The full Fock space is not automatically the irreducible basic module

The two finite checks. The shape $(4)$ is not $3$-restricted, because ${\lambda }_1-{\lambda }_2=4-0=4\ge 3$; its conjugate is $(1,1,1,1)$ (four columns of height one), not $(1,1,1)$, and it is not $3$-regular because the positive part $1$ occurs four times, and $4\ge 3$. By [F1], therefore, neither $(4)$ nor $(1,1,1,1)$ is a label of the basic component in its respective convention: $(4)$ is not a basic label in the symmetric-group model, and $(1,1,1,1)$ is not an $e$-regular label on the quantum-group side.

$\mathrm{\mid }1111⟩$ lies outside the field basic module. The integrable module of [F2] has the chosen type-one $P$-grading of the convention, with every weight in ${\mathrm{\Lambda }}_0-Q_{+}$. Since the simple roots are independent, a fixed label weight fixes all residue counts $m_i$ and hence the size ${\sum }_i{m}_i$; only finitely many partitions have that size, so its label-weight space is finite-dimensional. The weight ${\mathrm{\Lambda }}_0$ is maximal and its space is the single vacuum line. Thus, with the declared Choice, [F3] applies to $F_{\mathrm{Q}\mathrm{G}}$ with $N_0=M=U_q\mathrm{\mid }\mathrm{\varnothing }⟩=V({\mathrm{\Lambda }}_0)$. Its lattice splitting embeds $(F_0\cap M)\mathrm{/}q(F_0\cap M)$ into $F_0\mathrm{/}q{F}_0$; the corresponding crystal basis is the empty component, precisely the $e$-regular classes by [F2]. If $\mathrm{\mid }1111⟩\in M$, its residue would lie in their span. This contradicts [F5] and the nonregularity in step 1.1. Hence $\mathrm{\mid }1111⟩\mathrm{\notin }M$.

The ambient lower lift is excluded as well. By [F4] $G(1111)=\mathrm{\mid }1111⟩<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q{F}_0$, so $G(1111)$ has the same residue class as $\mathrm{\mid }1111⟩$; if $G(1111)\in M$, then $G(1111)\in F_0\cap M$ and its residue class would lie in the span of the $e$-regular classes, contradicting the independence of [F5] exactly as in step 2.1. Hence $G(1111)\mathrm{\notin }M$, i.e. the ambient lower global lift is outside the field basic module as well.

Transport to the symmetric-group convention. Applying $J^{-1}$ to the field module $M$ gives $M_{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}M$, and $J^{-1}\mathrm{\mid }1111⟩=M_{(1111{)}^{\mathrm{\prime }}}=M_{(4)}$; since $J^{-1}$ is a linear isomorphism, $M_{(4)}=J^{-1}\mathrm{\mid }1111⟩\mathrm{\notin }{M}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by step 2.1, and likewise $G_{\mathrm{s}\mathrm{y}\mathrm{m}}(4)=J^{-1}G(1111)\mathrm{\notin }{M}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by step 3.1. Since also $M_{(4)}\in F_{\mathrm{s}\mathrm{y}\mathrm{m}}$, the transported full Fock space is strictly larger than its cyclic basic module, refuting the false statement.

The full signature remains meaningful on these shapes. At $(4)$ the corners are the removable node $(1,4)$ of residue $3\equiv 0$ and the addable nodes $(1,5)$ of residue $4\equiv 1$ and $(2,1)$ of residue $-1\equiv 2$; the residue-$0$ word is the single $-$, so the modular signature has good node $(1,4)$ and ${\tilde{e}}_0(4)=(3)$, an edge of the full transported crystal. Both $(4)$ and $(3)$ are non-$3$-restricted, so this edge joins two vertices of other components, not a route into the empty component, in agreement with the component identification of [F1]; modular signature statements restricted to restricted labels do not apply to $(4)$ as a basic label.