is defined over the <integers> and remains a <basis> after reduction modulo three. Thus $\dim S^{(3,2)}=f^{(3,2)}=\binom52-\binom51=5$. The given <submodule> has dimension four by part (d), so the quotient is one-dimensional and part (c) leaves only the trivial or sign action.
A transposition has ordinary trace one on $S^{(3,2)}$: the <Murnaghan–Nakayama rule> has only the horizontal bottom-row two-strip, leaving shape $(3)$ with dimension one. Reduction of its integral action <matrix> preserves this trace in $F$. Its trace on $D^{(4,1)}=S^{(4,1)}$ is $3-1=2$, since the natural five-point <permutation> <representation> has three fixed coordinates and contains the trivial line. Therefore its trace on the one-dimensional quotient is $1-2=-1$, distinct from one in characteristic three. This gives the <characteristic-three sign quotient of the Specht module of shape three two>. The quotient is the <sign representation>:
$$
\boxed{\sigma\cdot\overline v=\operatorname{sgn}(\sigma)\,\overline v
\quad\text{on }S^{(3,2)}/D^{(4,1)}.}
$$
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