The empty partition generates an integrable highest weight submodule of highest weight ${\mathrm{\Lambda }}_0$

The vacuum data. By [F1] the vacuum is killed by all $E_i$, is a $K_i$-eigenvector with eigenvalue $q^{{\delta }_{i0}}$ and a $K_d$-eigenvector with eigenvalue $1$, so $\mathrm{w}\mathrm{t}(\mathrm{\varnothing })={\mathrm{\Lambda }}_0$ by the node-count pairing; in particular the vacuum is a highest weight vector of weight ${\mathrm{\Lambda }}_0$.

Integrability and the highest-vector relations. The nonzero cyclic submodule $M=U_q\mathrm{\mid }\mathrm{\varnothing }⟩$ inherits integrability and the label grading from $F$ by [F2]. Its generator has the highest-vector data of step 1.1, so it meets the integrable highest-vector hypothesis of [F3]. No assertion about the weight of a highest vector in an arbitrary submodule is needed.

Identification and the integral lattice. By [F3] the normalized assignment $v_0\mapsto \mathrm{\mid }\mathrm{\varnothing }⟩$ extends to a homomorphism $V({\mathrm{\Lambda }}_0)\to M$. It is surjective since the image contains the cyclic generator, and injective since its kernel is a submodule of the simple source and the map is nonzero. Thus $M\cong V({\mathrm{\Lambda }}_0)$, which proves simplicity as well as the highest weight identification. The divided node operators preserve the partition Laurent lattice by [F4]; hence $U_A^{-}\mathrm{\mid }\mathrm{\varnothing }⟩$ is an $A$-submodule of that lattice, and it spans $M$ over $\mathbb{Q}(q)$ because ordinary lower words are scalar multiples of divided lower words. This is the claimed integral lattice.

The lower lattice inside $M$. Transport $\mathcal{L}({\mathrm{\Lambda }}_0)$ to $M$ by step 3.1 and call its image $L_M$. By [F4]-[F6], its lowering-word generators lie in $F_0$, so $L_M\subseteq M\cap F_0$. Both lattices have top part $A_0\mathrm{\mid }\mathrm{\varnothing }⟩$. For the reverse inclusion, induct on $\mathrm{ht}({\mathrm{\Lambda }}_0-\mu )$. If $u\in (M\cap F_0{)}_{\mu }$ and $\mu \mathrm{\ne }{\mathrm{\Lambda }}_0$, every ${\tilde{e}}_{i}u$ lies in $L_M$ by lattice stability, naturality and the induction hypothesis. Since $L_M$ generates $M$ over $\mathbb{Q}(q)$, choose the least $n\ge 0$ with $q^{n}u\in L_M$. If $n>0$, its nonzero residue in $(L_M\mathrm{/}q{L}_M{)}_{\mu }$ is killed by every ${\tilde{e}}_i$, because $q^n{\tilde{e}}_{i}u\in q{L}_M$. This contradicts [F5]. Hence $n=0$ and $L_M=M\cap F_0$. All inductions concern finite root height and require no choice principle.

The crystal component. The equality in step 4.1 implies $L_M\cap q{F}_0=q{L}_M$: if $u=qv\in L_M$ with $v\in F_0$, then $v=q^{-1}u\in M\cap F_0=L_M$. Thus $L_M\mathrm{/}q{L}_M$ injects into $F_0\mathrm{/}q{F}_0$. By naturality and the lowering-word descriptions of [F4]-[F5], the canonical residue basis maps into the partition residue basis, its highest element maps to the empty partition, and every other image is reached by lowering arrows. The image is closed under both crystal operators, since the canonical basis is a crystal basis and the inclusion commutes with them. It is therefore exactly the connected component containing the empty partition. This proves the crystal assertion without identifying its graph vertices with standard field vectors.

Conclusion. The vacuum is a highest weight vector of weight ${\mathrm{\Lambda }}_0$, its cyclic submodule is the simple integrable $V({\mathrm{\Lambda }}_0)$ with integral lattice $U_A^{-}\mathrm{\mid }\mathrm{\varnothing }⟩$, and its canonical lower lattice is $M\cap F_0$, whose residue crystal is the empty graph component.