Field antisymmetrizers have rank-one own-shape image and detect dominance

For $\gamma ,\delta \in C_t$ the sign is multiplicative, so in the group algebra over any ring ${\kappa }_t\gamma =\mathrm{sgn}(\gamma ){\kappa }_t$ and in particular ${\kappa }_t^2=\mathrm{\mid }{C}_t\mathrm{\mid }{\kappa }_t$; and ${\kappa }_t$ acts $F$-linearly on $M_F^{\lambda }$ through the group action. Moreover the coefficient of $\{t\}$ in $e_t$ is $1$ by [F2], so $e_t\mathrm{\ne }0$ over $F$.

Let $\{s\}$ be a $\mu$-tabloid whose row contains two entries $x,y$ lying in one column of $t$, so that $\tau =(xy)\in C_t$. Writing $Z$ for a set of left coset representatives of $⟨\tau ⟩$ in $C_t$ gives the integral group-algebra identity ${\kappa }_t={\sum }_{z\in Z}\mathrm{sgn}(z)z(1-\tau )$; since $\tau \in R_s$ fixes the tabloid $\{s\}$ by [F3], applying this to $\{s\}$ gives ${\kappa }_t\{s\}={\sum }_{z\in Z}\mathrm{sgn}(z)(z\{s\}-z\tau \{s\})=0$. This is an identity between integral vectors, so it holds in $M_{\mathbb{Z}}^{\mu }$ and hence over $F$: the collision criterion of [F4] is field-independent.

Suppose ${\kappa }_t\{s\}\mathrm{\ne }0$ for a $\mu$-tabloid $\{s\}$. Then step 1.2 shows no row of $\{s\}$ contains two entries from one column of $t$, i.e. every row of a representing tableau meets every column of $t$ in at most one entry; by [F5] this gives $\lambda ⊵\mu$, and when $\lambda =\mu$ it gives $\rho \in R_s$, $\gamma \in C_t$ with $\rho \cdot s=\gamma \cdot t$. In the equal-shape case ${\kappa }_t\{s\}={\kappa }_t\{\rho \cdot s\}={\kappa }_t\gamma \cdot \{t\}=\mathrm{sgn}(\gamma ){\kappa }_t\{t\}=\mathrm{sgn}(\gamma )e_t$ by step 1.1.

Every element of $M_F^{\lambda }$ is an $F$-combination of $\lambda$-tabloids, and by step 2.1 each ${\kappa }_t\{s\}$ is either $0$ or $\pm e_t$; hence ${\kappa }_t{M}_F^{\lambda }\subseteq F{e}_t$. Since $e_t={\kappa }_t\{t\}\mathrm{\ne }0$ lies in the image, ${\kappa }_t{M}_F^{\lambda }=F{e}_t$, as asserted.

If ${\kappa }_t{M}_F^{\mu }\mathrm{\ne }0$, some $\mu$-tabloid $\{s\}$ satisfies ${\kappa }_t\{s\}\mathrm{\ne }0$, so step 2.1 gives $\lambda ⊵\mu$ in the order of [F6].

Over $\mathbb{C}$ the conclusions of steps 3.1 and 3.2 are exactly the published statements [F7]; the present proof rederives them over an arbitrary field from the integral collision identity of step 1.2 and the combinatorial lemma [F5], both of which involve only coefficients $0,\pm 1$, so the argument applies in characteristic two. For $n=0$ the empty tableau has ${\kappa }_t=1$ and $e_t=\{\mathrm{\varnothing }\}\mathrm{\ne }0$, the only partition is $\mathrm{\varnothing }$ with $\mathrm{\varnothing }⊵\mathrm{\varnothing }$, and the three displayed claims hold.