[proofplan]
We first prove the lifting statement across a square-zero ideal by writing an arbitrary lift $a$ of the idempotent and correcting it by an element $x$ of the ideal. The defect $d=a^2-a$ lies in the square-zero ideal and commutes with $a$, which makes the explicit correction $x=(1-2a)d$ work. For a general nilpotent ideal $I$, we lift successively through the finite tower of quotients $R/I^m$, where each intermediate kernel is square-zero.
[/proofplan]
[step:Reduce the problem to a square-zero lifting statement in an arbitrary unital ring]
We prove the following square-zero lifting statement. Let $A$ be a unital ring, let $J \trianglelefteq A$ be a two-sided ideal with $J^2=0$, and let $\bar p \in A/J$ satisfy $\bar p^2=\bar p$. Then there exists $p \in A$ such that $p^2=p$ and whose image in $A/J$ is $\bar p$.
Choose $a \in A$ whose image in $A/J$ is $\bar p$. Define
\begin{align*}
d := a^2-a \in A.
\end{align*}
Since the image of $a$ in $A/J$ is idempotent, $d \in J$. Moreover $a$ commutes with $d$, because
\begin{align*}
ad = a(a^2-a)=a^3-a^2
\end{align*}
and
\begin{align*}
da = (a^2-a)a=a^3-a^2.
\end{align*}
Define
\begin{align*}
x := (1_A-2a)d \in J.
\end{align*}
Since $J$ is a two-sided ideal and $d \in J$, this element lies in $J$. We claim that $p:=a+x$ is idempotent.
[guided]
The obstruction to $a$ already being idempotent is the element
\begin{align*}
d := a^2-a.
\end{align*}
The image of $a$ in $A/J$ is $\bar p$, and $\bar p$ is idempotent, so the image of $d$ in $A/J$ is
\begin{align*}
\bar p^2-\bar p=0.
\end{align*}
Thus $d \in J$. The important algebraic fact is that $d$ commutes with $a$. Indeed,
\begin{align*}
ad = a(a^2-a)=a^3-a^2
\end{align*}
and
\begin{align*}
da = (a^2-a)a=a^3-a^2.
\end{align*}
Hence $ad=da$.
We now look for a corrected lift of the form $a+x$, where $x \in J$. Choosing $x \in J$ ensures that $a+x$ still maps to $\bar p$ in $A/J$. The square-zero hypothesis $J^2=0$ will remove the quadratic term $x^2$ from the idempotence equation. Define
\begin{align*}
x := (1_A-2a)d.
\end{align*}
Because $d \in J$ and $J$ is a two-sided ideal, $x \in J$. Therefore $p:=a+x$ has the same image as $a$ in $A/J$, namely $\bar p$.
[/guided]
[/step]
[step:Verify that the square-zero correction is idempotent]
Since $x \in J$ and $J^2=0$, we have $x^2=0$. Expanding in the associative ring $A$ gives
\begin{align*}
(a+x)^2-(a+x)=a^2-a+ax+xa-x.
\end{align*}
Thus
\begin{align*}
(a+x)^2-(a+x)=d+ax+xa-x.
\end{align*}
Because $a$ commutes with $d$, it also commutes with $x=(1_A-2a)d$. Hence
\begin{align*}
ax+xa-x=(2a-1_A)x.
\end{align*}
Substituting the definition of $x$ gives
\begin{align*}
(2a-1_A)x=(2a-1_A)(1_A-2a)d.
\end{align*}
Therefore
\begin{align*}
(2a-1_A)x=-(2a-1_A)^2d.
\end{align*}
Now
\begin{align*}
(2a-1_A)^2=4a^2-4a+1_A=1_A+4d.
\end{align*}
Since $d \in J$ and $J^2=0$, we have $d^2=0$. Therefore
\begin{align*}
(2a-1_A)^2d=(1_A+4d)d=d.
\end{align*}
It follows that
\begin{align*}
ax+xa-x=-d.
\end{align*}
Consequently
\begin{align*}
(a+x)^2-(a+x)=d-d=0.
\end{align*}
Thus $p=a+x$ is idempotent and maps to $\bar p$ in $A/J$. This proves the square-zero lifting statement.
[guided]
We now verify the computation that makes the correction work. Since $x \in J$ and $J^2=0$, the product $x^2$ is zero. Therefore the idempotence defect of $a+x$ is
\begin{align*}
(a+x)^2-(a+x)=a^2+ax+xa+x^2-a-x.
\end{align*}
Using $x^2=0$ and $d=a^2-a$, this becomes
\begin{align*}
(a+x)^2-(a+x)=d+ax+xa-x.
\end{align*}
The goal is to choose $x$ so that the last three terms equal $-d$. Because $d$ commutes with $a$, the element $x=(1_A-2a)d$ also commutes with $a$: it is a product of $d$ with a polynomial in $a$. Hence
\begin{align*}
ax+xa-x=(2a-1_A)x.
\end{align*}
Substituting $x=(1_A-2a)d$ gives
\begin{align*}
(2a-1_A)x=(2a-1_A)(1_A-2a)d.
\end{align*}
Since $1_A-2a=-(2a-1_A)$, this is
\begin{align*}
(2a-1_A)x=-(2a-1_A)^2d.
\end{align*}
We compute the square:
\begin{align*}
(2a-1_A)^2=4a^2-4a+1_A.
\end{align*}
Using $d=a^2-a$, this is
\begin{align*}
(2a-1_A)^2=1_A+4d.
\end{align*}
Multiplying by $d$ gives
\begin{align*}
(2a-1_A)^2d=(1_A+4d)d.
\end{align*}
Since $d \in J$ and $J^2=0$, we have $d^2=0$, so
\begin{align*}
(2a-1_A)^2d=d.
\end{align*}
Therefore
\begin{align*}
ax+xa-x=-d.
\end{align*}
Substituting this into the defect computation gives
\begin{align*}
(a+x)^2-(a+x)=d-d=0.
\end{align*}
Thus $a+x$ is idempotent. Because $x \in J$, the element $a+x$ has the same image as $a$ in $A/J$, namely $\bar p$. This proves the square-zero case.
[/guided]
[/step]
[step:Apply the square-zero lifting statement to the nilpotent tower]
Return to the theorem. Let
\begin{align*}
\pi_m: R/I^{m+1} \longrightarrow R/I^m
\end{align*}
denote the quotient homomorphism for each $m \ge 1$ for which both quotients are defined. Since $I$ is nilpotent, choose $N \in \mathbb N$ such that $I^N=0$. For each $m \in \{1,\dots,N-1\}$, the kernel of $\pi_m$ is the ideal
\begin{align*}
I^m/I^{m+1} \trianglelefteq R/I^{m+1}.
\end{align*}
This ideal is square-zero, because
\begin{align*}
(I^m/I^{m+1})^2=I^{2m}/I^{m+1}=0
\end{align*}
inside $R/I^{m+1}$, since $2m \ge m+1$ for $m \ge 1$.
The quotient homomorphism $\pi_m$ induces the entrywise quotient homomorphism
\begin{align*}
M_n(R/I^{m+1}) \longrightarrow M_n(R/I^m).
\end{align*}
Its kernel is $M_n(I^m/I^{m+1})$, and this kernel is square-zero because the product of two matrices with entries in $I^m/I^{m+1}$ has entries in $(I^m/I^{m+1})^2=0$.
Starting with the given idempotent
\begin{align*}
e_1:=\bar e \in M_n(R/I),
\end{align*}
apply the square-zero lifting statement successively to lift $e_m \in M_n(R/I^m)$ to an idempotent
\begin{align*}
e_{m+1} \in M_n(R/I^{m+1})
\end{align*}
for $m=1,\dots,N-1$. The result is an idempotent
\begin{align*}
e_N \in M_n(R/I^N).
\end{align*}
Since $I^N=0$, the [quotient ring](/page/Quotient%20Ring) $R/I^N$ is naturally $R$. Under this identification, $e_N$ is an idempotent matrix $e \in M_n(R)$ whose image in $M_n(R/I)$ is the original idempotent $\bar e$. This proves the theorem.
[/step]
