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The James submodule theorem says that for every $F{S}_n$-submodule $U\le M^{\lambda }$, eitherwhere 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 identityIf $U⊈(S^{\lambda }{)}^{\perp }$, choose $u\in U$ and a tableau $t$ with $⟨u,e(t)⟩\ne 0$. 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 $F$.

The preceding part shows that $\theta (e(t^{\prime }))\ne 0$ in characteristic zero, so $S^{{\lambda }^{\prime }}⊈\ker \theta$. Apply the James submodule theorem to the proper submodule $\ker \theta \le M^{{\lambda }^{\prime }}$ to obtainThe Hook-length formula gives $\dim S^{\lambda }=\dim S^{{\lambda }^{\prime }}$. Surjectivity of $\theta$ therefore givesand hence

Fix the original tableaux $t,u$. For a $\lambda$-tableau $w$, let $g_w$ be the unique permutation satisfying $w=g_{w}t$ and put ${\epsilon }_w=\mathrm{sgn}(g_w)$. Sincethe quotient pairing $M^{{\lambda }^{\prime }}/(S^{{\lambda }^{\prime }}{)}^{\perp }\cong (S^{{\lambda }^{\prime }}{)}^{\ast }$ gives the explicit conjugate Specht module as a sign-twisted dual isomorphismfor every $\lambda$-tableau $w$.