Integral Specht lattice and base change

By [F3] the tabloids $\gamma \cdot \{t\}$ with $\gamma \in C_t$ are pairwise distinct, so in the expansion $e_t={\sum }_{\gamma }\mathrm{sgn}(\gamma )\{\gamma \cdot t\}$ no tabloid occurs twice; every coefficient is $0$ or $\pm 1$, the coefficient of $\{t\}$ is $1$, and $e_t\mathrm{\ne }0$. In particular $e_t$ is a genuine nonzero integral vector of $M_{\mathbb{Z}}^{\lambda }$.

For $\sigma \in S_n$ the column sets satisfy $C_{\sigma \cdot t}=\sigma C_t{\sigma }^{-1}$, and the substitution $\gamma =\sigma \delta {\sigma }^{-1}$ gives $e_{\sigma \cdot t}={\sum }_{\delta \in C_t}\mathrm{sgn}(\delta )\{\sigma \delta \cdot t\}=\sigma \cdot e_t$. This identity of two integral vectors holds over $\mathbb{C}$ by [F8], and by the uniqueness of tabloid coordinates it therefore holds in $M_{\mathbb{Z}}^{\lambda }$. Since every $\lambda$-tableau is $\sigma \cdot t_0$ for the tableau $t_0$ that lists $1,\dots ,n$ along the rows of $[\lambda ]$, the lattice $S_{\mathbb{Z}}^{\lambda }$ is generated over $\mathbb{Z}[S_n]$ by the single polytabloid $e_{t_0}$.

The Garnir element of [F5] is an integral group-algebra element, and $G_{X,Y}{e}_t={\sum }_{g\in T}{\sum }_{\gamma \in C_t}\mathrm{sgn}(g)\mathrm{sgn}(\gamma )\{g\gamma \cdot t\}$ has integer coefficients in the tabloid basis. The identity $G_{X,Y}{e}_t=0$ holds in $M_{\mathbb{C}}^{\lambda }$ between two integral vectors, so by uniqueness of tabloid coordinates it holds in $M_{\mathbb{Z}}^{\lambda }$: the Garnir relation is integral.

For $\lambda =\mathrm{\varnothing }$ there is exactly one tabloid and one tableau, with $C_t=\{1\}$, ${\kappa }_t=1$ and $e_t=\{\mathrm{\varnothing }\}$; the empty tableau is standard by [F11]. Thus $M_{\mathbb{Z}}^{\mathrm{\varnothing }}=\mathbb{Z}\{\mathrm{\varnothing }\}\cong \mathbb{Z}$ and $S_{\mathbb{Z}}^{\mathrm{\varnothing }}=\mathbb{Z}\{\mathrm{\varnothing }\}$ has rank one with the empty polytabloid as basis.

Run the published spanning argument inside $M_{\mathbb{Z}}^{\lambda }$: for column-standard $s$ that is not standard, the integral Garnir relation of step 1.3 with the explicit transversal containing $1$ isolates the identity term and gives $e_s=-{\sum }_{A\mathrm{\ne }X}\mathrm{sgn}(g_A)e_{g_A\cdot s}$ with each $g_A$ an explicit product of disjoint swaps; sorting columns and using the covariance of step 1.2 rewrites each $e_{g_A\cdot s}$ as $\pm e_u$ with $u$ column-standard and $s\prec u$ in the finite order [F4]; and any tableau is taken to column-standard form by a column permutation, again by step 1.2. Reverse induction along [F4] therefore gives, for every $\lambda$-tableau $t$, an identity $e_t={\sum }_{u\text{ standard}}{c}_{tu}{e}_u$ with integer coefficients $c_{tu}$. Hence the standard polytabloids span $S_{\mathbb{Z}}^{\lambda }$ over $\mathbb{Z}$.

Order the standard $\lambda$-tableaux $t_1,\dots ,t_d$ so that $\{t_1\}<\cdots <\{t_d\}$ in the tabloid order of [F4], possible since distinct standard tableaux have distinct tabloids and the order is total. By [F7], $e_{t_i}=\{t_i\}+{\sum }_{T<\{t_i\}}{c}_{T}T$ with integer $c_T$, so the coefficient of $\{t_j\}$ in $e_{t_i}$ is $0$ for $j>i$ and $1$ for $j=i$. A relation ${\sum }_i{a}_i{e}_{t_i}=0$ with integral $a_i$ therefore forces $a_d=0$, then $a_{d-1}=0$, and so on; the standard polytabloids are $\mathbb{Z}$-linearly independent and, with step 2.1, form a $\mathbb{Z}$-basis of $S_{\mathbb{Z}}^{\lambda }$.

Let $P:{\mathbb{Z}}^d\to M_{\mathbb{Z}}^{\lambda }$ send the $i$-th standard basis vector to $e_{t_i}$, so that $\mathrm{im}P=S_{\mathbb{Z}}^{\lambda }$ by step 3.1, and let $\pi :M_{\mathbb{Z}}^{\lambda }\to {\mathbb{Z}}^d$ take tabloid coordinates at $\{t_1\},\dots ,\{t_d\}$. By step 3.1 the matrix of $\pi \circ P$ is upper unitriangular with integer entries, hence invertible over $\mathbb{Z}$ by integer back-substitution, and $\rho :=(\pi \circ P{)}^{-1}\circ \pi$ satisfies $\rho \circ P=\mathrm{i}\mathrm{d}$. Thus $P$ is injective and split, $S_{\mathbb{Z}}^{\lambda }$ is a direct summand of the free $\mathbb{Z}$-module $M_{\mathbb{Z}}^{\lambda }$, and $M_{\mathbb{Z}}^{\lambda }\mathrm{/}{S}_{\mathbb{Z}}^{\lambda }$ is free: the lattice is saturated.

Let $R$ be a commutative ring and use the identification $R{\otimes }_{\mathbb{Z}}{M}_{\mathbb{Z}}^{\lambda }\cong M_R^{\lambda }$ of [F10]. The map ${\mathrm{i}\mathrm{d}}_R\otimes P$ has image $R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ and ${\mathrm{i}\mathrm{d}}_R\otimes \rho$ is a left inverse of it, so ${\mathrm{i}\mathrm{d}}_R\otimes P$ is injective; hence the canonical map $R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }\to M_R^{\lambda }$ is injective onto the $R$-span of the vectors $1\otimes e_{t_i}$, which is the $R$-span of the standard polytabloids. By the identity of step 2.1 every polytabloid is an integral combination of standard ones, so this span is exactly $S_R^{\lambda }$; therefore $S_R^{\lambda }\cong R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ for every commutative ring $R$, including rings of prime characteristic.

Taking $R=\mathbb{Z}$ in step 5.1 returns $S_{\mathbb{Z}}^{\lambda }$, and taking $\lambda =\mathrm{\varnothing }$ returns the rank-one lattice $\mathbb{Z}\{\mathrm{\varnothing }\}$ of step 1.4; the standard-polytabloid basis is step 3.1 and saturation is step 4.1. This proves the three asserted properties for all $n\ge 0$, including $n=0$.