= Solution
Let $b_t$ be the <Column antisymmetrizer of a Young tableau>. For a tabloid $\{u\}$, if two entries from one column of $t$ lie in one row of $u$, their column transposition fixes $\{u\}$ and pairs every term of $b_t\{u\}$ with its negative. Thus $b_t\{u\}=0$.
Otherwise each row of $u$ meets each column of $t$ at most once. Matching entries within columns then gives a column permutation $c\in C(t)$ for which $\{cu\}=\{t\}$. Reindexing the antisymmetrizer gives $b_t\{u\}=\operatorname{sgn}(c)e(t)$. Hence every basis tabloid maps into $\mathbb F e(t)$, while $b_t\{t\}=e(t)$, and therefore
$$
b_tM^\lambda=\mathbb F e(t).
$$
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