Modules generated by a primitive vector

(a). By [F7], a finite-dimensional $G$-submodule contains $v$; since $v$ generates $V$, that submodule is $V$. Let $Z$ be the $U^{-}$-submodule generated by $v$, defined as the span of the coefficients of its $U^{-}$-coaction. Coassociativity makes this span a subcomodule; equivalently it is the smallest subcomodule containing $v$. If a linear functional $f\in V^{\ast }$ vanishes on $Z$, the regular function $g\mapsto f(gv)$ restricts to zero on $U^{-}B$: universally over every base algebra, $ubv$ is the scalar by which $b$ acts on $kv$ times $uv$, and the latter has all its coefficients in $Z$. By [F1] and [F2] this regular function is zero on $G$. Thus $f$ vanishes on the coefficients of the $G$-coaction of $v$, whose span is $V$. In finite dimension the annihilator of $Z$ being zero implies $Z=V$, proving (a). No vector over a general base algebra is treated as an element of $V$.

(b), weights. By (a), $V$ is the coefficient span of the $U^{-}$-coaction of $v$. The ordered product of the negative root groups is $U^{-}$ by the root-group supplier. Applying [F3] successively to this universal product shows that all these coefficients lie in $kv+{\sum }_{\mu <\lambda }{V}_{\mu }$: each application of $u_{-\alpha }(c)$ replaces a weight $\mu$ by $\mu$ or by $\mu -i\alpha$ with $i\ge 1$, and $\lambda -\sum m_{\alpha }\alpha$ with $m_{\alpha }\ge 0$ is the corresponding form, the simple roots generating the positive roots. Distinct weights give independent eigenspaces, so the sum $kv\oplus {⨁}_{\mu <\lambda }{V}_{\mu }$ is direct, and the $V_{\lambda }$-component of every element of $V$ is a multiple of $v$; since $v\mathrm{\ne }0$ this gives $V=kv\oplus {⨁}_{\mu <\lambda }{V}_{\mu }$ and $V_{\lambda }=kv$ of dimension one.

(c). Let $\alpha \in \mathrm{\Delta }$ be a simple root and let $n$ represent the reflection $s_{\alpha }$. Since $v$ has weight $\lambda$, the vector $n\cdot v$ has weight $s_{\alpha }(\lambda )$ by [F6]; it is nonzero, so $s_{\alpha }(\lambda )$ is a weight of $V$, hence by step 2.1 of the form $\lambda -{\sum }_{\beta \in \mathrm{\Delta }}{m}_{\beta }\beta$ with $m_{\beta }\ge 0$. Writing $s_{\alpha }(\lambda )=\lambda -⟨\lambda ,{\alpha }^{\vee }⟩\alpha$ and comparing the expansions of $s_{\alpha }(\lambda )-\lambda$ in the basis $\mathrm{\Delta }$ of the root lattice, uniqueness of the coefficients [F5] gives $⟨\lambda ,{\alpha }^{\vee }⟩=m_{\alpha }\ge 0$. Hence $\lambda$ is dominant.

(d). Every proper $G$-stable subspace $W⊊V$ has $W\cap V_{\lambda }=0$: if $W$ contained a nonzero element of $V_{\lambda }=kv$, then it would contain $v$, hence by (a) the whole $U^{-}$-span of $v$, which is $V$, a contradiction. Since $W$ is $T$-stable, its weight components lie in $W$: applying the coordinate functional of each character basis element to its coaction extracts that component in $W$. Thus the vanishing of $W\cap V_{\lambda }$ forces every proper $G$-stable subspace to be contained in ${⨁}_{\mu <\lambda }{V}_{\mu }$, and so is their sum $M$; since $v\mathrm{\notin }M$ the sum is proper and is therefore the largest proper $G$-stable subspace. For any nonzero $G$-submodule $Q\subseteq V\mathrm{/}M$, its preimage in $V$ is a $G$-stable subspace strictly containing $M$, hence equal to $V$; so $V\mathrm{/}M$ is simple, and it is generated by the image of $v$.

Steps 1.1, 2.1, 3.1 and 3.2 prove (a), (b), (c) and (d).