= Absolute irreducibility of the Specht radical quotient
{title2=$D_F^\lambda=S_F^\lambda/(S_F^\lambda\cap(S_F^\lambda)^\perp)$}
The <James submodule theorem> follows from $\kappa_tv=\langle v,e_t\rangle e_t$. A nonzero restricted form therefore gives a simple radical quotient. If $m_j$ counts equal-length rows, all integral <polytabloid> pairings are divisible by $\prod_jm_j!$, while pairing with row reversal gives $\prod_j(m_j!)^j$. Thus the form is nonzero modulo $p$ exactly for <regular partitions>. The construction and the proof commute with every <field extension>, so the quotient is absolutely <irreducible>, including over the <prime field>.
Back to article page
