Nonzero maps into tabloid quotients force dominance

Since $L_{\lambda }$ and $U_{\lambda }$ are products of the same factorials $z_j!$ (in different multiplicities), a prime divides $L_{\lambda }$ if and only if it divides $U_{\lambda }$. As $\lambda$ is $p$-regular, [F1] and [F2] give $p\nmid g_{\lambda }$, hence $p\nmid L_{\lambda }$; if $p\mid U_{\lambda }$ then $p\mid L_{\lambda }$ by the first observation, a contradiction. Therefore $U_{\lambda }$ has nonzero image in $F$. In particular, for every $\lambda$-tableau $t$ the relation ${\kappa }_t{e}_{t^{\ast }}=U_{\lambda }{e}_t$ of [F2] has $U_{\lambda }\mathrm{\ne }0$ in $F$.

Let $N$ be an $F[S_n]$-module and $\phi :S_F^{\lambda }\to N$ an $F[S_n]$-homomorphism with $\phi (e_t)=0$ for one $\lambda$-tableau $t$. By [F3] every $\lambda$-polytabloid is $\sigma \cdot e_t$ for some $\sigma \in S_n$, so $\phi (\sigma \cdot e_t)=\sigma \cdot \phi (e_t)=0$ for all $\sigma$; since the polytabloids span $S_F^{\lambda }$, $\phi =0$. Hence a nonzero $\phi$ satisfies $\phi (e_s)\mathrm{\ne }0$ for every $\lambda$-tableau $s$.

For every partition $\nu ⊢n$ the space $(S_F^{\nu }{)}^{\perp }$ is an $F[S_n]$-submodule of $M_F^{\nu }$: if $x\in (S_F^{\nu }{)}^{\perp }$, $\sigma \in S_n$ and $s\in S_F^{\nu }$, then ${\beta }_F(\sigma x,s)={\beta }_F(x,{\sigma }^{-1}s)=0$ because ${\sigma }^{-1}s\in S_F^{\nu }$ by [F7].

Let $\phi :S_F^{\lambda }\to M_F^{\mu }\mathrm{/}U$ be a nonzero $F[S_n]$-homomorphism, and put $v:=\phi (e_{t^{\ast }})\in M_F^{\mu }\mathrm{/}U$ for a $\lambda$-tableau $t$. Using [F2], the $F[S_n]$-linearity of $\phi$ and steps 1.1-1.2, ${\kappa }_{t}v={\kappa }_t\phi (e_{t^{\ast }})=\phi ({\kappa }_t{e}_{t^{\ast }})=\phi (U_{\lambda }{e}_t)=U_{\lambda }\phi (e_t)\mathrm{\ne }0.$ Choose $w\in M_F^{\mu }$ with $v=w+U$. Then ${\kappa }_{t}w+U={\kappa }_{t}v\mathrm{\ne }0$, so ${\kappa }_{t}w\mathrm{\ne }0$ and hence ${\kappa }_t{M}_F^{\mu }\mathrm{\ne }0$; by [F4] this forces $\lambda ⊵\mu$. This is assertion 1.

Suppose now that $\lambda =\mu$ and let $\phi ,v,w$ be as in step 2.1, so ${\kappa }_{t}w\mathrm{\ne }0$ by that step. By [F4] one has ${\kappa }_t{M}_F^{\lambda }=F{e}_t$, so ${\kappa }_{t}w=c{e}_t$ for a unique $c\in F$, and $c\mathrm{\ne }0$ because ${\kappa }_{t}w\mathrm{\ne }0$. From $U_{\lambda }\phi (e_t)={\kappa }_{t}v={\kappa }_{t}w+U=c{e}_t+U$ and step 1.1 we get $\phi (e_t)=\frac{c}{U_{\lambda }}{e}_t+U=\frac{c}{U_{\lambda }}(e_t+U),$ a nonzero multiple of $e_t+U$ in $M_F^{\lambda }\mathrm{/}U$; hence $e_t+U\in \mathrm{Im}\phi$. Since $\mathrm{Im}\phi$ is an $F[S_n]$-submodule and $S_F^{\lambda }=F[S_n]e_t$ by [F3], this gives $(S_F^{\lambda }+U)\mathrm{/}U=F[S_n](e_t+U)\subseteq \mathrm{Im}\phi$. Conversely, every $x\in S_F^{\lambda }$ is a finite sum $x={\sum }_{\sigma }{a}_{\sigma }\sigma e_t$ with $a_{\sigma }\in F$, and then $\phi (x)={\sum }_{\sigma }{a}_{\sigma }\sigma \phi (e_t)={\sum }_{\sigma }{a}_{\sigma }\sigma (\frac{c}{U_{\lambda }}(e_t+U))=\frac{c}{U_{\lambda }}(x+U)\in (S_F^{\lambda }+U)\mathrm{/}U,$ so $\mathrm{Im}\phi \subseteq (S_F^{\lambda }+U)\mathrm{/}U$. Therefore $\mathrm{Im}\phi =(S_F^{\mu }+U)\mathrm{/}U$, which is nonzero because $\phi \mathrm{\ne }0$, and consequently $U$ does not contain $S_F^{\mu }$. This is assertion 2.

Let $\psi :D_F^{\lambda }\to M_F^{\mu }\mathrm{/}U$ be a nonzero $F[S_n]$-homomorphism and let $\pi :S_F^{\lambda }\twoheadrightarrow D_F^{\lambda }$ be the quotient map. Then $\phi :=\psi \circ \pi :S_F^{\lambda }\to M_F^{\mu }\mathrm{/}U$ is nonzero, so steps 2.1 and 3.1 apply to $\phi$ and give $\lambda ⊵\mu$; if $\lambda =\mu$, they give $\mathrm{Im}\psi =\mathrm{Im}\phi =(S_F^{\mu }+U)\mathrm{/}U$, which is nonzero, so $U⊉S_F^{\mu }$. This is assertion 3.

Suppose $\lambda$ and $\mu$ are both $p$-regular and let $\theta :D_F^{\lambda }\to D_F^{\mu }$ be an $F[S_n]$-isomorphism. Put $U:=(S_F^{\mu }{)}^{\perp }$, a submodule of $M_F^{\mu }$ by step 1.3. The natural map $j:S_F^{\mu }\to M_F^{\mu }\mathrm{/}U$ has kernel $S_F^{\mu }\cap (S_F^{\mu }{)}^{\perp }=R_F^{\mu }$, so it induces an injective $F[S_n]$-homomorphism $\iota :D_F^{\mu }\hookrightarrow M_F^{\mu }\mathrm{/}U$. Hence $\iota \circ \theta :D_F^{\lambda }\to M_F^{\mu }\mathrm{/}U$ is nonzero and step 4.1 gives $\lambda ⊵\mu$. By symmetry, with $U^{\mathrm{\prime }}:=(S_F^{\lambda }{)}^{\perp }$ let ${\iota }^{\mathrm{\prime }}:D_F^{\lambda }\hookrightarrow M_F^{\lambda }\mathrm{/}{U}^{\mathrm{\prime }}$ be the corresponding injection of step 1.3; then ${\iota }^{\mathrm{\prime }}\circ {\theta }^{-1}:D_F^{\mu }\to M_F^{\lambda }\mathrm{/}{U}^{\mathrm{\prime }}$ is nonzero, and step 4.1 with the roles of $\lambda$ and $\mu$ exchanged (both are $p$-regular) gives $\mu ⊵\lambda$. Antisymmetry of the dominance order [F6] yields $\lambda =\mu$. This is assertion 4 in the case that both labels are $p$-regular.

Steps 1.1-1.3, 2.1, 3.1, 4.1 and 5.1 prove assertions 1-4. For the final sentence of assertion 4, if $\lambda$ is $p$-regular, $\mu ⊢n$ and $D_F^{\lambda }\cong D_F^{\mu }$, then $D_F^{\lambda }\mathrm{\ne }0$ by [F5], so $D_F^{\mu }\mathrm{\ne }0$, hence $\mu$ is $p$-regular by [F5]; now step 5.1 gives $\mu =\lambda$. The boundary cases are included: for $n=0$ one has $\lambda =\mu =\mathrm{\varnothing }$, $S_F^{\mathrm{\varnothing }}=M_F^{\mathrm{\varnothing }}$ is one-dimensional, $R_F^{\mathrm{\varnothing }}=0$, $D_F^{\mathrm{\varnothing }}\cong F\mathrm{\ne }0$, and $M_F^{\mathrm{\varnothing }}\mathrm{/}U$ is nonzero only for $U=0$, in which case assertion 1 is trivial and assertion 2 reads $\mathrm{Im}\phi =S_F^{\mathrm{\varnothing }}$ for every nonzero $\phi$; for $U=0$ the equality case says a nonzero $S_F^{\lambda }\to S_F^{\lambda }$ is surjective, and for $U=M_F^{\mu }$ the codomain is zero so no nonzero $\phi$ exists. Characteristic $2$ is covered because at no point is the sign of a permutation used, and no step divides by a group order or averages over $S_n$.