[proofplan] We prove the basis theorem by separating spanning from [linear independence](/page/Linear%20Independence). For spanning, we invoke the Garnir straightening theorem in the precise form needed here: every polytabloid indexed by an arbitrary tableau is a finite $\mathbb C$-linear combination of standard polytabloids. For linear independence, we order tabloids compatibly with dominance and show that, in the expansion of a standard polytabloid, its own tabloid is the unique leading tabloid and appears with coefficient $1$. The resulting transition matrix from standard polytabloids to tabloids is triangular with nonzero diagonal entries. [/proofplan]

[step:Fix the tableau and tabloid orders used for straightening] Let $\mathcal T_\lambda$ denote the finite set of all bijective $\lambda$-tableaux with entries $\{1,\dots,n\}$. Let $\mathcal T_\lambda^{\operatorname{std}}\subset \mathcal T_\lambda$ denote the subset of standard tableaux, meaning tableaux whose entries increase strictly from left to right along every row and strictly from top to bottom along every column. Let $\mathcal R_\lambda\subset \mathcal T_\lambda$ denote the subset of row-standard tableaux, meaning tableaux whose entries increase strictly from left to right along every row. Let $\mathcal B_\lambda$ denote the finite set of all $\lambda$-tabloids. Define the row-standardization map \begin{align*} r:\mathcal B_\lambda&\to\mathcal R_\lambda \end{align*} by letting $r(\{u\})$ be the unique row-standard tableau obtained by sorting each row of the tabloid $\{u\}$ increasingly. For $a\in\mathcal R_\lambda$, $m\in\{1,\dots,n\}$, and $k\in\mathbb N$, define $N_k(a,m)$ to be the number of entries among $\{1,\dots,m\}$ that occur in one of the first $k$ rows of $a$, with rows below the last row of $\lambda$ contributing no entries. Define a partial order $\unlhd$ on $\mathcal R_\lambda$ as follows. For $a,b\in\mathcal R_\lambda$, write $a\unlhd b$ if \begin{align*} N_k(a,m)\le N_k(b,m) \end{align*} for every $m\in\{1,\dots,n\}$ and every $k\in\mathbb N$. Choose once and for all a total order $\le_{\lambda}$ on $\mathcal R_\lambda$ extending this partial order. We order tabloids by declaring \begin{align*} \{u\}\le_{\lambda}\{v\} \end{align*} when \begin{align*} r(\{u\})\le_{\lambda}r(\{v\}). \end{align*} Because $\mathcal R_\lambda$ is finite, every strictly increasing sequence for $\le_{\lambda}$ terminates. [/step]

[step:Declare the stabilizers and the polytabloid expansion]Let $t\in\mathcal T_\lambda$. By the theorem statement, $R_t\le S_n$ is the row stabilizer of $t$, $C_t\le S_n$ is the column stabilizer of $t$, and the associated polytabloid is \begin{align*} e_t=\sum_{\gamma\in C_t}\operatorname{sgn}(\gamma)\{\gamma t\}\in M^\lambda. \end{align*} Thus $S^\lambda$ is the $\mathbb C$-span of the vectors $e_t$ as $t$ ranges over all tableaux in $\mathcal T_\lambda$.[/step]

[guided]The theorem statement has now fixed all objects used in the proof. For each tableau $t\in\mathcal T_\lambda$, the subgroup $R_t\le S_n$ consists of the permutations preserving every row of $t$ setwise, and the subgroup $C_t\le S_n$ consists of the permutations preserving every column of $t$ setwise. The polytabloid is the alternating column sum \begin{align*} e_t=\sum_{\gamma\in C_t}\operatorname{sgn}(\gamma)\{\gamma t\}\in M^\lambda. \end{align*} This formula is the only expansion of $e_t$ used below. Since $S^\lambda$ is defined as the $\mathbb C$-span of all such $e_t$, proving the spanning half of the theorem means proving that every vector $e_t$ with $t\in\mathcal T_\lambda$ lies in the span of the standard polytabloids.[/guided]

[step:Straighten every polytabloid to standard polytabloids]We use the Garnir straightening theorem for Specht modules in the following form. For every $n\in\mathbb N$, every partition $\lambda\vdash n$, every complex permutation module $M^\lambda$ on $\lambda$-tabloids, and every polytabloid $e_t\in M^\lambda$ associated to a bijective $\lambda$-tableau $t\in\mathcal T_\lambda$, there are scalars $a_v\in\mathbb C$, indexed by finitely many standard tableaux $v\in\mathcal T_\lambda^{\operatorname{std}}$, such that \begin{align*} e_t=\sum_v a_v e_v. \end{align*} The hypotheses of this theorem match the present setting: the ground field is $\mathbb C$, $\lambda\vdash n$, $M^\lambda$ is the complex permutation module on $\lambda$-tabloids, and the vectors $e_t$ are precisely the alternating column sums defined above. Therefore every generator $e_t$ of $S^\lambda$ lies in the $\mathbb C$-span of the standard polytabloids. Since $S^\lambda$ is, by definition, the $\mathbb C$-span of all polytabloids $e_t$ with $t\in\mathcal T_\lambda$, the displayed straightening formula implies that the standard polytabloids span $S^\lambda$.[/step]

[guided]The spanning argument uses one external input: Garnir straightening. The version needed here says that, for the complex Specht module constructed from $\lambda$-tabloids, every polytabloid $e_t$ attached to an arbitrary bijective tableau $t$ can be rewritten as a finite $\mathbb C$-linear combination of polytabloids $e_v$ attached to standard tableaux $v$. Its hypotheses are exactly the objects already fixed: $n\in\mathbb N$, $\lambda\vdash n$, the complex tabloid module $M^\lambda$, and the polytabloids defined by alternating over the column stabilizer. Applying Garnir straightening to a fixed $t\in\mathcal T_\lambda$, we obtain scalars $a_v\in\mathbb C$, with only finitely many nonzero terms and with $v$ standard, such that \begin{align*} e_t=\sum_v a_v e_v. \end{align*} This matters because $S^\lambda$ is defined as the span of all polytabloids $e_t$, not only the standard ones. Since each generator $e_t$ is in the span of the standard polytabloids, the whole space $S^\lambda$ is contained in that span. The reverse containment is immediate because every standard polytabloid is one of the generators of $S^\lambda$. Hence the standard polytabloids span $S^\lambda$.[/guided]

[step:Identify the leading tabloid in each standard polytabloid] Let $t\in\mathcal T_\lambda^{\operatorname{std}}$. Expanding the polytabloid gives \begin{align*} e_t=\sum_{\gamma\in C_t}\operatorname{sgn}(\gamma)\{\gamma t\}. \end{align*} The theorem statement defines $1_{S_n}$ as the identity element of $S_n$. The term corresponding to $\gamma=1_{S_n}$ is $\{t\}$ and has coefficient $1$. We use the following dominance lemma for standard tableaux. If $t$ is standard and $\gamma\in C_t$, then \begin{align*} r(\{\gamma t\})\unlhd t, \end{align*} with equality only when $\gamma=1_{S_n}$. To see this, fix $m\in\{1,\dots,n\}$ and $k\in\mathbb N$. In the standard tableau $t$, the entries in $\{1,\dots,m\}$ form an order ideal in the Young diagram: if a box contains an entry at most $m$, then every box weakly above it in the same column and weakly to its left in the same row also contains an entry at most $m$. A column permutation can only move entries within their original columns, and row-standardization changes only the order inside rows, not the set of entries lying in the first $k$ rows. Hence the number of entries from $\{1,\dots,m\}$ lying in the first $k$ rows after applying $\gamma$ and row-standardizing is at most the corresponding number for $t$, which is precisely \begin{align*} N_k(r(\{\gamma t\}),m)\le N_k(t,m). \end{align*} If equality holds for every $k$ and $m$, then each initial set $\{1,…,m\}$ occupies the same collection of row positions before and after applying $\gamma$, so every entry remains in its original row. Since $\gamma\in C_t$ also keeps every entry in its original column, every entry remains in its original box, and therefore $\gamma=1_{S_n}$. Thus every tabloid occurring in $e_t$ is at most $\{t\}$ in the chosen order, and $\{t\}$ occurs with coefficient $1$ as the unique leading tabloid. [/step]

[step:Use triangularity to prove linear independence] Let $\mathcal S_\lambda=\mathcal T_\lambda^{\operatorname{std}}$, and suppose that a finite linear relation among standard polytabloids is given by \begin{align*} \sum_{t\in\mathcal S_\lambda} c_t e_t=0, \end{align*} with coefficients $c_t\in\mathbb C$. Since $\mathcal S_\lambda$ is finite, choose $t_0\in\mathcal S_\lambda$ maximal with respect to $\le_{\lambda}$ among those tableaux for which $c_{t_0}\ne 0$. By the previous step, the tabloid $\{t_0\}$ occurs in $e_{t_0}$ with coefficient $1$. If $t\in\mathcal S_\lambda$ and $t\ne t_0$, then $\{t_0\}$ cannot occur in $e_t$ when $t<_{\lambda}t_0$, because every tabloid occurring in $e_t$ is at most $\{t\}$. It also cannot occur for a distinct $t$ with $t=t_0$ as a row-standard tableau, since standard tableaux are uniquely determined by their tabloids after row-standardization. Therefore the coefficient of the basis tabloid $\{t_0\}$ in the left-hand side is exactly $c_{t_0}$. The tabloids of shape $\lambda$ form the defining basis of the permutation module $M^\lambda$, so the zero vector has coefficient $0$ on every tabloid. Hence $c_{t_0}=0$, contradicting the choice of $t_0$. Therefore all coefficients $c_t$ are zero, and the standard polytabloids are linearly independent. [/step]

[step:Conclude that the standard polytabloids form a basis] The standard polytabloids span $S^\lambda$ by straightening, and they are linearly independent by the triangular leading-tabloid argument. Hence \begin{align*} \{e_t : t\in\mathcal T_\lambda^{\operatorname{std}}\} \end{align*} is a $\mathbb C$-basis of $S^\lambda$. [/step]