Nonzero cellular heads classify the simple modules over a field

For $x={\sum }_t{a}_t{c}_t\in \mathrm{\Delta }(\lambda )$, the rank-one multiplication identity of [F1] gives$x{C}_{u,v}^{\lambda }={\sum }_t{a}_t⟨c_t,c_u{⟩}_{\lambda }{c}_v=⟨x,c_u{⟩}_{\lambda }{c}_v\mathrm{.}$ If the form is not identically zero, its radical is proper and $D(\lambda )\mathrm{\ne }0$. For any $x\mathrm{\notin }\mathrm{rad}\mathrm{\Delta }(\lambda )$ there is $c_u$ with $⟨x,c_u⟩\mathrm{\ne }0$. Every submodule of $\mathrm{\Delta }(\lambda )$ containing such an $x$ contains each basis vector $c_v$, because the action identity is exact and division by the nonzero field scalar gives $c_v=x{C}_{u,v}^{\lambda }\mathrm{/}⟨x,c_u⟩$. Thus every submodule not contained in the radical is all of $\mathrm{\Delta }(\lambda )$, and the quotient $D(\lambda )$ is simple.

Let $E\mathrm{/}F$ be any field extension. The Gram matrix of the form has the same rank over $E$ as over $F$: its rank is characterized by the largest nonzero minor, and a minor in $F$ vanishes in $E$ exactly when it already vanished in $F$. Hence $\mathrm{rad}(\mathrm{\Delta }(\lambda ){\otimes }_{F}E)=\mathrm{rad}\mathrm{\Delta }(\lambda ){\otimes }_{F}E$, and $D(\lambda ){\otimes }_{F}E$ is nonzero whenever $D(\lambda )$ is. Step 1.1, applied over $E$, proves it simple. Thus it remains simple after every field extension, which is absolute simplicity.

Suppose $D(\mu )\mathrm{\ne }0$, so the form on $\mathrm{\Delta }(\mu )$ is not identically zero. Let $M$ be a proper submodule of $\mathrm{\Delta }(\lambda )$ and suppose $\varphi :\mathrm{\Delta }(\mu )\to \mathrm{\Delta }(\lambda )\mathrm{/}M$ is nonzero. There is an $x\mathrm{\notin }\mathrm{rad}\mathrm{\Delta }(\mu )$ with $\varphi (x)\mathrm{\ne }0$: otherwise, for a fixed $x_0$ outside the radical, each radical vector $z$ is the difference $(x_0+z)-x_0$ of two vectors outside the radical, so $\varphi$ would vanish on the whole module. Choose $y={\sum }_s{b}_s{c}_s$ with $⟨x,y{⟩}_{\mu }=1$ and define $a_t={\sum }_s{b}_s{C}_{s,t}^{\mu }\in A^{\ge \mu }$. The rank-one identity gives $x{a}_t=c_t$, so $\varphi (c_t)=\varphi (x)a_t$. If $\lambda \ge ̸\mu$, then $A^{\ge \lambda }\cap A^{\ge \mu }\subseteq A^{>\lambda }$: a basis shape in the intersection is above both labels, and it cannot equal $\lambda$ without implying $\lambda \ge \mu$. For any cell basis representative of a vector in $\mathrm{\Delta }(\lambda )$, multiplying by $a_t$ lies in both ideals, so it acts as zero in $\mathrm{\Delta }(\lambda )\mathrm{/}M$. Thus every $\varphi (c_t)=0$, contradicting $\varphi \mathrm{\ne }0$. Therefore a nonzero map of this kind forces $\lambda \ge \mu$.

If $D(\lambda )$ and $D(\mu )$ are nonzero and isomorphic, compose the quotient map $\mathrm{\Delta }(\mu )\twoheadrightarrow D(\mu )$ with the isomorphism to obtain a nonzero map $\mathrm{\Delta }(\mu )\to \mathrm{\Delta }(\lambda )\mathrm{/}\mathrm{rad}\mathrm{\Delta }(\lambda )$. Step 2.2 gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$. Antisymmetry of the cell poset implies $\lambda =\mu$.

Choose a linear extension ${\lambda }_1,\dots ,{\lambda }_m$ of the finite poset in which every maximal remaining element is listed next. The prefix $U_k=\{{\lambda }_1,\dots ,{\lambda }_k\}$ is upward closed, so $J_k=\mathrm{span}\{C_{s,t}^{\mu }:\mu \in U_k\}$ is an ideal: it is the sum of the ideals $A^{\ge \mu }$ for $\mu \in U_k$. Also $J_k=J_{k-1}+A^{\ge {\lambda }_k}$, because every label strictly above ${\lambda }_k$ occurs earlier. Let $S$ be a simple right $A$-module. Since $J_m=A$, choose the least $k$ with $S{J}_k\mathrm{\ne }0$. Then $S{J}_{k-1}=0$ and $S{J}_k=S$, so $S{A}^{\ge {\lambda }_k}=S$ and $S{A}^{>{\lambda }_k}=0$. Choose $s_0\in S$ and $u$ such that $s_0{C}_{u,v}^{{\lambda }_k}\mathrm{\ne }0$ for some $v$; this is possible because $S{A}^{\ge {\lambda }_k}\mathrm{\ne }0$ and the cell basis spans that ideal. Define $\psi :\mathrm{\Delta }({\lambda }_k)\to S$ by $\psi (c_t)=s_0{C}_{u,t}^{{\lambda }_k}$. For $a\in A$, the cell axiom gives $C_{u,t}^{{\lambda }_k}a\equiv {\sum }_w{r}_{t,w}(a)C_{u,w}^{{\lambda }_k}$ modulo $A^{>{\lambda }_k}$; that higher ideal annihilates $S$, so $\psi (c_{t}a)=\psi (c_t)a$. The map is a nonzero module map and hence surjective. Choose $t$ with $\psi (c_t)\mathrm{\ne }0$. The annihilator $\{x\in S:x{A}^{\ge {\lambda }_k}=0\}$ is a submodule because $A^{\ge {\lambda }_k}$ is a two-sided ideal. It is proper since $S{A}^{\ge {\lambda }_k}=S\mathrm{\ne }0$, so simplicity makes it zero. Hence $\psi (c_t)A^{\ge {\lambda }_k}\mathrm{\ne }0$; since $S{A}^{>{\lambda }_k}=0$, some cell basis element $C_{p,q}^{{\lambda }_k}$ acts nontrivially on $\psi (c_t)$. If the form on $\mathrm{\Delta }({\lambda }_k)$ were zero, the rank-one action identity would give $c_t{C}_{p,q}^{{\lambda }_k}=0$, contradicting the module-map identity $\psi (c_t{C}_{p,q}^{{\lambda }_k})=\psi (c_t)C_{p,q}^{{\lambda }_k}\mathrm{\ne }0$. Thus the form is nonzero and its radical is proper. By step 1.1, every proper submodule of $\mathrm{\Delta }({\lambda }_k)$ is contained in its radical; since $\ker \psi$ is maximal, it equals the radical. Therefore $S\cong D({\lambda }_k)$.