The simple type-A Hecke modules are indexed by e-regular partitions

Use finite joint-primary projections without presupposing which labels survive. The regular right module has a cellular filtration with quotients copies of the cells; refining their common flags in [F3] gives a simultaneous triangular flag on it. Thus each $L_a$ is annihilated by a polynomial ${\prod }_b(X-b{)}^N$, with finitely many diagonal values $b\in F$ and $N\ge \dim H$. Polynomial Bezout and the Chinese remainder construction give $p_{a,b}$ congruent to one modulo $(X-b{)}^N$ and zero modulo the other factors. Substituting $L_a$ gives commuting idempotents. For a joint word $\xi$, their product $P_{\xi }$ is idempotent in $H$ and on a cell flag has diagonal one exactly at tableaux with word $\xi$, zero elsewhere. An idempotent has zero image if and only if all these diagonal entries vanish: its triangular matrix would otherwise be nilpotent and idempotent, hence zero. The vector-space functor $M\mapsto M{P}_{\xi }$ is exact on $H$-modules, because a lift $x$ of $y=y{P}_{\xi }$ can be replaced by $x{P}_{\xi }$, and kernels are preserved. No centrality or $H$-submodule assertion for this image is needed.

For finite $e$, equality of two diagonal values $[a{]}_q,[b{]}_q$ is equivalent to $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}e)$. Indeed $[a{]}_q-[b{]}_q=q^b[a-b{]}_q$ for all integers, including negative integers. When $q\mathrm{\ne }1$, $e$ is the multiplicative order of $q$, and when $q=1$, $e=\mathrm{char}F$ and the q-integer is the integer reduced to $F$. Thus a joint word is exactly a residue word in the node convention $c-r(\mathrm{m}\mathrm{o}\mathrm{d}e)$.

Prove the complete ladder-boundary comparison. Put $L(s,c)=(e-1)s+c-e$. Suppose $\alpha$ is restricted, every occupied node has $L<l$, and the addable ladder-$l$ set is nonempty. Set $C=l+e-1$, $T=\lfloor C\mathrm{/}(e-1)\rfloor$, $d=C-(e-1)T\in \{0,\dots ,e-2\}$ and $u_s=C-(e-1)s$. For $1\le s\le T$, $0\le {\alpha }_s\le u_s$, and no positive row occurs beyond $T$. The deficits $h_s=u_s-{\alpha }_s$ are nonnegative and satisfy $h_{s+1}-h_s={\alpha }_s-{\alpha }_{s+1}-(e-1)\le 0$. An addable ladder-$l$ row has deficit zero; thereafter every deficit is zero. Consecutive such rows have width difference $e-1>0$, so the ladder addables form a terminal interval $s_0,\dots ,T$. A first equality row that is not addable simply precedes this interval. Adding its bottommost $k$ nodes therefore adds in rows $T-k+1,\dots ,T$. The length of $\alpha$ is $T$ for $d>0$ and $T-1$ for $d=0$, including the empty case $T=1,d=0$. If $\beta ⊵\alpha$ has the same size, its length is at most that of $\alpha$, by the prefix inequality at that length. For $d=0$ all addable rows of $\beta$ are at most $T$. For $d>0$ the sole possible extra row is $T+1$, with residue $-T$, different from $i=l<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>e=d+1-T<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>e$ since $1\le d+1<e$. Hence every addable $i$-node of $\beta$ lies in a row at most $T$. If $\gamma$ adds any $k$ of these initial addables, at row cutoff $r$ its prefix increment is at least $\max (0,k-\max (T-r,0))$, exactly the increment for the bottommost-$k$ extension $\delta$ of $\alpha$. Adding $\beta$'s prefix inequalities proves $\gamma ⊵\delta$. This holds for every $k$ simultaneously and needs no same-content premise.

The integral intersection factorization. Work over $A=\mathbb{Z}[Q,Q^{-1}]$. Suppose a row-standard tableau $t$ already agrees with $t^{\eta }$ on rows before $j$, and let $B$ be the canonical label interval of row $j$. In rows $r\ge j$ let ${\beta }_r$ count its $B$-labels. First rearrange labels inside $B$ into increasing row-reading order, then do the same with all remaining tail labels, preserving the number of labels from each interval in every row. This gives a tableau $v$ whose row $r$ consists of a $B$-segment of length ${\beta }_r$ followed by a tail-segment of length ${\eta }_r-{\beta }_r$. The sorting removes inversions separately within the two label intervals, so $d(t)=d(v)wa$ with additive length, $w\in S_B$, and $a$ a permutation of the remaining tail labels. The intervals are disjoint, so $T_a$ commutes with $x_B$, and $T_w{x}_B=Q^{\mathrm{\ell }(w)}{x}_B$. Let $K^{\mathrm{\prime }}$ be the subgroup cutting each canonical row $r\ge j$ into lengths ${\beta }_r,{\eta }_r-{\beta }_r$, while retaining the earlier rows. Let $K$ cut $B$ into consecutive lengths ${\beta }_r$ and the tail into consecutive lengths ${\eta }_r-{\beta }_r$, again retaining earlier rows. The permutation $d(v)$ increases on these segments on both sides and transports $K^{\mathrm{\prime }}$ to $K$. Hence $x_{K^{\mathrm{\prime }}}{T}_{d(v)}=T_{d(v)}{x}_K$, by expansion into length-additive standard-basis words. If $E$ consists of the minimal representatives with $S_{\eta }={⨆}_{u\in E}u{K}^{\mathrm{\prime }}$, then $x_{\eta }={\sum }_{u\in E}{T}_u{x}_{K^{\mathrm{\prime }}}$. Write $\nu =({\eta }_1,\dots ,{\eta }_{j-1},{\eta }_j,{\eta }_j-{\beta }_j,\dots ,{\eta }_k-{\beta }_k)$, omitting zero parts. Multiplication of $x_K$ by $x_B$ gives $[\beta ]{!}_Q{x}_{\nu }$: the $B$-intersection subgroup contributes $\sum Q^{\mathrm{\ell }(b)}=[\beta ]{!}_Q$, while its tail and earlier-row factors commute with $x_B$. Consequently the exact identity is $x_{\eta }{T}_{d(t)}{x}_B=Q^{\mathrm{\ell }(w)}[\beta ]{!}_Q{\sum }_{u\in E}{T}_u{T}_{d(v)}{x}_{\nu }{T}_a$. No scalar is divided out. Sorting $\nu$ to a partition uses the invertible block-reordering identity in [F5], so this is in $[\beta ]{!}_{Q}H{x}_{{\nu }^{+}}H$. If ${\nu }^{+}⋬\eta$, the dominance-ideal assertion in [F5] makes its shape-$\eta$ part vanish. Otherwise the first $j+1$ largest parts of $\nu$ have sum at least ${\eta }_1+\cdots +{\eta }_j+{\eta }_j-{\beta }_j$; comparison with $\eta$ gives ${\beta }_j\ge {\eta }_j-{\eta }_{j+1}$.

Canonical-row support without factorial cancellation. The semistandard permutation-basis theorem Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations, followed by star, gives an exact coordinate condition for $H{x}_{\rho }$: every supported right Murphy tableau has semistandard type $\rho$. Apply it to the factor remaining after $[\beta ]{!}_Q$ in step 1.4, before putting back $T_a$. This factor belongs to $H{x}_{\nu }$. In shape $\eta$, semistandardness forces the first ${\eta }_1$ labels into the first row, the next ${\eta }_2$ into the second, and so on through $B$: a label cannot occur below its own row, and each earlier row is already full. Thus its right tableaux have the first $j$ canonical rows. The permutation $a$ uses only later labels. Expanding $T_a$ by the row action and straightening cannot alter these canonical rows: their prefix counts are maximal, and same-shape straightening increases every tableau-prefix count by that supplier's Statement. This proves the canonical-row assertion even for the quotient by the displayed factorial. Starting at $j=1$, repeat steps 1.4 and 2.1 for each surviving standard-tableau term. If ${\eta }_j-{\eta }_{j+1}\ge e$, every surviving term at that stage has ${\beta }_j\ge e$, so $[e]{!}_Q$ divides its factor $[\beta ]{!}_Q$. Later multiplication and integral straightening preserve this common divisor. This establishes the full repeated-symmetrizer support and divisibility used below.

For restricted $\eta$, number its nodes by increasing $L$, taking lower rows first on a tie, to obtain $t_{\eta }$. Moving right increases $L$ by one and moving down by $e-1$, so it is standard. The diagrams ${\alpha }_l$ of nodes with $L<l$ and ${\delta }_l$ with $L\le l$ are restricted: their row widths are the minimum of the restricted widths of $\eta$ and the positive part of the affine cutoff bound, each having adjacent difference at most $e-1$. A node on ladder $l$ forces every lower positive position on that ladder to be occupied, since ${\eta }_{s+1}\ge {\eta }_s-(e-1)$. Thus ${\delta }_l$ adds the bottommost $a_l$ ladder addables to ${\alpha }_l$. Any competing standard tableau with the same full residue word has, in the corresponding block, $a_l$ additions all of residue $l$. Every such node was addable before the block: adding an $i$-node creates new addable corners only of residues $i+1,i-1$, distinct from $i$ for $e\ge 2$. Starting with the empty diagrams, step 1.3 inductively shows its shape dominates ${\delta }_l$ at every completed block, and hence its final shape dominates $\eta$. No strictly lower shape has the word of $t_{\eta }$, and no same-shape tableau uniqueness is asserted.

Choose $P_{\xi }$ for the word of $t_{\eta }$. It has nonzero image on $C_{\mathrm{M}}(\eta )$ and zero image on every $C_{\mathrm{M}}(\rho )$ with $\rho ⊲\eta$, by steps 1.2 and 2.2. Therefore it has zero image on their form quotients. Apply the exact projection to a composition series of $C_{\mathrm{M}}(\eta )$. Some constituent has nonzero image; by [F1] and [F4] it is a nonzero $D_{\mathrm{M}}(\rho )$ with $\rho ⊴\eta$. The exclusion forces $\rho =\eta$, proving the restricted Murphy quotient nonzero over the original field. This is a joint-word and cellular-support proof, not an import of cellular/crystal label identification.

If ${\eta }_j-{\eta }_{j+1}\ge e$ for some $j$, every surviving term at that row has ${\beta }_j\ge e$ by steps 1.4 and 2.1 and is divisible by $[e]{!}_Q$. Later multiplication and straightening cannot remove this polynomial factor. Every Gram entry is the coefficient of $x_{\eta }$ in $x_{\eta }{T}_{d(s)}{T}_{d(t)}^{\ast }{x}_{\eta }$ modulo higher shapes; first expand its middle factor in the row-module basis and apply the proved row calculation. Thus every entry is divisible by $[e]{!}_Q$. Specializing $Q$ to $q$ makes that factor zero, so the entire Murphy form vanishes and $D_{\mathrm{M}}(\eta )=0$. This concerns a common divisor of entries, not a product-of-hooks formula for a determinant. Combining with step 3.1 proves the finite-$e$ Murphy criterion.

For $e>n$, including infinity, distinct standard tableaux of any shapes have distinct full diagonal words. At their first unequal step the preceding diagram has size $a-1$; its distinct addable-corner contents differ by a nonzero integer of absolute value at most $a\le n$, so their q-integers differ. Induction recovers every next node. Consequently every tableau word of $C_{\mathrm{M}}(\eta )$ occurs in no strictly lower cell. Exact projections as in step 3.1 and diagonal multiplicity one in [F4] show that all $\dim C_{\mathrm{M}}(\eta )$ flag factors occur in $D_{\mathrm{M}}(\eta )$; thus this quotient has the full cell dimension and its form radical is zero. In particular at $e=\mathrm{\infty }$ every partition is restricted and has nonzero quotient. At $n=0$ the unique cell is $F$ with norm one.

Establish the complete semisimplicity clause. Put $f_{\eta }=\dim C_{\mathrm{M}}(\eta )$ and $d_{\eta }=\dim D_{\mathrm{M}}(\eta )$, with zero for a zero quotient. The Murphy basis gives $\dim H={\sum }_{\eta }{f}_{\eta }^2$. On each nonzero quotient its nondegenerate form and the cell law supply all rank-one operators. Since the cells are right modules, their action gives a surjective algebra map $H\to {\mathrm{End}}_F(D_{\mathrm{M}}(\eta ){)}^{\mathrm{o}\mathrm{p}}$, a full matrix algebra after transpose. Its kernels are distinct maximal two-sided ideals: equality would give the same matrix-algebra simple module, contradicting [F1]. They are pairwise comaximal. Their finite product map is surjective: for each kernel $I_a$ and each other $I_b$ choose $u_{ab}\in I_b$ congruent to one modulo $I_a$, multiply over $b\mathrm{\ne }a$, and use these elements to isolate each prescribed matrix component. Its kernel is the radical by the complete simple list and [F6]. Hence $\dim (H\mathrm{/}\mathrm{rad}H)={\sum }_{\eta }{d}_{\eta }^2$. Step 4.2 makes this equal to $\dim H$ when $e>n$, giving an isomorphism with a product of full matrix algebras over $F$. When finite $e\le n$, the one-dimensional Murphy row cell has norm $[n]{!}_q=0$, since $x_{(n)}^2=[n]{!}_q{x}_{(n)}$. Its quotient dimension is zero while its cell dimension is one; every other $d_{\eta }\le f_{\eta }$. The dimension sum is therefore strictly smaller and $\mathrm{rad}H\mathrm{\ne }0$. This proves both directions of part 3 without any claim that some Specht module must be reducible.

Apply the same exact involution to every Murphy quotient: $D^{\lambda }=D_{\mathrm{M}}({\lambda }^{\mathrm{\prime }}{)}^{\theta }$ is nonzero exactly when ${\lambda }^{\mathrm{\prime }}$ is restricted, equivalently when $\lambda$ is regular. The nonzero quotients remain pairwise distinct absolutely simple and complete by [F1] and [F2]. This proves parts 1 and 2, and step 5.1 gives part 3. At $q=1$, [F6] gives precisely the ordinary $p$-regular group classification, including characteristic two; no splitting or semisimplicity assumption was made. All selections and polynomial computations were finite, so no choice principle is used.