Idempotents and Module Structure

Definitions and Basic Properties of Idempotents

Definition

Let $R$ be a ring with identity. An idempotent in $R$ is a non-zero element $e$ with $e^2=e$.

Facts.

• $eRe$ is a ring with identity $e$, where $eRe=\{eae:a\in R\}$.
• For any $x\in Re$, $x=re$ for some $r\in R$ and then $xe=ree=re=x$.

Definition

Two idempotents $e_1,e_2$ are called orthogonal if $e_1{e}_2=0=e_2{e}_1$. More generally, $n$ idempotents $e_1,\cdots ,e_n$ are called pairwise orthogonal if $e_i{e}_j=0=e_j{e}_i$ for any $i\ne j$.

• An idempotent $e$ is called primitive, if it is not a sum of two orthogonal idempotents.
• An idempotent decomposition of an idempotent $e$ is a representation $e=e_1+\cdots +e_n$, where $e_1,\cdots ,e_n$ are pairwise orthogonal. If moreover $e_1,\cdots ,e_n$ are primitive, then the decomposition is called a primitive idempotent decomposition.

Remarks.

• If $e_1,\cdots ,e_n$ are orthogonal idempotents, then $e_1+\cdots +e_n$ is also an idempotent.
• For $e\in R$ with $e^2=e$, $1=e+(1-e)$ is a idempotent decomposition.

Modules Generated by Idempotents

Theorem

Let $e$ be an idempotent of $R$.

• If $e=e_1+\cdots +e_n$ is an idempotent decomposition, then $Re=R{e}_1\oplus \cdots \oplus R{e}_n$ as left ideals or left $R$-modules.
• Conversely, if $Re$ is a direct sum of left ideals of $R$, $Re=I_1\oplus \cdots \oplus I_n$, then there is an idempotent decomposition $e=e_1+\cdots +e_n$ with $e_i\in I_i$ and $I_i=R{e}_i$.

In this way, the idempotent decompositions of $e$ are in bijection with the direct sum decompositions of the left ideals $Re$.

\begin{proof} Let $e=e_1+\cdots +e_n$ be an idempotent decomposition. Then $e_{i}e=e_i$ yields that $R{e}_i\subseteq Re$. For any $a\in R$, $ae=a{e}_1+\cdots +a{e}_n\in R{e}_1+\cdots +R{e}_n$. So $Re=R{e}_1+\cdots +R{e}_n$. For another expression $ae=a_1{e}_1+\cdots +a_n{e}_n$, note that $a{e}_i=ae{e}_i=a_i{e}_i$. So the expression is unique.

Conversely, let $Re=I_1\oplus \cdots \oplus I_n$. Since $e\in Re$, $e$ can be written as $e=e_1+\cdots +e_n$ with $e_i\in I_i$. For any $a_i\in I_i\subseteq Re$, note that

$$a_i=a_{i}e=a_i{e}_1+\cdots +a_i{e}_n$$

where $a_i{e}_i\in I_i$. Since $I_1\oplus \cdots \oplus I_n$ is a direct sum, the decomposition of $a_i$ is unique and so $a_i{e}_i=a_i$ and $a_i{e}_j=0$ for any $i\ne j$. Take $a_i=e_i$, then $e_i$ is idempotent and $e_i{e}_j=e_j{e}_i=0$. Therefore, $e=e_1+\cdots +e_n$ is a idempotent decomposition.
\end{proof}

Corollary

Let $e$ be an idem of $R$. TFAE:

• $e$ is primitive;
• ${}_R(Re)$ is indecomposable as a left $R$-module;
• $e$ is the unique idempotent in $eRe$.

\begin{proof} i)←>ii) is easy.

i)→iii) Suppose $e^{\prime }\in eRe$ such that $e^{\prime }\ne 0$ and $e^{\prime 2}=e^{\prime }$. Then $e=e^{\prime }+(e-e^{\prime })$ is a decomposition, which contradicts with $e$ primitive.

iii)→i) If $e=e_1+e_2$ is an imprimitive idempotent in $R$, then $e_1,e_2\in eRe$, because $e{e}_{1}e=e_1$ and $e{e}_{2}e=e_2$, which is impossible. \end{proof}

Theorem

Let $e,f$ be idempotents of $R$, and let $V$ be an $R$-module.

• ${\mathrm{Hom}}_R(Re,V)\simeq eV$. In particular, ${\mathrm{Hom}}_R(Re,Rf)\simeq eRf$ as abelian groups.
• ${\mathrm{End}}_R(Re)\simeq (eRe{)}^{\mathrm{op}}$ as rings.

\begin{proof} i) Let $\phi \in {\mathrm{Hom}}_R(Re,V)$, then $\phi (e)=\phi (ee)=e\phi (e)\in eV$. Consider the map $g:{\mathrm{Hom}}_R(Re,V)\to eV$, $\phi \mapsto \phi (e)$.

Note that

• $\phi (ae)=a\phi (e)$ for any $a\in R$. So $\phi$ is determined by $\phi (e)$ and so $g$ is injective.
• Take arbitrary $v\in eV$. For a $R$-homomorphism $\psi :Re\to V,xe\mapsto xev=xv$, $g(\psi )=v$. So $g$ is surjective.

Therefore, ${\mathrm{Hom}}_R(Re,V)\simeq eV$.

ii) Define $V=Re$. For $\sigma ,\tau \in {\mathrm{End}}_R(Re)$, we have

$$\sigma (\tau (e))=\sigma (\tau (e)e)=\tau (e)\sigma (e)$$

and so ${\mathrm{End}}_R(Re)\simeq (eRe{)}^{\mathrm{op}}$. \end{proof}

Theorem

Let $e,f$ be idempotent of $R$. TFAE:

• $Re\simeq Rf$ as left $R$-modules.
• $eR\simeq fR$ as right $R$-modules.
• There exist $b\in fRe$ and $a\in eRf$ with $ab=e$, $ba=f$.

\begin{proof} Here we only prove i) and iii) are equivalent.

i)→iii) Let $\phi :Re\to Rf$ be an $R$-isomorphism. Set $a=\phi (e)$ and $b={\phi }^{-1}(f)$. Then we have

$$a=\phi (ee)=e\phi (e)\in eRf,\text{ and }b\in fRe.$$

Note that

$$f=\phi ({\phi }^{-1}(f))=\phi (b)=\phi (be)=b\phi (e)=ba$$

and we can similarly prove that

$$e={\phi }^{-1}(\phi (e))={\phi }^{-1}(af)=a{\phi }^{-1}(f)=ab.$$

iii)→i) Assume that there exist $b\in fRe$ and $a\in eRf$ with $ab=e$ and $ba=f$. Since $a\in eRf$, there is $eaf=a$. Define ${\phi }_a:Re\to Rf,xe\mapsto xa$ and ${\phi }_b:Rf\to Re,xf\to xb$. Then we can check that ${\phi }_a\circ {\phi }_b={\mathrm{id}}_{Re}$ and ${\phi }_b\circ {\phi }_a={\mathrm{id}}_{Rf}$.

• For any $xe\in Re$, we have $({\phi }_b\circ {\phi }_a)(xe)=xa=xaf=xab=xe$.
• For any $xf\in Rf$, we have $({\phi }_a\circ {\phi }_b)(xf)=xb=xbe=xba=xf$.

Now we finish the proof. \end{proof}

Lifting Idempotents Over Nilpotent Ideals

Definition

Let $I$ be a $2$-sided ideal of $R$, and let $f$ is an idempotent of $R$. If $e=f+I$ is an idempotent in $R/I$, we say $f$ lifts $e$.

If there is an idempotent $f\in R$ lifting $e$ for every idempotent $e$ of $R/I$, then we say we can lift idempotents from $R/I$ to $R$.

Theorem

If $I$ is a nilpotent ideal of $R$, then we can lift idempotents from $R/I$ to $R$. Moreover, if $e\in R/I$  is a primitive idempotent, then any idempotent lift $f\in R$ of $e$ is also primitive.

\begin{proof} Suppose $\bar{r}$ is idempotent in $R/I$. Consider its preimage $r\in R$ and define $s=1-r$. Note that $\overline{s}$ is also an idempotent in $R/I$.

We have that $rs=r-r^2\in I$ and $rs=sr$. So there is some $k$ such that $r^k{s}^k=0$. Since

$${\bar{r}}^k+{\bar{s}}^k=\bar{r}+\bar{s}=\overline{1}\in R/I,$$

we have $x=1-r^k-s^k\in I$ is nilpotent and then $1-x=r^k+s^k$ is invertible in $R$, whose inverse is $u=1+x+x^2+\dots +x^{l-1}$ with $x^l=0$. Notice that $\bar{u}=\overline{1}\in R/I$, and $u$ commutes with $r$ and $s$ (since $x$ does).

Finally, we have $u(r^k+s^k)=1$. It deduces that

$$(u{r}^k{)}^2+u^2{r}^k{s}^k=u{r}^k,$$

where $u^2{r}^k{s}^k=0$ by definition of $k$. Also note that $\overline{u{r}^k}=\overline{r^k}=\overline{r}$. Therefore, $u{r}^k$ is what we desire.

Suppose $e$ is primitive and $f$ has an idempotent decomposition $f=f_1+f_2$. Then $e=e_1+e_2$ with $e_i=\overline{f_i}$. Since $e$ is primitive, one of $e_1,e_2$ is $0$ and so WLOG $\overline{f_1}=0$ and so $f_1\in I$. It contradicts with $f_1$ idempotent. \end{proof}

Corollary

Let $I$ be a two-sided nilpotent ideal or $R$, and let $1=e_1+\cdots +e_n$ be an idempotent decomposition in $R/I$. Then there is an idempotent decomposition $1=f_1+\cdots +f_n$ in $R$ with $e_i=f_i+I$. If $e_i$ are primitive, so are $f_i$.

\begin{proof} Induction on $n$. When $n=1$, it is trivial, and now we assume that it holds for $m<n$.

Define $e_2+\cdots +e_n=E$, then $E$ is idempotent and $1=e_1+E$ in $R/I$. We lift $e_1$ to $f_1\in R$ and define $F=1-f_1$. Then $F$ is a lift of $E$.

Note that $F$ is the identity of the ring $FRF$, and $FIF$ is a nilpotent ideal of $FRF$. Consider the composition homomorphism

$$FRF\hookrightarrow R\twoheadrightarrow R/I,$$

whose kernel is $FRF\cap I$. There is $FRF\cap I=FIF$, because

• $FIF\subseteq FRF\cap I$
• for any $x\in FRF\cap I$, $FxF\subseteq FIF$.

So we get an inclusion $FRF/FIF\hookrightarrow R/I$, and its image is $E(R/I)E$. As $E(R/I)E$ has identity $E=e_2+\cdots +e_n$, by induction hypothesis one have $F=f_2+\cdots +f_n$ with $f_i+I=e_i$ for $i⩾2$. Thus $1=f_1+\cdots +f_n$ and we finish the proof. \end{proof}

Corollary

Let $f$ be an idempotent in a ring $R$ that has a nilpotent ideal $I$. Then $f$ is primitive in $R$ iff $f+I$ is primitive in $R/I$.

Structure of Artinian Rings and their Modules

Lemma

Let $A$ be a semisimple Artinian ring.

• There is a primitive idempotent decomposition $1=e_1+\cdots +e_n$. This corresponds $A=A{e}_1\oplus \cdots \oplus A{e}_n$ where $A{e}_i$ is simple.
• For every simple $A$-module $S$, there is a primitive idempotent $e$ with $S\simeq Ae$.
• for any simple $A$-module $S$, there exists $1⩽i⩽n$, $e_{i}S\ne 0$. If $T$ is a simple $A$-module with $T\not\simeq S$, then $e_{i}T=0$.

\begin{proof} i) Recall that there is a bijection between the idempotent decomposition of $e$ are in bijection with the direct sum decomposition of the left ideals $Ae$. Since $A$ is semisimple and Artinian, there is a decomposition $1=e_1+\cdots +e_n$ and $A={\oplus }_{i=1}^{n}A{e}_i$. By ^tv7l9r, $A{e}_i$ is simple.

ii) Since each simple $A$-module is a quotient of the left regular module ${}_{A}A$ by ^09efb3, $S$ is isomorphic to $Ae$ for some primitive idempotent $e$.

iii) Notice that $S=1S=(e_1+\cdots +e_n)S$, and there exists $i$ such that $e_{i}S\ne 0$. For any given simple $A$-module $T$, there exists $e$ such that $T\simeq Ae$. By ^yr2ffd, one have

$$e_{i}T=e_{i}Ae\simeq {\mathrm{Hom}}_R(Ae,A{e}_i)=0$$

if $Ae\not\simeq A{e}_i$. Now we finish the proof. \end{proof}

Theorem

Let $A$ be an Artinian ring, and let $S$ be a simple $A$-module.

• There is an indecomposable projective $A$-module $P_S$ with $P_S/\mathrm{Rad}(P_S)\simeq S$ of the form $P_S\simeq Af$ where $f$ is a primitive idempotent in $A$.
• The idempotent $f$ has the property that $fS\ne 0$, and if $T$ is any simple module with $T\not\simeq S$ then $fT=0$.
• $P_S$ is the projective cover of $S$, it is uniquely determined up to isomorphism by $S$.

\begin{proof} Let $e\in A/J(A)$ be a primitive idempotent element such that $eS\ne \varnothing$, and let $f$ be any lift of $e$ to $A$. Then $S\simeq (A/J(A))e$ and $f$ is primitive by ^84f4de. Since $J(A)$ annihilator all simple modules, by ^drcz17 we know $fS=eS\ne 0$ and $fT=eT=0$ for any simple module $T\not\simeq S$.

Since $1=f+(1-f)$, by ^1dezlm one can check $A=Af\oplus A(1-f)$ and $Af$ is a direct sum of a free module $A$. Hence, $Af$ is a indecomposable projective module by ^3rkmd6. Set $P_S=Af$, and notice that

$$\begin{array}{rl}Af& \to (A/J(A))\cdot (f+J(A))=S\\ af& \mapsto (a+J(A))\cdot (f+J(A))\end{array}$$

is an $A$-morphism with kernel $J(A)Af=\mathrm{Rad}(Af)=\mathrm{Rad}(P_S)$. Then $P_S/\mathrm{Rad}{P}_S\simeq S$ and $P_S$ is a projective cover of $S$. \end{proof}

Theorem

Let $A$ be an Artinian ring.

Up to isomorphism, the indecomposable projective $A$-modules are exactly the module $P_S$ that are the projective covers of simple modules, and $P_S\simeq P_T$ iff $S\simeq T$.

Each $P_S$ appears as direct summand of ${}_{A}A$ (left regular module) with multiplicity equal to the multiplicity of $S$ as a summand of ${}_{A/J(A)}A/J(A)$. Precisely, ${}_{A}A\simeq {\oplus }_S(P_S{)}^{n_S}$, where $S$ runs through simple $A$-modules up to isomorphism, and $n_S={\dim }_{D}S$ with $D={\mathrm{End}}_A(S)$.

\begin{proof} Let $\bar{A}=A/J(A)$. Since $A$ is Artinian, $\overline{A}$ is an Artinian semisimple ring. Hence by ^drcz17, one have a primitive idempotent decomposition $1=e_1+\cdots +e_n$ in $\overline{A}$. By ^3rmbjt, lift each $e_i$ to an idempotent $f_i\in A$ and so $1=f_1+\cdots +f_n$ in $A$. Thus, the regular module $A^A$ decomposes as ${}_{A}A=A{f}_1\oplus \cdots \oplus A{f}_n$, where each $A{f}_i$ is an indecomposable projective module. By ^koarlo, we know $A{e}_i\simeq A{f}_i/\mathrm{Rad}\left(A{f}_i\right)$ is a simple $\bar{A}$-module and hence a simple $A$-module. So each $A{f}_i$ is the projective cover $P_S$ of some simple module $S$, and every such cover arises this way.

Now let $P$ be any indecomposable projective $A$-module. Then $P/\mathrm{Rad}(P)\simeq S_1\oplus \cdots \oplus S_t$, where each $S_i$ is a simple module. On the other hand, $P_{S_1}\oplus \cdots \oplus P_{S_t}\to S_1\oplus \cdots \oplus S_t$ is also a projective cover. By the uniqueness of projective covers, it follows that

$$P\simeq P_{S_1}\oplus \cdots \oplus P_{S_t}.$$

Since $P$ is indecomposable, we must have $t=1$, and so $P\simeq P_S$ for some simple module $S$. Therefore, every indecomposable projective is isomorphic to some $P_S$.

Finally, since $A/J(A)$ is semisimple, we have $A/J(A)\simeq {\oplus }_S{S}^{n_s}$ and ${}_{A}A\simeq {⨁}_S{\left(P_S\right)}^{n_S}$, where $n_S={\dim }_{{\mathrm{End}}_A(S)}S$ by ^8omj6z, as desired. Now we finish the proof. \end{proof}

Cartan Matrix

Theorem

Let $A$ be an Artinian ring, and let $f$ be a primitive idempotent of $A$. Define $e=f+J(A)\in A/J(A)=\overline{A}$. In any finitely generated $A$-module $V$, the multiplicity of $\overline{A}e$ in $V$ is equal to the composition length of $fV$ as an $fAf$-module.

\begin{proof} Since $A$ is an Artinian ring and $V$ is a finitely generated $A$-module, we know $V$ has a finite length composition series by ^u3of0p. Notice that $e$ is primitive by ^lz95r0.

Let $V=V_0\supseteq V_1\supseteq \cdots \supseteq V_n=0$ be a composition series of $V$. Consider the chain of $fAf$-modules

$$fV=f{V}_0\supseteq \cdots \supseteq f{V}_n=0,$$

where $V_i/V_{i+1}\simeq \overline{A}e$ iff $f{V}_i/f{V}_{i+1}\ne 0$. Furthermore,

$$e\overline{A}e\simeq f\left(V_i/V_{i+1}\right)\cong f{V}_i/\left(f{V}_i\cap V_{i+1}\right)=f{V}_i/f{V}_{i+1}.$$

By ^yr2ffd, $e\overline{A}e\simeq ({\mathrm{End}}_{\overline{A}}(\overline{A}e){)}^{\mathrm{op}}$ is a division ring by Schur’s lemma. Since $fAf/J(fAf)\simeq e\overline{A}e$, we know $e\overline{A}e$ is a simple $fAf$-module. Therefore, the multiplicity of $\overline{A}e$ in any finitely generated $A$-module $V$ is equal to the composition length of $fV$ as an $fAf$-module. \end{proof}

Remark. For a primitive idempotent $f\in A$ with Artinian ring $A$, $fAf$ is a local ring.

• We claim that $J(fAf)=fJ(A)f$.
  • Since $J:=J(A)$ is an ideal of $A$, $fJf$ is an ideal of $fAf$.
  • For any $x\in fJf$, there is $x=faf$ for some $a\in J$. Since $A$ is Artinian, $J$ is nilpotent and so $fJf$ is nilpotent. It deduces that $fJf\subseteq J(fAf)$.
  • On the other hand, $fAf/fJf=f\overline{A}f$ is semisimple and so $fJf\supseteq J(fAf)$.
• We claim that $fAf$ is local.
  • By ^yr2ffd, $e\overline{A}e\simeq ({\mathrm{End}}_{\overline{A}}(\overline{A}e){)}^{\mathrm{op}}$ is a division ring by Schur’s lemma.
  • Since $fAf/J(fAf)\simeq e\overline{A}e$, we know $J(fAf)$ is a maximal ideal and $fAf$ is local.

Let $1=e_1+\cdots +e_n$ be a primitive idempotent decomposition in $A/J(A)$, and let $1=f_1+\cdots +f_n$ be a primitive idempotent decomposition in $A$ with $e_i=f_i+J(A)$.

Definition

We denote by $c_{ij}$ the multiplicity of $(A/J(A))e_i$ in $A{f}_j$, which is called the Cartan variant in $A$. The $n\times n$ matrix $(c_{ij}{)}_{n\times n}$ is called the Cartan matrix of $A$.

Theorem

$c_{ij}\ne 0$ iff $f_{i}A{f}_j\ne 0$.

\begin{proof} It is a direct corollary of ^3hxo3d. \end{proof}

For convenience, if $S$ and $T$ are two simple $A$-modules, we denote the corresponding Cartan invariant by $c_{ST}$, i.e., the multiplicity of $S$ in $P_T$.

Proposition

Let $A$ be a finitely generated $k$-algebra over a field $k$, and let $S$ be a simple $A$-module with projective over $P_S$. Let $U$ be a finitely dimensional $A$-module.

• If $T$ is a simple $A$-module, then

  $${\dim }_k{\mathrm{Hom}}_A(P_S,T)=\{\begin{array}{ll}\dim {\mathrm{End}}_A(S)& \text{if }S\simeq T,\\ 0& \text{otherwise.}\end{array}$$

• The multiplicity of $S$ in $U$ is $\dim {\mathrm{Hom}}_A(P_S,U)/\dim {\mathrm{End}}_A(S)$.

• If $e\in A$ is an idempotent, then $\dim \mathrm{Hom}(Ae,U)=\dim eU$.

• Let $e_S$ and $e_T$ be idempotents so that $P_S=A{e}_S$ and $P_T=A{e}_T$ are projective covers of $S$ and $T$. Then

  $$c_{ST}=\dim {\mathrm{Hom}}_A(P_S,P_T)/\dim {\mathrm{End}}_A(S)=\dim e_{S}A{e}_T/\dim {\mathrm{End}}_A(S).$$

  If moreover $k=\overline{k}$, then $c_{ST}=\dim {\mathrm{Hom}}_A(P_S,P_T)=\dim e_{S}A{e}_T$.

\begin{proof} i) If $P_S\to T$ is nonzero, then the kernel must be a maximal submodule, containing $\mathrm{Rad}(P_S)$. But $P_S/\mathrm{Rad}(P_S)\simeq S$, so $S\simeq T$. Note that there is a commutative diagram

one have ${\mathrm{Hom}}_A(S,T)\simeq {\mathrm{End}}_A(S)$.

ii) is trivial by i).

iii) is trivial by ^yr2ffd.

iv) is trivial by iii) and note that ${\dim }_k{\mathrm{End}}_A(S)=1$ by Schur’s lemma. \end{proof}

From Residue Field to Algebra: Lifting Decompositions

Let $R$ be a complete discrete valuation ring with valuation $\nu$, and let $F=R/P$ with $P=(\pi )$ be residue field of $\nu$ respectively.

Let $A$ be an $R$-algebra, which is finitely generated as an $R$-module. Then $A^{\ast }=A/A\pi$ is a finitely generated $F$-algebra. The following theorem shows that $J(A)$ and $A\pi$ are topologically equivalent (see here and here), since $J(A{)}^n\subseteq A\pi \subseteq J(A)$.

Theorem

Let $A$ be finitely generated $R$-algebra. Then

• $A\pi \subseteq J(A)$;
• $J(A{)}^n\subseteq A\pi$ for some $n>0$.

\begin{proof} i) Let $V$ be a simple $A$-module. Then either $\pi V=V$ or $\pi V=0$. Since $V=Av$ for any $0\ne v\in V$ and $A$ is finitely generated over $R$, we know $V$ is finitely generated over $R$. Thus if $\pi V=V$, then $V=0$ by Nakayama lemma. So $\pi V=A\pi V=0$ and $A\pi \subseteq J(A)$.

ii) Define $A^{\ast }=A/A\pi$. Then $J(A^{\ast })=J(A)/A\pi$. Since $A^{\ast }$ is a finite-dimensional $F$-algebra, we have $A^{\ast }$ is Artinian and $J(A^{\ast })$ is nilpotent. It deduces that $J(A{)}^n\subseteq A\pi$ for some $n>0$. \end{proof}

lifting of isomorphism

Let $A$ be a finitely generated $R$-algebra. Let $I$ be an ideal of $A$ with $I\subseteq J(A)$ and $\overline{A}=A/I$.

• Let $\overline{c}$ be an idempotent in $\overline{A}$, and let $\overline{c}={\overline{c}}_1+\cdots +{\overline{c}}_n$ be an idempotent decomposition in $\overline{A}$. Then $\overline{c}$ lifts to an idempotent $e$ of $A$ and there exists orthogonal idempotent $e_1,\cdots ,e_n$ of $A$ satisfying ${\overline{e}}_i={\overline{c}}_i$ and $e=e_1+\cdots +e_n$.
• An idempotent $e$ of $A$ is primitive iff $\overline{e}$ is primitive in $\overline{A}$.
• Let $e,f$ be primitive idempotent of $A$, then $Ae\simeq Af$ iff $\overline{A}\overline{e}\simeq \overline{A}\overline{f}$.

\begin{proof} Note that $I\subseteq J(A)$ yields $I^m\subseteq A\pi$ for some $m$. Hence $I^{mr}\subseteq A{\pi }^r$ for all $r⩾1$, and set Set $\overline{A^{(r)}}=A/I^{mr}$.

i) Since $I/I^m$ is an nilpotent ideal in $\overline{A^{(1)}}$, by ^84f4de, $\overline{c}$ lifts to a idempotent of $\overline{A^{(1)}}$, namely, there exists $c^{(1)}\in A$ such that $(c^{(1)}{)}^2\equiv c^{(1)}(\mathrm{mod}{I}^m)$ and $c^{(1)}\equiv c(\mathrm{mod}I)$. Next, $I^m/I^{2m}$ is a nilpotent ideal in $\overline{A^{(2)}}$, and there exists $c^{(2)}\in A$ such that $(c^{(2)}{)}^2\equiv c^{(2)}(\mathrm{mod}{I}^{2m})$ and $c^{(2)}\equiv c^{(1)}(\mathrm{mod}{I}^m)$. Repeat this procedure, then $\{c^{(r)}{\}}_r$ is a Cauchy sequence and $e={\lim }_{r\to \infty }{c}^{(r)}$ exists. One can check that $e$ is an idempotent of $A$ that lifts $\overline{c}$ to $A$.

Similar as the above arguments, there exists $c_i^{(r)}\in A$ for each $r$ such that $c_i^{(r)}\equiv c_i^{(r-1)}(\mathrm{mod}{I}^{(r-1)m})$, and $e\equiv c_1^{(r)}+\cdots +c_n^{(r)}(\mathrm{mod}{I}^{rm})$ is an idempotent decomposition in $\overline{A^{(r)}}$. Each $(c_i^{(r)}{)}_r$ is a Cauchy sequence. Let $e_i={\lim }_{r\to \infty }{c}_i^{(r)}$. Then $e_1,\cdots ,e_n$ satisfy our requirement.

ii) Similar as

If $I$ is a nilpotent ideal of $R$, then we can lift idempotents from $R/I$ to $R$. Moreover, if $e\in R/I$  is a primitive idempotent, then any idempotent lift $f\in R$ of $e$ is also primitive.

Link to original

iii) One can prove that $Ae$ is the projective cover of $\overline{A}\overline{e}$ (omitted), then by ^9so7ao we finish the proof. \end{proof}

Remark. Let $I=A\pi$, then $\overline{A}=A^{\ast }$ where $A$ is an $R$-algebra, $A^{\ast }$ is an $F$-algebra and $k{\otimes }_{R}A$ is a $k$-algebra.