We had a totally real field $F$ of even degree and a central division algebra $D/F$ non-split at exactly the infinite places. For the coefficients, we had $L / \mathbb{Q}_ \ell$ a finite extension with $\mathcal{O} \twoheadrightarrow \mathcal{O} / \lambda = \mathbb{F}$. Let $R$ be a finite set of finite places of $F$ not diving $\ell$. For each $v \in R$, we had a subgroup $\Delta_v \subseteq k(v)^\times$ and a character $\chi_v \colon \Delta_v \to \mathcal{O}^\times$, and wrote $\Delta = \prod_{v \in R} \Delta_v$ and $\chi = \prod_{v \in R} \chi_v$. Then we had $$ U_\Delta(R) = \biggl\lbrace \begin{pmatrix} a & b \br c & d \end{pmatrix} \in \GL_2(\hat{\mathcal{O}}_ F) : v \mid c, a / d \in \Delta_v \biggr\rbrace \xrightarrow{\chi} \mathcal{O}^\times, $$ where the map is given by $\prod_{v \in R} \chi_v(a/d)$. Then for every $\mathcal{O}$-algebra $A$, we looked at the space of automorphic forms $$ \begin{align} S(U_\Delta(R), A)_ \chi &= \lbrace \varphi \colon D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / (\mathbb{A}_ F^\infty)^\times \to A : \varphi(gu) = \chi(u) \varphi(g) \text{ for } u \in U_\Delta(R) \rbrace \br &= \bigoplus_{g \in D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / \mathbb{A}_ F^{\infty\times} U_\Delta(R)} A(\chi)^{(\mathbb{A}_ F^\infty)^\times U_\Delta(R) \cap g^{-1} D^\times g}, \end{align} $$ which was a finite free $A$-module because the indexing set is finite and this group $(\mathbb{A}_ F^\infty)^\times U_\Delta(R) \cap g^{-1} D^\times g / F^\times$ has finite order prime to $\ell$. For coefficients in $\bar{\mathbb{Q}}_ \ell$, we also have $$ S(U_\Delta(R), \bar{\mathbb{Q}}_ \ell) = \mathcal{A}(D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times)_ 0^{U_\Delta(R),\chi} = \bigoplus_\pi \pi^{U_\Delta(R),\chi}. $$
Lemma 1. Suppose $R = R_1 \amalg R_2$ and $\Delta_1 \subseteq \Delta_2$ where $\chi \colon \Delta_2 \to \mathcal{O}^\times$ satisfies $\chi_v = 1$ for $v \in R_2$ and $\Delta_{1,v} = \Delta_{2,v}$ and $\lvert \Delta_2 / \Delta_1 \rvert$ is a power of $\ell$. Then $S(U_{\Delta_1}(R), A)_ \chi$ is a free $A[\Delta_2 / \Delta_1]$-module.
What is this action? Given any $\delta \in \Delta_2 / \Delta_1$, we find a $u_\delta \in \prod_{v \in R_2} U_R(\Delta_2)_ v$ lifting $\delta$. If we act by it to define $(\delta \varphi)(g) = \varphi(g u_\delta)$, this is going to be independent of the choice of $u_\delta$. We can write $$ S(U_{\Delta_1}(R), A) = \bigoplus_{g \in D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / \mathbb{A}_ F^{\infty\times} U_{\Delta_2}(R)} \bigoplus_{h \in D^\times \backslash D^\times g \mathbb{A}_ F^{\infty\times} U_{\Delta_2}(R) / \mathbb{A}_ F^{\infty\times} U_{\Delta_1}(R)} A(\chi)^{h^{-1} D^\times h \cap \mathbb{A}_ F^{\infty\times} U_{\Delta_1}(R) / F^\times}. $$ Writing $h = gu$, we can rewrite this inner direct sum as $$ \bigoplus_{u \in \mathbb{A}_ F^{\infty\times} U_{\Delta_2}(R) / (g^{-1} D^\times g \cap U_{\Delta_2}(R)) U_{\Delta_1}(R) \mathbb{A}_ F^{\infty\times}} A(\chi)^{\cdots} $$ where this $\cdots$ is a conjugate of $$ g^{-1} D^\times \cap \mathbb{A}_ F^{\infty\times} U_{\Delta_1}(R) / F^\times = g^{-1} D^\times g \cap \mathbb{A}_ F^\times U_{\Delta_2}(R) / F^\times $$ by $u$, where this equality follows from the fact that this inclusion is a subgroup of a group of order prime to $\ell$ by a subgroup of order that is a power of $\ell$. Conjugation by $u$ does not matter, because at the end we are looking at its image under $\chi$. On the other hand, again we have $g^{-1} D^\times g \cap U_{\Delta_2}(R) \mathbb{A}_ F^{\infty\times}$ order prime to $\ell$ modulo $F^\times$. This shows that $u$ lies in $$ \mathbb{A}_ F^{\infty\times} U_{\Delta_2}(R) / (g^{-1} D g \cap U_{\Delta_2}(R) \mathbb{A}_ F^{\infty\times}) U_{\Delta_1}(R) \mathbb{A}_ F^{\infty\times} = \Delta_2 / \Delta_1, $$ because the left side is a quotient of an $\ell$-power group by some prime to $\ell$ subgroup. So at the end, we get $$ \bigoplus_{g \in D^\times \backslash (D \otimes \mathbb{A}^\infty)^\times / \mathbb{A}_ F^{\infty\times} U_{\Delta_2}(R)} \bigoplus_{\delta \in \Delta_2 / \Delta_1} A(\chi)^{g^{-1} D^\times g \cap U_{\Delta_2}(R) \mathbb{A}_ F^{\infty\times}}, $$ where the action of $\Delta_2 / \Delta_1$ is by translation.Proof.
Assume that $A$ is Noetherian from now on. The space $$ S(U_\Delta(R), A) $$ has an action of the Hecke operator $T_v$ for $v \notin R \cup \lbrace v \mid \ell \rbrace$ coming from the matrix $U_\Delta(R) \operatorname{diag}(\pi_v, 1) U_\Delta(R)$.
Definition 2. We define $\mathbb{T}(U_\Delta(R), A) \subseteq \End_A(S(U_\Delta(R), A))$ the $A$-subalgebra generated by the $T_v$ for $v \nmid \ell$ and $v \notin R$. This is commutative, finite over $A$, and $A$-torsion free.
We always have a surjective map $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi \otimes_\mathcal{O} A \twoheadrightarrow \mathbb{T}(U_\Delta(R), A), $$ because the right hand side is generated by Hecke operators by definition. We also claim that its kernel is killed by a power of $\ell$. This is because if we write $$ 0 \to \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi \hookrightarrow \mathcal{O}^{\oplus n^2} \to Q \to 0, $$ then we have $$ \Tor_1^\mathcal{O}(Q, A) \to \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi \otimes A \to A^{\oplus 2}, $$ where the image of the second map is $\mathbb{T}(U_\Delta(R), A)_ \chi$. Because $Q$ is the direct sum of a free part and a torsion part, we see that $\Tor_1^\mathcal{O}(Q, A)$ is killed by some power of $\ell$.
Because $\mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi$ is finite over $\mathcal{O}$, it decomposes as a product of local rings $$ \mathbb{T}(U_\Delta(R), \mathcal{O})_ \chi = \prod_\mathfrak{m} \mathbb{T}(U_\Delta(R), \mathcal{O})_ {\chi,\mathfrak{m}}. $$ Then we will be able to write $$ S(U_\Delta(R), A)_ \chi = \bigoplus_{\mathfrak{m}} S(U_\Delta(R), A)_ {\chi,\mathfrak{m}}. $$