Lemma 10.31.6. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Then $R \to R/I$ induces a bijection on idempotents.

**First proof of Lemma 10.31.6.**
As $I$ is locally nilpotent it is contained in every prime ideal. Hence $\mathop{\mathrm{Spec}}(R/I) = V(I) = \mathop{\mathrm{Spec}}(R)$. Hence the lemma follows from Lemma 10.20.3.
$\square$

**Second proof of Lemma 10.31.6.**
Suppose $\overline{e} \in R/I$ is an idempotent. We have to lift $\overline{e}$ to an idempotent of $R$.

First, choose any lift $f \in R$ of $\overline{e}$, and set $x = f^2 - f$. Then, $x \in I$, so $x$ is nilpotent (since $I$ is locally nilpotent). Let now $J$ be the ideal of $R$ generated by $x$. Then, $J$ is nilpotent (not just locally nilpotent), since it is generated by the nilpotent $x$.

Now, assume that we have found a lift $e \in R$ of $\overline{e}$ such that $e^2 - e \in J^ k$ for some $k \geq 1$. Let $e' = e - (2e - 1)(e^2 - e) = 3e^2 - 2e^3$, which is another lift of $\overline{e}$ (since the idempotency of $\overline{e}$ yields $e^2 - e \in I$). Then

by a simple computation.

We thus have started with a lift $e$ of $\overline{e}$ such that $e^2 - e \in J^ k$, and obtained a lift $e'$ of $\overline{e}$ such that $(e')^2 - e' \in J^{2k}$. This way we can successively improve the approximation (starting with $e = f$, which fits the bill for $k = 1$). Eventually, we reach a stage where $J^ k = 0$, and at that stage we have a lift $e$ of $\overline{e}$ such that $e^2 - e \in J^ k = 0$, that is, this $e$ is idempotent.

We thus have seen that if $\overline{e} \in R/I$ is any idempotent, then there exists a lift of $\overline{e}$ which is an idempotent of $R$. It remains to prove that this lift is unique. Indeed, let $e_1$ and $e_2$ be two such lifts. We need to show that $e_1 = e_2$.

By definition of $e_1$ and $e_2$, we have $e_1 \equiv e_2 \mod I$, and both $e_1$ and $e_2$ are idempotent. From $e_1 \equiv e_2 \mod I$, we see that $e_1 - e_2 \in I$, so that $e_1 - e_2$ is nilpotent (since $I$ is locally nilpotent). A straightforward computation (using the idempotency of $e_1$ and $e_2$) reveals that $(e_1 - e_2)^3 = e_1 - e_2$. Using this and induction, we obtain $(e_1 - e_2)^ k = e_1 - e_2$ for any positive integer $k$. Since all high enough $k$ satisfy $(e_1 - e_2)^ k = 0$ (since $e_1 - e_2$ is nilpotent), this shows $e_1 - e_2 = 0$, so that $e_1 = e_2$, which completes our proof. $\square$

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