 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 $b_t\{u\}=\mathrm{sgn}(c)e(t)$. Hence every basis tabloid maps into $\mathbb{F}e(t)$, while $b_t\{t\}=e(t)$, and therefore