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*}

10.31 Locally nilpotent ideals

Here is the definition.

Definition 10.31.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. We say $I$ is locally nilpotent if for every $x \in I$ there exists an $n \in \mathbf{N}$ such that $x^ n = 0$. We say $I$ is nilpotent if there exists an $n \in \mathbf{N}$ such that $I^ n = 0$.

Example 10.31.2. Let $R = k[x_ n | n \in \mathbf{N}]$ be the polynomial ring in infinitely many variables over a field $k$. Let $I$ be the ideal generated by the elements $x_ n^ n$ for $n \in \mathbf{N}$ and $S = R/I$. Then the ideal $J \subset S$ generated by the images of $x_ n$, $n \in \mathbf{N}$ is locally nilpotent, but not nilpotent. Indeed, since $S$-linear combinations of nilpotents are nilpotent, to prove that $J$ is locally nilpotent it is enough to observe that all its generators are nilpotent (which they obviously are). On the other hand, for each $n \in \mathbf{N}$ it holds that $x_{n + 1}^ n \not\in I$, so that $J^ n \not= 0$. It follows that $J$ is not nilpotent.

Lemma 10.31.3. Let $R \to R'$ be a ring map and let $I \subset R$ be a locally nilpotent ideal. Then $IR'$ is a locally nilpotent ideal of $R'$.

Proof. This follows from the fact that if $x, y \in R'$ are nilpotent, then $x + y$ is nilpotent too. Namely, if $x^ n = 0$ and $y^ m = 0$, then $(x + y)^{n + m - 1} = 0$. $\square$

Lemma 10.31.4. Let $R$ be a ring and let $I \subset R$ be a locally nilpotent ideal. An element $x$ of $R$ is a unit if and only if the image of $x$ in $R/I$ is a unit.

Proof. If $x$ is a unit in $R$, then its image is clearly a unit in $R/I$. It remains to prove the converse. Assume the image of $y \in R$ in $R/I$ is the inverse of the image of $x$. Then $xy = 1 - z$ for some $z \in I$. This means that $1\equiv z$ modulo $xR$. Since $z$ lies in the locally nilpotent ideal $I$, we have $z^ N = 0$ for some sufficiently large $N$. It follows that $1 = 1^ N \equiv z^ N = 0$ modulo $xR$. In other words, $x$ divides $1$ and is hence a unit. $\square$

slogan

Lemma 10.31.5. Let $R$ be a Noetherian ring. Let $I, J$ be ideals of $R$. Suppose $J \subset \sqrt{I}$. Then $J^ n \subset I$ for some $n$. In particular, in a Noetherian ring the notions of “locally nilpotent ideal” and “nilpotent ideal” coincide.

Proof. Say $J = (f_1, \ldots , f_ s)$. By assumption $f_ i^{d_ i} \in I$. Take $n = d_1 + d_2 + \ldots + d_ s + 1$. $\square$

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. $\square$

Lemma 10.31.7. Let $A$ be a possibly noncommutative algebra. Let $e \in A$ be an element such that $x = e^2 - e$ is nilpotent. Then there exists an idempotent of the form $e' = e + x(\sum a_{i, j}e^ ix^ j) \in A$ with $a_{i, j} \in \mathbf{Z}$.

Proof. Consider the ring $R_ n = \mathbf{Z}[e]/((e^2 - e)^ n)$. It is clear that if we can prove the result for each $R_ n$ then the lemma follows. In $R_ n$ consider the ideal $I = (e^2 - e)$ and apply Lemma 10.31.6. $\square$

Lemma 10.31.8. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $n \geq 1$ be an integer which is invertible in $R/I$. Then

  1. the $n$th power map $1 + I \to 1 + I$, $1 + x \mapsto (1 + x)^ n$ is a bijection,

  2. a unit of $R$ is a $n$th power if and only if its image in $R/I$ is an $n$th power.

Proof. Let $a \in R$ be a unit whose image in $R/I$ is the same as the image of $b^ n$ with $b \in R$. Then $b$ is a unit (Lemma 10.31.4) and $ab^{-n} = 1 + x$ for some $x \in I$. Hence $ab^{-n} = c^ n$ by part (1). Thus (2) follows from (1).

Proof of (1). This is true because there is an inverse to the map $1 + x \mapsto (1 + x)^ n$. Namely, we can consider the map which sends $1 + x$ to

\begin{align*} (1 + x)^{1/n} & = 1 + {1/n \choose 1}x + {1/n \choose 2}x^2 + {1/n \choose 3}x^3 + \ldots \\ & = 1 + \frac{1}{n} x + \frac{1 - n}{2n^2}x^2 + \frac{(1 - n)(1 - 2n)}{6n^3}x^3 + \ldots \end{align*}

as in elementary calculus. This makes sense because the series is finite as $x^ k = 0$ for all $k \gg 0$ and each coefficient ${1/n \choose k} \in \mathbf{Z}[1/n]$ (details omitted; observe that $n$ is invertible in $R$ by Lemma 10.31.4). $\square$


Comments (2)

Comment #2998 by Herman Rohrbach on

I want to contribute some useful examples in the spirit of the to-do list, so here is an example of an ideal that is locally nilpotent but not nilpotent.

Let be the polynomial ring in infinitely many variables over . Let be the ideal generated by and . Then the ideal generated by the image in of the set is locally nilpotent, but not nilpotent. Indeed, since -linear combinations of nilpotents are nilpotent, to prove that is locally nilpotent it is enough to observe that all its generators are nilpotent (which they obviously are). On the other hand, for each it holds that , so that . It follows that is not nilpotent.

Comment #3121 by on

Thanks very much. Next time, can you also email the latex to stacks.project (at) gmail.com please? The corresponding changes are here.


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 0AMF. Beware of the difference between the letter 'O' and the digit '0'.