Matrix units lift through a finite free split formal deformation

If $\mathrm{\Lambda }$ is empty, then $A\mathrm{/}hA=0$; since a finite free module of rank $n$ reduces to $k^n$ and $k\mathrm{\ne }0$ by [F3, F4], this forces $n=0$ and hence $A=0$, so the empty product conclusion holds. Otherwise, for each standard diagonal matrix idempotent ${\bar{e}}_{\lambda ,a}$ in the given reduction choose the finite complete orthogonal family of all such idempotents. By [F1] it lifts to pairwise orthogonal idempotents $e_{\lambda ,a}\in A$ whose sum is $1$.

For any two lifted diagonal idempotents $e_i,e_j$, the map $p_{ij}:A\to A$, $x\mapsto e_{i}x{e}_j$, is an idempotent $R$-linear projection with image $e_{i}A{e}_j$. The decomposition $A=\mathrm{im}(p_{ij})\oplus \ker (p_{ij})$ reduces modulo $h$ to the corresponding image/kernel decomposition of ${\bar{p}}_{ij}$; hence $(e_{i}A{e}_j)\mathrm{/}h(e_{i}A{e}_j)\cong {\bar{e}}_i(A\mathrm{/}hA){\bar{e}}_j$. In the specified product of matrix algebras this corner has dimension $1$ when $i,j$ belong to the same block and dimension $0$ when they belong to different blocks. Applying [F2] after identifying the finite free module $A$ with $R^n$ shows that same-block corners are free of rank $1$ and cross-block corners are zero.

Fix one block $\lambda$ and its first diagonal idempotent $e_{\lambda ,1}$. For each $a$ choose lifts $x_{\lambda ,a}\in e_{\lambda ,a}A{e}_{\lambda ,1}$ and $y_{\lambda ,a}\in e_{\lambda ,1}A{e}_{\lambda ,a}$ of the standard matrix units $E_{a1}$ and $E_{1a}$. Each corner is free of rank $1$ by step 2.1; an element whose reduction is nonzero is a basis, because its coefficient relative to any basis has nonzero constant term and is therefore a unit by [F5]. Thus $y_{\lambda ,a}{x}_{\lambda ,a}=c_a{e}_{\lambda ,1}$ with $c_a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}h)$; replacing $y_{\lambda ,a}$ by $c_a^{-1}{y}_{\lambda ,a}$ gives $y_{\lambda ,a}{x}_{\lambda ,a}=e_{\lambda ,1}$. Also $x_{\lambda ,a}{y}_{\lambda ,a}=d_a{e}_{\lambda ,a}$ with $d_a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}h)$. This product is idempotent because $y_{\lambda ,a}{x}_{\lambda ,a}=e_{\lambda ,1}$, so $d_a^2=d_a$; since $d_a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}h)$ is a unit by [F5], $d_a=1$. Hence $x_{\lambda ,a}{y}_{\lambda ,a}=e_{\lambda ,a}$.

Define $E_{\lambda ,ab}:=x_{\lambda ,a}{y}_{\lambda ,b}$. If $b=c$ then $y_{\lambda ,b}{x}_{\lambda ,c}=e_{\lambda ,1}$, and if $b\mathrm{\ne }c$ then this product is zero by orthogonality of $e_{\lambda ,b},e_{\lambda ,c}$. Therefore $E_{\lambda ,ab}{E}_{\lambda ,cd}={\delta }_{bc}{E}_{\lambda ,ad}$. Products between distinct blocks vanish by the zero cross-block corners of step 2.1, and ${\sum }_{\lambda ,a}{E}_{\lambda ,aa}={\sum }_{\lambda ,a}{e}_{\lambda ,a}=1$. Their reductions are exactly all standard matrix units.

The total number of elements $E_{\lambda ,ab}$ is ${\dim }_k(A\mathrm{/}hA)={\mathrm{rank}}_R(A)$ by [F3, F4]. Relative to an $R$-basis of $A$, their coordinate matrix $C$ reduces to an invertible matrix $C_0$ over $k$, so $C=C_0+h{C}_1$ for some matrix $C_1$. The factorization $C=C_0(I+h{C}_0^{-1}{C}_1)$ is invertible: the partial sums of ${\sum }_{m\ge 0}(-h{C}_0^{-1}{C}_1{)}^m$ converge entrywise by completeness in [F3], and the finite geometric identity plus separatedness in [F3] makes the limit the two-sided inverse of the second factor. Thus the lifted matrix units form an $R$-basis by [F4]. The linear map from the product of matrix algebras sending standard matrix units to $E_{\lambda ,ab}$ is multiplicative by step 4.1, is unital, and maps a basis to a basis; it is the claimed algebra isomorphism.