= Solution
The <James submodule theorem> says that for every $FS_n$-submodule $U\leq M^\lambda$, either
$$
S^\lambda\subseteq U
\qquad\text{or}\qquad
U\subseteq(S^\lambda)^\perp,
$$
where orthogonality is taken with respect to the <tabloid bilinear form>.
Fix a $\lambda$-tableau $t$. Part a shows that for every tabloid $\{s\}$, the vector $b_t\{s\}$ is either zero or a signed copy of the <polytabloid> $e(t)$. Comparing the coefficient of $\{t\}$ gives the precise identity
$$
b_tu=\langle u,e(t)\rangle e(t)
\qquad (u\in M^\lambda).
$$
If $U\nsubseteq(S^\lambda)^\perp$, choose $u\in U$ and a tableau $t$ with $\langle u,e(t)\rangle\ne0$. Since $U$ is a submodule, the identity puts $e(t)$ in $U$. Every polytabloid of shape $\lambda$ is an $S_n$-translate of $e(t)$, so their span $S^\lambda$ lies in $U$. If no such $u,t$ exist, then by definition $U\subseteq(S^\lambda)^\perp$. This proves the theorem over the arbitrary field $F$.
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