= Solution
A <tabloid> records the sets of entries in each row of a tableau. Let $C_t$ be the subgroup permuting entries independently within each column. The <polytabloid> is
$$
e_t=\sum_{\pi\in C_t}\operatorname{sgn}(\pi)\{\pi t\},
$$
and the <Specht module> $S^\lambda$ is spanned by these <vectors>. A <standard polytabloid> is one whose tableau increases along rows and down columns.
Order <tabloids> by first sorting each row increasingly, then comparing the resulting row-reading words lexicographically. For a standard tableau $t$, every nonidentity column <permutation> produces a strictly larger word. In the first row affected by that <permutation>, each replacement comes from farther down the same column and is larger than the entry it replaces; all earlier rows remain unchanged. Sorting the affected row therefore increases its word. Also no nontrivial column <permutation> fixes its row sets, since a cell is the unique intersection of its row and column. Hence the least <tabloid> in $e_t$ is $\{t\}$ with coefficient one.
Distinct standard tableaux have distinct row <tabloids>. In a nonzero <linear relation> among their <polytabloids>, choose the least leading <tabloid> with a nonzero coefficient. None of the <vectors> with larger leading <tabloids> can contribute to it, so its coefficient cannot cancel. This contradiction proves \b[<linear independence> of the <standard polytabloids>].
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