Orthogonal idempotents lift over a complete formal power-series ring

By [F2], a finite $R$-basis identifies $A$ with $R^n$, so [F1] makes $A$ complete and separated. Multiplication is continuous: when $x^{\mathrm{\prime }}-x,y^{\mathrm{\prime }}-y\in h^{r}A$, expansion gives $x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}-xy=(x^{\mathrm{\prime }}-x)y+x(y^{\mathrm{\prime }}-y)+(x^{\mathrm{\prime }}-x)(y^{\mathrm{\prime }}-y)\in h^{r}A$. For any idempotent $q$, the map $x\mapsto qxq$ is continuous and idempotent; its image is the kernel of the continuous map $1-(x\mapsto qxq)$, hence is closed in the separated module $A$ and is therefore complete.

In any such complete algebra $B$ with identity $q$, start with $a\in B$ satisfying $\delta :=a^2-a\in hB$. The element $u:=2a-q$ commutes with $\delta$ and satisfies $u^2=q+4\delta$. The partial sums $S_N:={\sum }_{r=0}^N(-4\delta {)}^r$ are Cauchy because ${\delta }^r\in h^{r}B$; let $S$ be their limit. The finite geometric identity $(q+4\delta )S_N=q-(-4\delta {)}^{N+1}=S_N(q+4\delta )$ and separatedness give $(q+4\delta )S=S(q+4\delta )=q$, so $u^{-1}=uS=Su$. Set $a^{\mathrm{\prime }}:=a-u^{-1}\delta$. The terms commute, and expansion gives $(a^{\mathrm{\prime }}{)}^2-a^{\mathrm{\prime }}=\delta -u(u^{-1}\delta )+(u^{-1}\delta {)}^2=(u^{-1}\delta {)}^2$. Thus one correction squares the error and leaves the reduction of $a$ unchanged.

Starting with any lift $a_0$ of an idempotent in $B\mathrm{/}hB$, apply step 2.1 repeatedly. If $a_n^2-a_n\in h^{r_n}B$, then $a_{n+1}-a_n\in h^{r_n}B$ and $a_{n+1}^2-a_{n+1}\in h^{2r_n}B$. Starting with $r_0=1$, we have $r_n=2^n$, so the sequence is Cauchy. Its limit $e$ exists by step 1.1; continuity of multiplication and the error tending to zero give $e^2=e$, and every correction lies in $hB$, so $e$ has the prescribed reduction.

Lift a complete orthogonal family by induction on $m$. If $m=0$, completeness means $1=0$ in $A\mathrm{/}hA$; since $A$ is free of finite rank and $R\mathrm{/}hR\cong k$ is a field, its reduction is $k^n$, so $A\mathrm{/}hA=0$ forces its rank $n$ to be zero and hence $A=0$. The empty family then lifts. For $m=1$, the only member is $1$, which already lifts the family. For $m\ge 2$, lift ${\bar{e}}_1$ by step 3.1 to an idempotent $e_1$ and set $q:=1-e_1$. The remaining reductions lie in the corner $\bar{q}(A\mathrm{/}hA)\bar{q}$ and form a complete orthogonal family there. The reduction map $qAq\to \bar{q}(A\mathrm{/}hA)\bar{q}$ is onto because any lift $x$ of a corner element can be replaced by $qxq$. Its kernel is $h(qAq)$: if $y\in qAq$ lies in $hA$, write $y=ha$; then $y=qyq=hqaq$ since $h$ is central by [F3], and the reverse inclusion is immediate. Thus the reduction identifies $qAq\mathrm{/}h(qAq)$ with the indicated corner. The corner is complete by step 1.1, so induction lifts the remaining family in $qAq$; its members are orthogonal to $e_1$ since $q{e}_1=e_{1}q=0$, and their sum is $q$. This proves the claim for $m$.

The construction makes only finitely many initial lift choices, one for each member of the input finite family, and then applies the deterministic Newton recurrence in step 2.1. It uses no maximality principle or arbitrary family of choices. Steps 3.1 and 4.1 produce the required exact idempotents and preserve every prescribed reduction.