Definition 10.32.1. 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 $I^ n = 0$.
10.32 Locally nilpotent ideals
Here is the definition.
Example 10.32.2. 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 10.32.3. 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 10.32.4. 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. $\square$
Lemma 10.32.5. 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 10.32.6. 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 10.32.6. 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 10.21.3. $\square$
Second proof of Lemma 10.32.6. 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 odd 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 10.32.7. 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 10.32.6. $\square$
Lemma 10.32.8. 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
the $n$th power map $1 + I \to 1 + I$, $1 + x \mapsto (1 + x)^ n$ is a bijection,
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 10.32.4) 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 10.32.4). $\square$
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