Filtered projectives prove canonical basic-set reciprocity

First work over $k_0$ and write $H_0=H_{k_0}$. Corestrict $\theta$ to ${\theta }_0:A\to k_0=\mathrm{Frac}(\theta (A))$; it has the same characteristic as $k$ and every ${\theta }_0(f_E)$ is nonzero whenever $\theta (f_E)$ is, so [F4] applies and $H_0$ is split. For a simple $M$, take a projective cover $\pi :P\to M$ with superfluous kernel $K$ by [F5]. A nonzero vector of $M$ gives a surjection $H_0\to M$. Projectivity of the free module $H_0$ lifts it to $s:H_0\to P$; the image of $s$ together with $K$ is $P$, so superfluity makes $s$ surjective. Projectivity of $P$ splits $s$, realizing $P$ as a direct summand of $H_0$. Its projection is right multiplication by an idempotent $e$, since a left-linear endomorphism of $H_0$ is determined by its value at $1$. Hence $P\cong H_{0}e$. Every maximal submodule of $P$ contains $K$: otherwise its sum with $K$ is $P$, contradicting superfluity. Since $K$ itself is maximal, it is the unique maximal submodule, and the head of $P$ is the single copy $M$.

We justify extension to $k$ without assuming an unproved radical base-change rule. For any finite-dimensional split algebra $T$ over $k_0$, choose representatives $S_1,\dots ,S_r$ of its simple modules; there are finitely many because every simple is a quotient of the regular module and thus a factor of a composition series of $T$. The ideal $N={\bigcap }_j{\mathrm{Ann}}_T(S_j)$ is nilpotent: it annihilates each factor of such a series, so its successive powers lower the series and $N^{\dim T}=0$. The action map $T\to {⨁}_j{\mathrm{End}}_{k_0}(S_j)$ is surjective. Indeed choose bases in the $S_j$ and consider the image of $T$ in the direct sum of the corresponding tuples. If this image were proper, semisimplicity of the direct sum of simples would give a nonzero map to some $S_j$ killing the image. Maps between distinct simple modules are zero, and ${\mathrm{End}}_T(S_j)=k_0$, so this would yield a nontrivial scalar relation among the selected basis vectors of $S_j$ by evaluation at $1$. This is impossible. Thus $T\mathrm{/}N$ is the displayed product of full matrix algebras. Under extension to $k$, the ideal $k\otimes N$ stays nilpotent, hence annihilates every simple: if it acted surjectively on a simple, nilpotence would contradict its nonzeroness. The quotient is still a product of full matrix algebras. It follows that all simples are exactly the extensions of the $S_j$, and extending a composition series preserves its simple factors and multiplicities. For completeness, a finite direct sum of simple modules is semisimple by induction: split a submodule's intersection with the previous summands, quotient that intersection, and project to the final simple summand. The projection is zero or onto; in the latter case its remaining submodule is the graph of a homomorphism from that simple into the complementary previous summands, and that graph has those previous summands as a complement. This justifies the semisimplicity used in the simultaneous action-map argument.

For a simple $M^{\mathrm{\prime }}$, every nonzero map $P\to M^{\mathrm{\prime }}$ has that unique maximal kernel $K$, so ${\mathrm{Hom}}_{H_0}(P,M^{\mathrm{\prime }})$ is $k_0$ when $M^{\mathrm{\prime }}=M$ and zero otherwise, using [F4]. The evaluation map ${\mathrm{Hom}}_{H_0}(H_{0}e,X)\to eX$, $f\mapsto f(e)$, is an isomorphism, with inverse $x\mapsto (he\mapsto hx)$. Thus $\dim e{M}^{\mathrm{\prime }}=1$ for $M^{\mathrm{\prime }}=M$ and zero otherwise. Multiplication by $e$ is exact, so applying it along any composition series of a finite-dimensional module $X$ yields $[X:M]={\dim }_{k_0}eX$.

By [F3] lift $e$ to an idempotent $q\in H_R$. The model $L_E$ is free by the explicit integral construction in [F2]. If $L$ is any other explicitly $R$-free stable model of $E_F$, then $qL$ is a direct summand of $L$, hence free with equal generic and residue ranks. Therefore $[L\mathrm{/}{\mathfrak{m}}_{R}L:M]=\dim e(L\mathrm{/}{\mathfrak{m}}_{R}L)={\dim }_{F}q{E}_F$. The last expression is independent of $L$. In particular it applies to the free $J_R$-standard model $L_E$ from [F2], proving $d_{E,M}=[\ast V_E:M]$ over $k_0$ and agreement with every other free model. If a chosen model was initially free over $A_{\ker \theta }$, extension to $R$ is a free model of this same generic representation and its residue is the original specialized module, so the same equality applies. The explicit global model in [F2] itself gives a free $A_{\ker \theta }$-model. No assertion about arbitrary nonfree torsion-free lattices, or exactness of their reduction, is made.

Filter $H_{R}q$ by $I_{i}q$. Exact projection gives its layer $B_{i,R}q$, which is finite free over $R$ and is a left $J_R$-module because right $H_R$ commutes with left $J_R$. By [F2] and [F3] its $E$-multiplicity is the same in the generic and residue fibres. In the generic fibre, [F1] identifies its $J_F$-simples with the Hecke simples $E_F$, and only layer $i={\alpha }_E$ can contain $E_F$. Additivity of composition multiplicities along the finite filtration therefore says that its multiplicity in that layer is its multiplicity in $H_{F}q$. On the matrix block ${\mathrm{End}}_F(E_F)$, right multiplication by $q$ has image consisting of matrices with $\mathrm{rank}(q\mathrm{\mid }{E}_F)$ independent columns after a basis change; as a left module this is exactly that many copies of $E_F$. Consequently the multiplicity is ${\dim }_{F}q{E}_F=d_{E,M}$ by step 3.1. Reduction gives the asserted reciprocity over $k_0$, with zero multiplicity in all other degrees.

Apply step 1.2 to $T=H_0$ and to the split algebra $J_{k_0}$. Their simple labels, the reductions defining $d$, and the decompositions of the layers all extend with unchanged multiplicities. For $P=H_{0}e$, let $N$ be the nilpotent ideal from step 1.2 and $K$ the kernel of its projective-cover map. Since $N$ annihilates $M$, $NP\subseteq K$, so $P\mathrm{/}NP$ is nonzero. This quotient is semisimple over $H_0\mathrm{/}N$, and every simple quotient of it is a simple quotient of $P$, hence is $M$ by step 2.1. Moreover, ${\mathrm{Hom}}_{H_0}(P\mathrm{/}NP,M)\cong {\mathrm{Hom}}_{H_0}(P,M)=k_0$ by step 2.1, so $P\mathrm{/}NP\cong M$ and $K=NP$. After extension, the kernel is $(k\otimes N)(k\otimes P)$ and the quotient is the extended simple $M$. This kernel is superfluous: if a submodule together with it generates $k\otimes P$, the quotient by that submodule equals its image under the nilpotent ideal $k\otimes N$, and iteration forces the quotient to be zero. Thus the extended idempotent again gives a projective cover with simple head $M$. Therefore reciprocity from step 4.1 holds over $k$ for these idempotents. Any other projective cover of $M$ is isomorphic to it by [F5]; an isomorphism $H_{k}e\cong H_k{e}^{\mathrm{\prime }}$ is represented by mutually inverse corner elements, so right multiplication by those elements identifies $B_{i}e$ and $B_i{e}^{\mathrm{\prime }}$ as $J_k$-modules. The formula therefore holds for every idempotent allowed in the statement, completing the proof.