The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

\[ (e')^2 - e' = (4e^2 - 4e - 3)(e^2 - e)^2 \in J^{2k} \]

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. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.31: Locally nilpotent ideals

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00J9. Beware of the difference between the letter 'O' and the digit '0'.