Locally nilpotent sets of derivations

TL;DR

This paper investigates the properties of locally nilpotent sets of derivations on an algebra, exploring their structure and implications for Lie subalgebras composed of such derivations.

Contribution

It introduces the concept of locally nilpotent sets of derivations and examines their fundamental properties and structural consequences.

Findings

Established basic results about locally nilpotent sets

Analyzed conditions under which Lie subalgebras of locally nilpotent derivations are nilpotent

Explored implications for the structure of derivation Lie algebras

Abstract

Let B be an algebra over a field k and let Der(B) be the set of k-derivations from B to B. We define what it means for a subset of Der(B) to be a locally nilpotent set. We prove some basic results about that notion and explore the following questions. Let L be a Lie subalgebra of Der(B); if every element of L is a locally nilpotent derivation then does it follow that L is a locally nilpotent set? Does it follow that L is a nilpotent Lie algebra?

Equations

$$\text{Δ is uniformly locally nilpotent}\Rightarrow \text{Δ is locally nilpotent}\Rightarrow \Delta \subseteq \mathrm{LND}(B)$$

$$[x_3,x_2,x_1,x_0]=[x_3,[x_2,[x_1,x_0]]]=x_3\cdot (x_2\cdot (x_1\cdot x_0))=x_3\cdot x_2\cdot x_1\cdot x_0.$$

$$[x_{n},\dots,x_{0}]=\big{(}\operatorname{ad}(x_{n})\circ\cdots\circ\operatorname{ad}(x_{1})\big{)}(x_{0})\;.$$

$$A=\operatorname{Span}_{\mathbf{k}}\big{\{}\,x_{m}\cdots x_{0}\,\mid\,\text{$m\in\mathbb{N}$ and $(x_{0},\dots,x_{m})\in H^{m+1}$}\,\big{\}}.$$

$$\max\big{\{}\,n\in\mathbb{N}\,\mid\,\text{there exists $(F_{1},\dots,F_{n})\in\Delta^{n}$ satisfying $(F_{n}\circ\cdots\circ F_{1})(x)\neq 0$}\,\big{\}}$$

$$\mathrm{Nil}(\bar{\Delta })\subseteq \mathrm{Nil}({\Delta }_{\circ })\subseteq \mathrm{Nil}(\Delta )$$

$$\mathrm{UNil}(\bar{\Delta })\subseteq \mathrm{UNil}({\Delta }_{\circ })\subseteq \mathrm{UNil}(\Delta )$$

$$x\in V$$

$$H_i={\sum }_{j=0}^{m_i}{\lambda }_{i,j}{G}_{i,j}.$$

$$(G_{n,j_n}\circ \cdots \circ G_{0,j_0})(x)=0\text{for all (j0,…,jn)∈Im0×⋯×Imn}$$

$$\sum _{(j_0,\dots ,j_n)\in {\mathbb{I}}_{m_0}\times \cdots \times {\mathbb{I}}_{m_n}}{\mu }_{(j_0,\dots ,j_n)}(G_{n,j_n}\circ \cdots \circ G_{0,j_0})(x)$$

$$(D_n\circ \cdots \circ D_0)(x\cdot y)=\sum _{\wp }{D}_I(x)\cdot D_{{\mathbb{I}}_n\setminus I}(y).$$

$$Z=\big{\{}\,n\in\mathbb{N}\,\mid\,\text{there exists a subset $I$ of $\mathbb{I}_{n}$ such that $I\in X$ and $\mathbb{I}_{n}\setminus I\in X$}\,\big{\}}$$

$$(D_n\circ \cdots \circ D_0)(x\cdot y)=\sum _{\wp }{D}_I(x)\cdot D_{{\mathbb{I}}_n\setminus I}(y)$$

$$P(n):\qquad\forall_{x,y\in A\setminus\{0\}}\ \big{(}\deg_{\Delta}(x)+\deg_{\Delta}(y)\leq n\implies\deg_{\Delta}(x\cdot y)\leq n\big{)}.$$

$${\mathrm{deg}}_{\Delta }(D_1(x)\cdot y)\le n\text{and}{\mathrm{deg}}_{\Delta }(x\cdot D_1(y))\le n.$$

$${\mathrm{deg}}_{\Delta }(D_1(x\cdot y))$$

$$\displaystyle\leq\max\big{(}\deg_{\Delta}(D_{1}(x)\cdot y),\deg_{\Delta}(x\cdot D_{1}(y))\big{)}\leq n$$

$$(D_{n+2}\circ \cdots \circ D_1)(x\cdot y)=(D_{n+2}\circ \cdots \circ D_2)(D_1(x\cdot y))=0,$$

$$F_n(e_i)=0$$

$$F_n\circ F_m=0\text{for all m,n such that m≤n.}$$

$$(F_n\circ \cdots \circ F_1)(x)=0\text{for all n≥N and all (F1,…,Fn)∈Δn.}$$

$$[G_{n},\dots,G_{1}]\in\operatorname{Span}_{\mathbf{k}}\big{\{}\,F_{n}\circ\cdots\circ F_{1}\,\mid\,(F_{1},\dots,F_{n})\in\Delta^{n}\,\big{\}},$$

$$\big{(}(F_{0},\dots,F_{n}),(F_{0},\dots,F_{n},F_{n+1})\big{)}\;.$$

$$[F_n,\dots ,F_1](x)=0\text{for all n≥N and all (F1,…,Fn)∈Δn},$$

$$[F_n,\dots ,F_1](x)=0\text{for all n≥N and all (F1,…,Fn)∈Δ~n}.$$

$$[H_n,\dots ,H_1](x)=0\text{for all (H1,…,Hn)∈Δ~n}.$$

$$[G_n,\dots ,G_1](x)=0\text{for all (G1,…,Gn)∈(Δ□)n}.$$

$$G_i=[F_{i,k_i},\dots ,F_{i,2},F_{i,1}]$$

$$\max\big{\{}\,n\in\mathbb{N}\,\mid\,\text{there exists $(a_{1},\dots,a_{n})\in H^{n}$ satisfying $a_{n}\cdots a_{1}\cdot x\neq 0$}\,\big{\}}\;.$$

$${\mathrm{Nil}}^{\prime }(H)=\mathrm{Nil}(\phi (H))$$

Full text

††footnotetext: 2010 Mathematics Subject Classification. Primary: 14R20, 13N15. Secondary: 17B30, 17B65, 17B66.

In this article, an algebra over a field $\mathbf{k}$ is a pair $(B,\cdot)$ where $B$ is a $\mathbf{k}$-vector space and “$\cdot$” is an arbitrary $\mathbf{k}$-bilinear map $B\times B\to B$, $(x,y)\mapsto x\cdot y$. Let $B$ be an algebra over a field $\mathbf{k}$ and let $\operatorname{Der}_{\mathbf{k}}(B)$ be the set of all $\mathbf{k}$-derivations $D:B\to B$. A derivation $D\in\operatorname{Der}_{\mathbf{k}}(B)$ is said to be locally nilpotent if for each $x\in B$ there exists an $n>0$ such that $D^{n}(x)=0$. We write $\operatorname{LND}(B)$ for the set of locally nilpotent derivations of $B$. We say that a subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$ is locally nilpotent if for each $x\in B$ the following holds:

for every infinite sequence $(D_{1},D_{2},\dots)$ of elements of $\Delta$, there exists $n$

such that $(D_{n}\circ\cdots\circ D_{1})(x)=0$.

We say that $\Delta$ is uniformly locally nilpotent if for each $x\in B$ there exists $n$ such that $(D_{n}\circ\cdots\circ D_{1})(x)=0$ for all $(D_{1},\dots,D_{n})\in\Delta^{n}$. Then it is clear that the implications

[TABLE]

are true, and it is easy to see that both converses are false. Since $\Delta\subseteq\operatorname{LND}(B)$ does not imply that $\Delta$ is locally nilpotent, it is natural to ask whether the stronger condition $\operatorname{Span}_{\mathbf{k}}(\Delta)\subseteq\operatorname{LND}(B)$ would imply that $\Delta$ is locally nilpotent; it turns out that the answer is negative:

{Example*}

Let $\mathbf{k}$ be a field and $B=\mathbf{k}[x,y,z]$ (polynomial ring in $3$ variables). Let $\Delta=\{D,E\}$ where $D=x\frac{\partial}{\partial y}+y\frac{\partial}{\partial z}$ and $E=y\frac{\partial}{\partial x}-z\frac{\partial}{\partial y}$. Then $\operatorname{Span}_{\mathbf{k}}(\Delta)\subseteq\operatorname{LND}(B)$. However, we have $D(E(x))=x$, so $\Delta$ is not a locally nilpotent set.

In the above Example the derivation $[D,E]=D\circ E-E\circ D$ is not locally nilpotent (it sends $x$ to itself), so the Lie algebra generated by $\Delta$ is not included in $\operatorname{LND}(B)$. So it makes sense to ask: if we make the stronger assumption that the Lie algebra generated by $\Delta$ is included in $\operatorname{LND}(B)$, then does it follow that $\Delta$ is locally nilpotent? This is equivalent to asking: {Question} Let $L$ be a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. If $L\subseteq\operatorname{LND}(B)$, does it follow that $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$?

This question leads naturally to another question: can we characterize the Lie algebras that can be embedded in $\operatorname{LND}(B)$? This question may be too ambitious, so we shall restrict ourselves to the following version of it:

{Question}

Let $L$ be a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. If $L$ satisfies one of:

$\bullet$ $L\subseteq\operatorname{LND}(B)$,

$\bullet$ $L$ is a locally nilpotent (or uniformly locally nilpotent) subset of $\operatorname{Der}_{\mathbf{k}}(B)$,

$\bullet$ $L$ is a Lie-locally nilpotent (or uniformly so) subset of $\operatorname{Der}_{\mathbf{k}}(B)$,

then does it follow that $L$ is a nilpotent Lie algebra, or that $L$ satisfies some other nilpotency condition that makes sense for abstract Lie algebras? (The notion of a Lie-locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ is defined in Sec. 3.)

The aim of this article is to develop the basic theory of locally nilpotent sets and to explore what can be said about Questions 1 and 2 (where in fact Question 2 contains several questions). The basic theory is developed in Section 2; the first steps are done in the more general context of linear endomorphisms of vector spaces, and the later part of the Section is devoted to derivations. Section 3 deals with a variant of the notion of locally nilpotent set that is relevant in the context of Lie algebras of derivations. Section 4 studies five notions of nilpotency for Lie or associative algebras, needed in order to address Question 2. Three of them are classical (an algebra can be nilpotent, nil, or locally nilpotent). The other two are those of a sequentially nilpotent algebra and of a locally nil algebra; as far as we know these two have not been studied previously. Sections 5 and 6 are mainly devoted to Questions 1 and 2. Section 5 shows that if $B$ is the polynomial ring in $|\mathbb{N}|$ variables over a field $\mathbf{k}$ then the following hold:

1. (1)

(Ex. 5) There exists a Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ such that

1. (a)

every finitely generated Lie subalgebra of $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ (so $L\subseteq\operatorname{LND}(B)$); 2. (b)

$L$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

So Question 1 has a negative answer. 2. (2)

(Ex. 5) For each integer $m\geq 2$ there exists an $m$-generated Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

every $(m-1)$-generated Lie subalgebra of $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ (so $L\subseteq\operatorname{LND}(B)$); 2. (b)

$L$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

So Question 1 has a negative answer even when $L$ is a finitely generated Lie algebra. 3. (3)

(Ex. 5) There exists a Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ such that

1. (a)

$L$ is a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$; 2. (b)

$L$ is the free Lie algebra on a countably infinite set (so $L$ is as non-nilpotent as a Lie algebra can be).

So all questions that are part of Question 2 have negative answers. 4. (4)

(Cor. 5.9) There exists an infinite subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

$\Delta$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$; 2. (b)

every finite subset of $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. 5. (5)

(Cor. 5.9) For each integer $m\geq 2$, there exists an $m$-subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

$\Delta$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$; 2. (b)

every proper subset of $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

Facts (4) and (5) show just how capriciously the notion of locally nilpotent set can behave. Note that (2) and (5) are based on Golod’s famous example of an associative algebra that is nil and finitely generated but not nilpotent.

The above facts suggest that if we do not assume that $B$ satisfies some finiteness condition then we cannot expect our questions to have affirmative answers. Section 6 re-examines Question 2 under the additional assumption that $B$ is “derivation-finite” (this is more general than $B$ being finitely generated, see Def. 6). Cor. 6.4 and Prop. 6.6 give some positive answers to Question 2, but Question 1 is left open.

In future work, we hope to apply these ideas to the following questions. Let $B$ be a commutative algebra over a field $\mathbf{k}$ of characteristic zero, let $\Delta$ be a subset of $\operatorname{LND}(B)$ and consider the subset Y=\big{\{}\,\exp(D)\,\mid\,D\in\Delta\,\big{\}} of $\operatorname{Aut}_{\mathbf{k}}(B)$. How are the properties of $\Delta$ related to those of $Y$? When is $Y$ a group? When is $Y$ included in an algebraic group?

1 Preliminaries

{nothing*}

Directed trees. Consider a pair $(S,E)$ where $S$ is a set and $E$ is a subset of $S\times S$ such that every element $(u,v)$ of $E$ satisfies $u\neq v$. Then $(S,E)$ can be viewed as a directed graph where $S$ is the vertex-set and $E$ is the edge-set (if $(u,v)\in E$ then $(u,v)$ is a directed edge from $u$ to $v$). We allow $S$ and $E$ to be infinite sets. By a path in the graph $(S,E)$, we mean a sequence $\gamma=(x_{0},x_{1},x_{2},\dots)$ (finite or infinite) of elements of $S$ such that $(x_{i},x_{i+1})\in E$ for all $i\geq 0$. If $\gamma=(x_{0},x_{1},\dots,x_{n})$ is a finite path, we say that $\gamma$ is a path from $x_{0}$ to $x_{n}$. If $x\in S$ then we regard $(x)$ as a path from $x$ to $x$.

Now suppose that $(S,E)$ is a pair of the above type. Then it is not hard to see that there exists at most one vertex $v_{0}\in S$ satisfying: for every $v\in S$ there exists exactly one path from $v_{0}$ to $v$. If such a vertex $v_{0}$ exists then we say that $(S,E)$* is a directed tree with root $v_{0}$*. We shall use the following fact several times; its proof is left to the reader.

Theorem 1.1**.**

Let $(S,E)$ be a directed tree with root $v_{0}$, and suppose:

• $\bullet$

for each $v\in S$, there exist finitely many $v^{\prime}\in S$ such that $(v,v^{\prime})\in E$;

• $\bullet$

there exists no infinite path $(x_{0},x_{1},x_{2},\dots)$ in $(S,E)$ satisfying $x_{0}=v_{0}$.

Then $S$ is a finite set.

{nothing*}

Algebras. Let $\mathbf{k}$ be a field.

(a)   A $\mathbf{k}$-algebra is a pair $(A,\cdot)$ where $A$ is a $\mathbf{k}$-vector space and “$\cdot$” is an arbitrary $\mathbf{k}$-bilinear map $A\times A\to A$, $(x,y)\mapsto x\cdot y$. We call $\cdot$ the multiplication of $A$.

Let $A$ be a $\mathbf{k}$-algebra.

A subalgebra of $A$ is a $\mathbf{k}$-subspace $A^{\prime}$ of $A$ closed under the multiplication of $A$. If $\Delta$ is a subset of $A$ then the subalgebra of $A$ generated by $\Delta$ is the intersection of all subalgebras $A^{\prime}$ of $A$ satisfying $\Delta\subseteq A^{\prime}$.

A $\mathbf{k}$-derivation of $A$ is a $\mathbf{k}$-linear map $D:A\to A$ satisfying $D(x\cdot y)=D(x)\cdot y+x\cdot D(y)$ for all $x,y\in A$.

A homomorphism $\varphi:A\to B$ of $\mathbf{k}$-algebras is a $\mathbf{k}$-linear map satisfying $\varphi(x\cdot y)=\varphi(x)\cdot\varphi(y)$ for all $x,y\in A$.

(b)  An associative algebra over $\mathbf{k}$ is a $\mathbf{k}$-algebra whose multiplication is associative.

(c)  A Lie algebra over $\mathbf{k}$ is a $\mathbf{k}$-algebra $(A,\cdot)$ satisfying $x\cdot x=0$ and the Jacobi identity $x\cdot(y\cdot z)+y\cdot(z\cdot x)+z\cdot(x\cdot y)=0$ for all $x,y,z\in A$.

For associative algebras and Lie algebras, the notions of subalgebra, subalgebra generated by a set, homomorphism and $\mathbf{k}$-derivation are those defined in part (a) for general algebras.

{notation}

(a)  If $(A,\cdot)$ is an associative algebra and if we define $a_{1}*a_{2}=a_{1}\cdot a_{2}-a_{2}\cdot a_{1}$ for all $a_{1},a_{2}\in A$ then $(A,*)$ is a Lie algebra that we denote $A_{\text{\rm L}}$.

(b)  If $V$ is a vector space over $\mathbf{k}$ then $\mathcal{L}_{\mathbf{k}}(V)$ denotes the set of all $\mathbf{k}$-linear maps $V\to V$. We always regard $\mathcal{L}_{\mathbf{k}}(V)$ as an associative algebra, the multiplication being the composition of maps. We write $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$ for the corresponding Lie algebra, as explained in part (a).

(c)  Let $A$ be an algebra over a field $\mathbf{k}$. The set of all $\mathbf{k}$-derivations of $A$ is denoted $\operatorname{Der}_{\mathbf{k}}(A)$, and is a $\mathbf{k}$-subspace of $\mathcal{L}_{\mathbf{k}}(A)$. If $D,E\in\operatorname{Der}_{\mathbf{k}}(A)$ then $D\circ E-E\circ D\in\operatorname{Der}_{\mathbf{k}}(A)$, so $\operatorname{Der}_{\mathbf{k}}(A)$ is a subalgebra of the Lie algebra $\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}$.

{notation}

Let $A$ be an associative algebra over a field $\mathbf{k}$. If $\Delta$ is any subset of $A$ then $\bar{\Delta}$ denotes the subalgebra of $A$ generated by $\Delta$ and $\tilde{\Delta}$ denotes the subalgebra of $A_{\text{\rm L}}$ generated by $\Delta$. We have $\Delta\subseteq\tilde{\Delta}\subseteq\bar{\Delta}$.

Note that $\bar{\Delta}$ is the intersection of all $\mathbf{k}$-subspaces $A^{\prime}$ of $A$ that are closed under the multiplication of $A$ and satisfy $\Delta\subseteq A^{\prime}$. We also point out that $\bar{\Delta}=\operatorname{Span}_{\mathbf{k}}(\Delta_{\circ})$ where $\Delta_{\circ}$ is the set of nonempty products of elements of $\Delta$.

The notation $\bar{\Delta}$ has the above meaning even when $A$ happens to have a unity $1$. For instance, if $A$ is the commutative polynomial ring $\mathbf{k}[x,y]$ and $\Delta=\{x,y\}$ then $\bar{\Delta}$ is the ideal generated by $x,y$.

{nothing*}

We adopt the right-associativity convention, which stipulates that any unparenthesized product $a_{n}\cdots a_{1}$ (where $n>2$ and $a_{1},\dots,a_{n}$ are elements of some algebra) is to be interpreted as meaning $a_{n}\cdot(a_{n-1}\cdots(a_{3}\cdot(a_{2}\cdot a_{1}))\dots)$. For instance, $a\cdot b\cdot c\cdot d=a\cdot(b\cdot(c\cdot d))$. This convention is in effect throughout the paper, in all types of algebras.

{nothing*}

Let $(A,\cdot)$ be a Lie algebra. It is customary to use the bracket notation for multiplication, i.e., to define $[x,y]=x\cdot y$ for all $x,y\in A$. We shall use both notations. If $x_{0},x_{1},\dots,x_{n}\in A$, the element $[x_{n},[x_{n-1},\dots[x_{2},[x_{1},x_{0}]]\dots]]$ of $A$ is simply denoted $[x_{n},\dots,x_{0}]$. This agrees with the right-associativity convention; for instance, going back and forth between the two notations,

[TABLE]

In general we have $[x_{n},\cdots,x_{0}]=x_{n}\cdots x_{0}$, where we allow the case $n=0$: $[x_{0}]=x_{0}$.

If $x\in A$ then the map $y\mapsto[x,y]$ is denoted $\operatorname{ad}(x):A\to A$. The following properties of $\operatorname{ad}$ are well known, and easily verified:

(i) $\operatorname{ad}:A\to\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}$ is a homomorphism of Lie algebras;

(ii) $\operatorname{ad}(A)\subseteq\operatorname{Der}_{\mathbf{k}}(A)$.

Note that if $n>0$ and $x_{0},\dots,x_{n}\in A$ then

[TABLE]

We leave it to the reader to check the following fact (which is trivial in the associative case):

Lemma 1.2**.**

Let $(A,\cdot)$ be an associative algebra or a Lie algebra over a field $\mathbf{k}$.

1. (a)

Let $x_{1},\dots,x_{n}\in A$ and let $x\in A$ be any parenthesization of the product $x_{n}\cdots x_{1}$. Then $x$ is a finite sum $\sum_{i}r_{i}\,x_{i,n}\cdots x_{i,1}$ where, for each $i$, $r_{i}$ belongs to the prime subring of $\mathbf{k}$ and $(x_{i,1},\dots,x_{i,n})$ is a permutation of $(x_{1},\dots,x_{n})$. 2. (b)

If $H$ is a generating set for $A$ then

[TABLE]

2 Locally nilpotent sets of linear maps

Throughout this section, we consider a vector space $V$ over a field $\mathbf{k}$ and we let $\mathcal{L}_{\mathbf{k}}(V)$ denote the set of all $\mathbf{k}$-linear maps $V\to V$. An element $F$ of $\mathcal{L}_{\mathbf{k}}(V)$ is said to be locally nilpotent if for each $x\in V$ there exists $n>0$ such that $F^{n}(x)=0$. We write $\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$ for the set of all locally nilpotent elements of $\mathcal{L}_{\mathbf{k}}(V)$.

If $\Delta$ is a set then $\Delta^{\mathbb{N}}$ denotes the set of all infinite sequences $(a_{0},a_{1},a_{2},\dots)$ of elements of $\Delta$.

{Definition}

Given a subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$, we define the subsets $\operatorname{Nil}(\Delta)$ and $\operatorname{UNil}(\Delta)$ of $V$ as follows. If $\Delta=\varnothing$, we set $\operatorname{Nil}(\Delta)=V=\operatorname{UNil}(\Delta)$. If $\Delta\neq\varnothing$,

• $\bullet$

$\operatorname{Nil}(\Delta)$ is the set of all $x\in V$ satisfying:

for every sequence $(F_{0},F_{1},\dots)\in\Delta^{\mathbb{N}}$ there exists $n\in\mathbb{N}$ such that

$(F_{n}\circ\cdots\circ F_{0})(x)=0$;

• $\bullet$

$\operatorname{UNil}(\Delta)$ is the set of all $x\in V$ satisfying:

there exists $n\in\mathbb{N}$ such that for every $(F_{1},\dots,F_{n})\in\Delta^{n}$ we have $(F_{n}\circ\cdots\circ F_{1})(x)=0$.

Note that $\operatorname{UNil}(\Delta)\subseteq\operatorname{Nil}(\Delta)$ are linear subspaces of $V$. If $\operatorname{Nil}(\Delta)=V$, we say that $\Delta$ is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ (or simply, that $\Delta$ is locally nilpotent); if $\operatorname{UNil}(\Delta)=V$, we say that $\Delta$ is uniformly locally nilpotent.

If $\Delta$ is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ then $\Delta\subseteq\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$; however, the converse is not true.

If $\Delta$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ then $\Delta$ is locally nilpotent, but the converse is not true (see Ex. 6). However, note the following:

Lemma 2.1**.**

Let $\Delta$ be a finite subset of $\mathcal{L}_{\mathbf{k}}(V)$. Then $\operatorname{Nil}(\Delta)=\operatorname{UNil}(\Delta)$. In particular, $\Delta$ is locally nilpotent if and only if it is uniformly locally nilpotent.

Proof 2.2**.**

It suffices to show that $\operatorname{Nil}(\Delta)\subseteq\operatorname{UNil}(\Delta)$. Let $x\in\operatorname{Nil}(\Delta)\setminus\{0\}$. Consider the set $S$ of finite sequences $(F_{1},\dots,F_{n})$ of elements of $\Delta$ satisfying $(F_{n}\circ\cdots\circ F_{1})(x)\neq 0$ (where the empty sequence $()$ belongs to $S$). Let $E\subseteq S\times S$ be the set of ordered pairs of the form \big{(}(F_{1},\dots,F_{n}),(F_{1},\dots,F_{n},F_{n+1})\big{)}. Then $(S,E)$ is a directed tree with root $()$ (see 1). For each vertex $v\in S$, the number of $v^{\prime}\in S$ satisfying $(v,v^{\prime})\in E$ is at most $|\Delta|$, which is finite. The fact that $x\in\operatorname{Nil}(\Delta)$ implies that there is no infinite path \big{(}(),(F_{1}),(F_{1},F_{2}),\dots\big{)} in this tree. It follows from Thm 1.1 that $S$ is a finite set, and this implies that $x\in\operatorname{UNil}(\Delta)$. ∎

{Definition}

Given a nonempty subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$ we define a map $\deg_{\Delta}:V\to\mathbb{N}\cup\{-\infty,\infty\}$ by declaring that $\deg_{\Delta}(0)=-\infty$, if $x\in V\setminus\operatorname{UNil}(\Delta)$ then $\deg_{\Delta}(x)=\infty$, and if $x\in\operatorname{UNil}(\Delta)\setminus\{0\}$ then $\deg_{\Delta}(x)$ is equal to:

[TABLE]

(to be clear, if $x\neq 0$ and $F(x)=0$ for all $F\in\Delta$ then we define $\deg_{\Delta}(x)=0$). If $\Delta=\varnothing$ then we define $\deg_{\Delta}(0)=-\infty$ and $\deg_{\Delta}(x)=0$ for all $x\in V\setminus\{0\}$.

Lemma 2.3**.**

Let $\Delta$ be a subset of $\mathcal{L}_{\mathbf{k}}(V)$. Then the map $\deg_{\Delta}:V\to\mathbb{N}\cup\{\pm\infty\}$ has the following properties, where $x,y$ are arbitrary elements of $V$.

1. (a)

$\deg_{\Delta}(x)<0$* if and only if $x=0$.* 2. (b)

$\deg_{\Delta}(x)\leq 0$* if and only if $F(x)=0$ for all $F\in\Delta$.* 3. (c)

$\deg_{\Delta}(x)<\infty$* if and only if $x\in\operatorname{UNil}(\Delta)$.* 4. (d)

If $x\in\operatorname{UNil}(\Delta)\setminus\{0\}$ and $F\in\Delta$ then $\deg_{\Delta}(F(x))<\deg_{\Delta}(x)$. 5. (e)

$\deg_{\Delta}(x+y)\leq\max(\deg_{\Delta}(x),\deg_{\Delta}(y))$**

Verification of Lemma 2.3 is left to the reader.

The notations $\bar{\Delta}$ and $\tilde{\Delta}$ are defined in 1.

Proposition 2.4**.**

Let $\Delta$ be a subset of $\mathcal{L}_{\mathbf{k}}(V)$, let $\bar{\Delta}$ be the associative subalgebra of $\mathcal{L}_{\mathbf{k}}(V)$ generated by $\Delta$ and let $\tilde{\Delta}$ be the Lie subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$ generated by $\Delta$. Then $\Delta\subseteq\tilde{\Delta}\subseteq\bar{\Delta}\subseteq\mathcal{L}_{\mathbf{k}}(V)$ and the following hold.

1. (a)

$\operatorname{Nil}(\bar{\Delta})=\operatorname{Nil}(\tilde{\Delta})=\operatorname{Nil}(\Delta)$** 2. (b)

$\deg_{\bar{\Delta}}=\deg_{\tilde{\Delta}}=\deg_{\Delta}$* and $\operatorname{UNil}(\bar{\Delta})=\operatorname{UNil}(\tilde{\Delta})=\operatorname{UNil}(\Delta)$* 3. (c)

If $\Delta$ is locally nilpotent then so are $\bar{\Delta}$ and $\tilde{\Delta}$. 4. (d)

If $\Delta$ is uniformly locally nilpotent then so are $\bar{\Delta}$ and $\tilde{\Delta}$.

Proof 2.5**.**

We may assume that $\Delta\neq\varnothing$. Let $\Delta_{\circ}$ be as in 1 and observe that $\Delta\subseteq\Delta_{\circ}\subseteq\bar{\Delta}$ and $\Delta\subseteq\tilde{\Delta}\subseteq\bar{\Delta}$. These inclusions have the following trivial consequences:

[TABLE]

Let $x\in\operatorname{Nil}(\Delta)$. If $(G_{0},G_{1},\dots)\in\Delta_{\circ}^{\mathbb{N}}$ then there exist $(F_{0},F_{1},\dots)\in\Delta^{\mathbb{N}}$ and $1\leq n_{0}<n_{1}<\cdots$ in $\mathbb{N}$ such that $G_{0}=F_{n_{0}-1}\circ\cdots\circ F_{0}$, $G_{1}=F_{n_{1}-1}\circ\cdots\circ F_{n_{0}}$, $G_{2}=F_{n_{2}-1}\circ\cdots\circ F_{n_{1}}$, and so on. Since $(F_{k}\circ\cdots\circ F_{0})(x)=0$ for some $k\in\mathbb{N}$, it follows that $(G_{m}\circ\cdots\circ G_{0})(x)=0$ for some $m\in\mathbb{N}$. This shows that $x\in\operatorname{Nil}(\Delta_{\circ})$ and hence that $\operatorname{Nil}(\Delta)\subseteq\operatorname{Nil}(\Delta_{\circ})$. By (1), we get $\operatorname{Nil}(\Delta_{\circ})=\operatorname{Nil}(\Delta)$.

We continue to consider $x\in\operatorname{Nil}(\Delta)=\operatorname{Nil}(\Delta_{\circ})$. Let $(H_{0},H_{1},\dots)\in\bar{\Delta}^{\mathbb{N}}$; let us prove that there exists $n$ such that $(H_{n}\circ\cdots\circ H_{0})(x)=0$. Recall that $\bar{\Delta}=\operatorname{Span}_{\mathbf{k}}(\Delta_{\circ})$; for each $i\in\mathbb{N}$, we choose $m_{i}\in\mathbb{N}$, $\lambda_{i,0},\dots,\lambda_{i,m_{i}}\in\mathbf{k}$ and $G_{i,0},\dots,G_{i,m_{i}}\in\Delta_{\circ}$ satisfying

[TABLE]

The element $x$, the sequence $(H_{0},H_{1},\dots)$ and the relations (4) determine a directed tree $T$ that we now define. The elements of the vertex-set $S$ are all finite sequences $(G_{0,j_{0}},G_{1,j_{1}},\dots,G_{n,j_{n}})$ such that $(G_{n,j_{n}}\circ\cdots\circ G_{0,j_{0}})(x)\neq 0$ (where $0\leq j_{i}\leq m_{i}$ for all $i=0,\dots,n$ and where the $G_{i,j}$ are the same as in (4)). The edge-set of $T$ is the subset $E$ of $S\times S$ whose elements are the pairs of the form \big{(}(G_{0,j_{0}},\dots,G_{n,j_{n}}),\,(G_{0,j_{0}},\dots,G_{n,j_{n}},G_{n+1,j_{n+1}})\big{)}. The root of $T$ is the empty sequence $()\in S$. Note that if \big{(}(),\,(G_{0,j_{0}}),\,(G_{0,j_{0}},G_{1,j_{1}}),\,(G_{0,j_{0}},G_{1,j_{1}},G_{2,j_{2}}),\,\dots\big{)} is an infinite path in $T$ then the sequence $(G_{0,j_{0}},G_{1,j_{1}},G_{2,j_{2}},\dots)\in\Delta_{\circ}^{\mathbb{N}}$ satisfies $(G_{n,j_{n}}\circ\cdots\circ G_{0,j_{0}})(x)\neq 0$ for all $n\in\mathbb{N}$, which contradicts the fact that $x\in\operatorname{Nil}(\Delta_{\circ})$. So, there is no infinite path in $T$ starting at the root. We also note that if $v=(G_{0,j_{0}},\dots,G_{n,j_{n}})$ is any vertex then the number of vertices $v^{\prime}\in S$ satisfying $(v,v^{\prime})\in E$ is at most $m_{n+1}$, which is finite. It follows from Thm 1.1 that $T$ has finitely many vertices. So there exists $n\in\mathbb{N}$ satisfying:

[TABLE]

where we write \mathbb{I}_{m}=\big{\{}\,i\in\mathbb{N}\,\mid\,0\leq i\leq m\,\big{\}}. Let $n$ be as in (5) and note that $(H_{n}\circ\cdots\circ H_{0})(x)$ is a linear combination

[TABLE]

where $\mu_{(j_{0},\dots,j_{n})}\in\mathbf{k}$ for all $(j_{0},\dots,j_{n})\in\mathbb{I}_{m_{0}}\times\cdots\times\mathbb{I}_{m_{n}}$. By (5), all terms of the sum (6) are zero, so $(H_{n}\circ\cdots\circ H_{0})(x)=0$. This proves that $x\in\operatorname{Nil}(\bar{\Delta})$ and hence that $\operatorname{Nil}(\Delta)\subseteq\operatorname{Nil}(\bar{\Delta})$. Then (1) gives $\operatorname{Nil}(\bar{\Delta})=\operatorname{Nil}(\tilde{\Delta})=\operatorname{Nil}(\Delta)$, so (a) is proved.

Let us prove (b). If $(G_{1},\dots,G_{n})\in\Delta_{\circ}^{n}$ then $G_{n}\circ\cdots\circ G_{1}=F_{N}\circ\cdots\circ F_{1}$ for some $N\geq n$ and $(F_{1},\dots,F_{N})\in\Delta^{N}$. This implies that $\deg_{\Delta_{\circ}}(x)\leq\deg_{\Delta}(x)$ for all $x\in V$, so we obtain $\deg_{\Delta_{\circ}}=\deg_{\Delta}$ by (3).

Consider $x\in V\setminus\{0\}$ and let us prove that $\deg_{\bar{\Delta}}(x)\leq\deg_{\Delta_{\circ}}(x)$. We may assume that $\deg_{\Delta_{\circ}}(x)=n\in\mathbb{N}$. If $(H_{1},\dots,H_{n+1})\in\bar{\Delta}^{n+1}$ then $(H_{n+1}\circ\cdots\circ H_{1})(x)$ is a finite sum $\sum_{i}\lambda_{i}t_{i}$ where $\lambda_{i}\in\mathbf{k}$ and $t_{i}=(G_{i,n+1}\circ\cdots\circ G_{i,1})(x)$ with $G_{i,1},\dots,G_{i,n+1}\in\Delta_{\circ}$. Since $\deg_{\Delta_{\circ}}(x)=n$ we have $t_{i}=0$ for all $i$, so $(H_{n+1}\circ\cdots\circ H_{1})(x)=0$. Consequently, $\deg_{\bar{\Delta}}(x)\leq n=\deg_{\Delta_{\circ}}(x)$. We get $\deg_{\bar{\Delta}}=\deg_{\Delta_{\circ}}$ by (3), and since $\deg_{\Delta_{\circ}}=\deg_{\Delta}$ it follows that $\deg_{\bar{\Delta}}=\deg_{\Delta}$. Using (3) again gives $\deg_{\Delta}=\deg_{\tilde{\Delta}}=\deg_{\bar{\Delta}}$, and hence $\operatorname{UNil}(\Delta)=\operatorname{UNil}(\tilde{\Delta})=\operatorname{UNil}(\bar{\Delta})$. So (b) is proved. Assertions (c) and (d) follow from (a) and (b). ∎

Corollary 2.6**.**

Let $A$ be either an associative subalgebra of $\mathcal{L}_{\mathbf{k}}(V)$ or a Lie subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$. If $A$ is finitely generated then $\operatorname{Nil}(A)=\operatorname{UNil}(A)$ and, consequently, $A$ is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ if and only if it is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$.

Proof 2.7**.**

Let $\Delta$ be a finite generating set for the algebra $A$. Then $\operatorname{Nil}(A)=\operatorname{Nil}(\Delta)=\operatorname{UNil}(\Delta)=\operatorname{UNil}(A)$ by Prop. 2.4 and Lemma 2.1. ∎

A special case: sets of derivations

Let $(A,\cdot)$ be an algebra over a field $\mathbf{k}$, in the sense of 1(a). Recall (from 1) that this determines a Lie subalgebra $\operatorname{Der}_{\mathbf{k}}(A)$ of $\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}$. If $\Delta$ is a subset of $\operatorname{Der}_{\mathbf{k}}(A)$ then $\Delta\subseteq\mathcal{L}_{\mathbf{k}}(A)$, so it makes sense to consider the subsets $\operatorname{Nil}(\Delta)$ and $\operatorname{UNil}(\Delta)$ of $A$, and the map $\deg_{\Delta}:A\to\mathbb{N}\cup\{\pm\infty\}$.

The rest of the section is devoted to the proof of:

Theorem 2.8**.**

Let $(A,\cdot)$ be an algebra over a field $\mathbf{k}$. If $\Delta$ is a subset of $\operatorname{Der}_{\mathbf{k}}(A)$ then the following hold.

1. (a)

The sets $\operatorname{Nil}(\Delta)$ and $\operatorname{UNil}(\Delta)$ are subalgebras of $(A,\cdot)$. 2. (b)

The map $\deg_{\Delta}:A\to\mathbb{N}\cup\{\pm\infty\}$ satisfies $\deg_{\Delta}(x\cdot y)\leq\deg_{\Delta}(x)+\deg_{\Delta}(y)$ for all $x,y\in\operatorname{UNil}(\Delta)$.

We already know that $\operatorname{Nil}(\Delta)$ and $\operatorname{UNil}(\Delta)$ are linear subspaces of $A$. So, to prove (a), it suffices to show that those two sets are closed under the multiplication of $A$.

{Remark}

If $\mathbf{k}\subseteq A$ then $\mathbf{k}\subseteq\operatorname{UNil}(\Delta)\subseteq\operatorname{Nil}(\Delta)$, because $\mathbf{k}\subseteq\ker D$ for all $D\in\operatorname{Der}_{\mathbf{k}}(A)$.

Until the end of the section, we assume that $(A,\cdot)$ is a $\mathbf{k}$-algebra. The Theorem follows from Lemmas 2.13 and 2.15 and Cor. 2.17.

{notation}

If $n\in\mathbb{N}$, we write \mathbb{I}_{n}=\big{\{}\,i\in\mathbb{N}\,\mid\,0\leq i\leq n\,\big{\}}. Let $n\in\mathbb{N}$ and $F_{0},\dots,F_{n}\in\mathcal{L}_{\mathbf{k}}(A)$. If $I=\{i_{1},\dots,i_{m}\}$ is a subset of $\mathbb{I}_{n}$ and $i_{1}<\cdots<i_{m}$, we write $F_{I}=F_{i_{m}}\circ\cdots\circ F_{i_{1}}$ (if $I=\varnothing$ then $F_{I}=\operatorname{id}_{A}$). If $S$ is a set then \raisebox{0.7pt}{\large\wp}(S) denotes the set of all subsets of $S$.

Lemma 2.9**.**

Let $x,y\in A$, $n\in\mathbb{N}$ and $D_{0},\dots,D_{n}\in\operatorname{Der}_{\mathbf{k}}(A)$. Then

[TABLE]

Proof 2.10**.**

This can be proved by induction on $n$. The straightforward argument is left to the reader. ∎

{Definition}

We define a partial order $\preceq$ on the set \raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) of finite subsets of $\mathbb{N}$ by declaring that the condition $I\preceq J$ (where I,J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N})) is equivalent to:

$I\subseteq J$ and for every $i\in I$ and $j\in J\setminus I$ we have $i<j$.

Note in particular that $\varnothing\preceq J$ for all J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}). Also observe that if J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) and $m\in\mathbb{N}$ then $\mathbb{I}_{m}\cap J\preceq J$.

Lemma 2.11**.**

Let $X$ be a subset of \raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) such that

• $\bullet$

for all I,J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) satisfying $I\preceq J$ and $J\in X$, we have $I\in X$;

• $\bullet$

$(X,\preceq)$* satisfies the ascending chain condition (ACC).*

Then the set

[TABLE]

is finite.

Proof 2.12**.**

By contradiction, assume that $Z$ is an infinite set. For each $n\in Z$, choose I_{n}\in\raisebox{0.7pt}{\large\wp}(\mathbb{I}_{n}) such that $I_{n}\in X$ and $\mathbb{I}_{n}\setminus I_{n}\in X$. We inductively define an infinite sequence of sets $Z_{0},Z_{1},\dots$. Choose an infinite subset $Z_{0}$ of $Z$ such that \big{(}\mathbb{I}_{0}\cap I_{n}\big{)}_{n\in Z_{0}} is a constant sequence. Let $m\in\mathbb{N}$ and suppose that $Z_{0},\dots,Z_{m}$ satisfy:

• $\bullet$

$Z\supseteq Z_{0}\supseteq Z_{1}\supseteq\cdots\supseteq Z_{m}$* are infinite sets;*

• $\bullet$

for each $j\in\{0,\dots,m\}$, $\min Z_{j}\geq j$ and \big{(}\mathbb{I}_{j}\cap I_{n}\big{)}_{n\in Z_{j}} is a constant sequence.

Since \big{(}\mathbb{I}_{m+1}\cap I_{n}\big{)}_{n\in Z_{m}} is an infinite sequence of elements of the finite set \raisebox{0.7pt}{\large\wp}(\mathbb{I}_{m+1}), we may choose an infinite subset $Z_{m+1}$ of $Z_{m}$ such that \big{(}\mathbb{I}_{m+1}\cap I_{n}\big{)}_{n\in Z_{m+1}} is a constant sequence and $\min Z_{m+1}\geq m+1$. By induction, it follows that there exists an infinite sequence $Z_{0},Z_{1},Z_{2},\dots$ satisfying:

1. (i)

$Z\supseteq Z_{0}\supseteq Z_{1}\supseteq Z_{2}\supseteq\cdots$* are infinite sets;* 2. (ii)

for each $j\in\mathbb{N}$, $\min Z_{j}\geq j$ and \big{(}\mathbb{I}_{j}\cap I_{n}\big{)}_{n\in Z_{j}} is a constant sequence.

For each $j\in\mathbb{N}$, define $B_{j}=\mathbb{I}_{j}\cap I_{n}$ for any $n\in Z_{j}$, and let $B_{j}^{*}=\mathbb{I}_{j}\setminus B_{j}$. We shall now prove that the two sequences \big{(}B_{j}\big{)}_{j\in\mathbb{N}} and \big{(}B_{j}^{*}\big{)}_{j\in\mathbb{N}} eventually stabilize; this is absurd, because $B_{j}\cup B_{j}^{*}=\mathbb{I}_{j}$.

Let $j\in\mathbb{N}$. Given any $n\in Z_{j}$ we have $B_{j}=\mathbb{I}_{j}\cap I_{n}\preceq I_{n}$ and $I_{n}\in X$, so $B_{j}\in X$. Also, $\mathbb{I}_{j}\cap(\mathbb{I}_{n}\setminus I_{n})\preceq\mathbb{I}_{n}\setminus I_{n}\in X$ implies $\mathbb{I}_{j}\cap(\mathbb{I}_{n}\setminus I_{n})\in X$; since $\mathbb{I}_{j}\cap(\mathbb{I}_{n}\setminus I_{n})=(\mathbb{I}_{j}\cap\mathbb{I}_{n})\setminus(\mathbb{I}_{j}\cap I_{n})=\mathbb{I}_{j}\setminus B_{j}=B_{j}^{*}$ (where we used $n\in Z_{j}\Rightarrow n\geq j\Rightarrow\mathbb{I}_{j}\cap\mathbb{I}_{n}=\mathbb{I}_{j}$), we have $B_{j}^{*}\in X$. Thus \big{(}B_{j}\big{)}_{j\in\mathbb{N}} and \big{(}B_{j}^{*}\big{)}_{j\in\mathbb{N}} are sequences in $X$.

Again, let $j\in\mathbb{N}$. Pick $n\in Z_{j+1}$ (so $n\in Z_{j}$), then $\mathbb{I}_{j}\cap B_{j+1}=\mathbb{I}_{j}\cap(\mathbb{I}_{j+1}\cap I_{n})=\mathbb{I}_{j}\cap I_{n}=B_{j}$, so $B_{j}\preceq B_{j+1}$. Thus $B_{0}\preceq B_{1}\preceq B_{2}\preceq\cdots$, and since $(X,\preceq)$ satisfies ACC there exists $M\in\mathbb{N}$ such that \big{(}B_{j}\big{)}_{j\geq M} is a constant sequence. For $j\geq M$ we have $B_{j}=B_{j+1}=B_{M}$, so $\mathbb{I}_{j}\cap B_{j+1}^{*}=\mathbb{I}_{j}\cap(\mathbb{I}_{j+1}\setminus B_{j+1})=\mathbb{I}_{j}\cap(\mathbb{I}_{j+1}\setminus B_{M})=\mathbb{I}_{j}\setminus B_{M}=\mathbb{I}_{j}\setminus B_{j}=B_{j}^{*}$, so $B_{j}^{*}\preceq B_{j+1}^{*}$. Thus $B_{M}^{*}\preceq B_{M+1}^{*}\preceq\cdots$, so the sequence \big{(}B_{j}^{*}\big{)}_{j\geq M} stabilizes. As we already explained, this is absurd. ∎

Recall that $A$ is an algebra over a field $\mathbf{k}$.

Lemma 2.13**.**

For every subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(A)$, $\operatorname{Nil}(\Delta)$ is closed under the multiplication of $A$.

Proof 2.14**.**

We may assume that $\Delta\neq\varnothing$. Let $x,y\in\operatorname{Nil}(\Delta)$ and $(D_{0},D_{1},\dots)\in\Delta^{\mathbb{N}}$. To prove the Lemma, it suffices to show that there exists $n\in\mathbb{N}$ such that $(D_{n}\circ\cdots\circ D_{0})(x\cdot y)=0$.

Define X_{x}=\big{\{}\,I\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N})\,\mid\,D_{I}(x)\neq 0\,\big{\}}. Let us prove:

1. (i)

for all I,J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) satisfying $I\preceq J$ and $J\in X_{x}$, we have $I\in X_{x}$; 2. (ii)

$(X_{x},\preceq)$* satisfies the ACC.*

Consider I,J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) satisfying $I\preceq J$ and $J\in X_{x}$. Then $0\neq D_{J}(x)=(D_{J\setminus I}\circ D_{I})(x)$, so $D_{I}(x)\neq 0$, so $I\in X_{x}$. This proves (i).

Consider an infinite sequence $I_{0}\preceq I_{1}\preceq\cdots$ in $X_{x}$. By contradiction, assume that \big{(}I_{n}\big{)}_{n\in\mathbb{N}} does not stabilize. Then $I=\bigcup_{n\in\mathbb{N}}I_{n}$ is an infinite subset of $\mathbb{N}$. Let $i_{0}<i_{1}<i_{2}<\cdots$ be the elements of $I$ and consider $(D_{i_{0}},D_{i_{1}},\dots)\in\Delta^{\mathbb{N}}$. Given any $k\in\mathbb{N}$ there exists $n\in\mathbb{N}$ such that $\{i_{0},i_{1},\dots,i_{k}\}\subseteq I_{n}$; then $\{i_{0},i_{1},\dots,i_{k}\}\preceq I_{n}\in X_{x}$, so $\{i_{0},i_{1},\dots,i_{k}\}\in X_{x}$ by (i), so $(D_{i_{k}}\circ\cdots\circ D_{i_{0}})(x)\neq 0$. Since this holds for all $k\in\mathbb{N}$, we have $x\notin\operatorname{Nil}(\Delta)$, a contradiction. So (ii) is proved.

Let X_{y}=\big{\{}\,I\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N})\,\mid\,D_{I}(y)\neq 0\,\big{\}}; then it is clear that (i) and (ii) are true with “$X_{y}$” in place of “$X_{x}$”. Consequently, if we define $X=X_{x}\cup X_{y}$ then:

1. (iii)

for all I,J\in\raisebox{0.7pt}{\large\wp}_{\!\mbox{\rm\tiny fin}}(\mathbb{N}) satisfying $I\preceq J$ and $J\in X$, we have $I\in X$; 2. (iv)

$(X,\preceq)$* satisfies the ACC.*

By (iii), (iv) and Lemma 2.11, the set

[TABLE]

is finite. Choose $n\in\mathbb{N}\setminus Z$. Then for each subset $I$ of $\mathbb{I}_{n}$ we have $I\notin X$ (i.e., $D_{I}(x)=0=D_{I}(y)$) or $\mathbb{I}_{n}\setminus I\notin X$ (i.e., $D_{\mathbb{I}_{n}\setminus I}(x)=0=D_{\mathbb{I}_{n}\setminus I}(y)$). So all terms are zero in the right-hand-side of

[TABLE]

(the equality follows from Lemma 2.9). Thus $(D_{n}\circ\cdots\circ D_{0})(x\cdot y)=0$, as desired. ∎

Lemma 2.15**.**

Let $\Delta$ be a subset of $\operatorname{Der}_{\mathbf{k}}(A)$. Then the map $\deg_{\Delta}:A\to\mathbb{N}\cup\{\pm\infty\}$ satisfies $\deg_{\Delta}(x\cdot y)\leq\deg_{\Delta}(x)+\deg_{\Delta}(y)$ for all $x,y\in\operatorname{UNil}(\Delta)$.

Proof 2.16**.**

The case $\Delta=\varnothing$ is trivial, so let us assume that $\Delta\neq\varnothing$. It suffices to show that $P(n)$ is true for all $n\in\mathbb{N}$, where we define

[TABLE]

Suppose that $x,y\in A\setminus\{0\}$ satisfy $\deg_{\Delta}(x)+\deg_{\Delta}(y)\leq 0$. Then $\deg_{\Delta}(x)=0=\deg_{\Delta}(y)$, so for all $D\in\Delta$ we have $D(x)=0=D(y)$ and hence $D(x\cdot y)=D(x)\cdot y+x\cdot D(y)=0$, so $\deg_{\Delta}(x\cdot y)\leq 0$, showing that $P(0)$ is true.

Let $n\in\mathbb{N}$ and assume that $P(n)$ is true. To prove that $P(n+1)$ is true, consider $x,y\in A\setminus\{0\}$ satisfying $\deg_{\Delta}(x)+\deg_{\Delta}(y)\leq n+1$. We have to show that $\deg_{\Delta}(x\cdot y)\leq n+1$. To do this, it suffices to show that $(D_{n+2}\circ\cdots\circ D_{1})(x\cdot y)=0$ for all $(D_{1},\dots,D_{n+2})\in\Delta^{n+2}$. So let us consider $(D_{1},\dots,D_{n+2})\in\Delta^{n+2}$. As a first step, we claim that

[TABLE]

To see this, first note that if $D_{1}(x)=0$ then $\deg_{\Delta}(D_{1}(x)\cdot y)=-\infty\leq n$; so, to prove the first part of (7), we may assume that $D_{1}(x)\neq 0$. Note that $\deg_{\Delta}(D_{1}(x))<\deg_{\Delta}(x)$ by part (d) of Lemma 2.3, so $\deg_{\Delta}(D_{1}(x))+\deg_{\Delta}(y)\leq n$; since both $D_{1}(x)$ and $y$ belong to $A\setminus\{0\}$ and $P(n)$ is true, we get $\deg_{\Delta}(D_{1}(x)\cdot y)\leq n$. By the same argument we have $\deg_{\Delta}(x\cdot D_{1}(y))\leq n$, so (7) is proved. Then

[TABLE]

by part (e) of Lemma 2.3 and (7). Now (8) implies that

[TABLE]

which is the desired conclusion. So $\deg_{\Delta}(x\cdot y)\leq n+1$ and hence $P(n+1)$ is true. ∎

Corollary 2.17**.**

Let $\Delta$ be a subset of $\operatorname{Der}_{\mathbf{k}}(A)$. Then $\operatorname{UNil}(\Delta)$ is closed under the multiplication of $A$.

Proof 2.18**.**

This follows from Lemma 2.15 and Part (c) of Lemma 2.3. ∎

This completes the proof of Thm 2.8.

3 Lie-locally nilpotent sets of linear maps

Let $V$ be a vector space over a field $\mathbf{k}$.

{caution}

Let $x\in V$ and $F_{0},\dots,F_{n+1}\in\mathcal{L}_{\mathbf{k}}(V)$. It is clear that $(F_{n}\circ\cdots\circ F_{0})(x)=0$ implies $(F_{n+1}\circ F_{n}\circ\cdots\circ F_{0})(x)=0$ and that $[F_{n},\dots,F_{0}]=0$ implies $[F_{n+1},F_{n},\dots,F_{0}]=0$, but it is not the case that $[F_{n},\dots,F_{0}](x)=0$ implies $[F_{n+1},F_{n},\dots,F_{0}](x)=0$. (This should be kept in mind while reading this Section.)

{Definition}

Given a subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$, we define the subsets $\operatorname{Nil}^{\text{\rm L}}(\Delta)$ and $\operatorname{UNil}^{\text{\rm L}}(\Delta)$ of $V$ as follows. If $\Delta=\varnothing$, we set $\operatorname{Nil}^{\text{\rm L}}(\Delta)=V=\operatorname{UNil}^{\text{\rm L}}(\Delta)$. If $\Delta\neq\varnothing$,

• $\bullet$

$\operatorname{Nil}^{\text{\rm L}}(\Delta)$ is the set of all $x\in V$ satisfying:

for each $(F_{0},F_{1},\dots)\in\Delta^{\mathbb{N}}$ there exists $N\in\mathbb{N}$ such that for all $m\geq 0$ and $(G_{1},\dots,G_{m})\in\Delta^{m}$ we have $[G_{m},\dots,G_{1},F_{N},\dots,F_{0}](x)=0$.111We compute $[G_{m},\dots,G_{1},F_{N},\dots,F_{0}]$ in the Lie algebra $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$ (see 1 and 1).

• $\bullet$

$\operatorname{UNil}^{\text{\rm L}}(\Delta)$ is the set of all $x\in V$ satisfying:

there exists $N\in\mathbb{N}$ such that for every $n\geq N$ and $(F_{1},\dots,F_{n})\in\Delta^{n}$ we have $[F_{n},\dots,F_{1}](x)=0$.

Note that $\operatorname{UNil}^{\text{\rm L}}(\Delta)\subseteq\operatorname{Nil}^{\text{\rm L}}(\Delta)$ are linear subspaces of $V$. If $\operatorname{Nil}^{\text{\rm L}}(\Delta)=V$, we say that the set $\Delta$ is Lie-locally nilpotent; if $\operatorname{UNil}^{\text{\rm L}}(\Delta)=V$, we say that $\Delta$ is uniformly Lie-locally nilpotent.

{Remark}

The condition that $\Delta$ is a Lie-locally nilpotent (or a uniformly Lie-locally nilpotent) subset of $\mathcal{L}_{\mathbf{k}}(V)$ does not imply that $\Delta\subseteq\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$. For instance, if $F\in\mathcal{L}_{\mathbf{k}}(V)\setminus\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$ then $\Delta=\{F\}$ is Lie-locally nilpotent.

{Example}

In general there are no inclusion relations between $\operatorname{Nil}(\Delta)$ and $\operatorname{Nil}^{\text{\rm L}}(\Delta)$. The $\Delta$ of 3 satisfies $\operatorname{Nil}(\Delta)\nsupseteq\operatorname{Nil}^{\text{\rm L}}(\Delta)$. We now give an example satisfying $\operatorname{Nil}(\Delta)\nsubseteq\operatorname{Nil}^{\text{\rm L}}(\Delta)$.

Let $(e_{i})_{i\in\mathbb{N}}$ be a basis of a vector space $V$ over a field $\mathbf{k}$. Let $\Delta=\{F_{1},F_{2},F_{3},\dots\}$ where for each $n\geq 1$ we define the $\mathbf{k}$-linear map $F_{n}:V\to V$ by

[TABLE]

Then we have:

[TABLE]

We claim that $\operatorname{Nil}(\Delta)=V$. Indeed, consider a sequence $(F_{i_{0}},F_{i_{1}},F_{i_{2}},\dots)\in\Delta^{\mathbb{N}}$. Then $(i_{0},i_{1},i_{2},\dots)$ cannot be strictly decreasing, so there must exist $n\geq 1$ such that $i_{n-1}\leq i_{n}$. Then $F_{i_{n}}\circ F_{i_{n-1}}=0$ by (9), so $F_{i_{n}}\circ\cdots\circ F_{i_{0}}=0$. This implies in particular that $\operatorname{Nil}(\Delta)=V$.

Now consider the sequence $(F_{1},F_{2},F_{3},\dots)\in\Delta^{\mathbb{N}}$. Using (9), it is easy to show by induction that $[F_{n},\dots,F_{1}]=(-1)^{n+1}F_{1}\circ F_{2}\circ\cdots F_{n}$ for all $n\geq 1$. Since $(F_{1}\circ F_{2}\circ\cdots F_{n})(e_{0})=e_{1}$ for all $n\geq 1$, we have $[F_{n},\dots,F_{1}](e_{0})\neq 0$ for all $n\geq 1$, so $e_{0}\notin\operatorname{Nil}^{\text{\rm L}}(\Delta)$. So $\operatorname{Nil}(\Delta)\nsubseteq\operatorname{Nil}^{\text{\rm L}}(\Delta)$.

In contrast with Ex. 3, we have:

Lemma 3.1**.**

For every subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$, we have $\operatorname{UNil}(\Delta)\subseteq\operatorname{UNil}^{\text{\rm L}}(\Delta)$.

Proof 3.2**.**

Let $x\in\operatorname{UNil}(\Delta)$. Then there exists $N\geq 1$ such that $(F_{N}\circ\cdots\circ F_{1})(x)=0$ for all $(F_{1},\dots,F_{N})\in\Delta^{N}$. So,

[TABLE]

Now given any $n\geq N$ and $(G_{1},\dots,G_{n})\in\Delta^{n}$ we have

[TABLE]

so $[G_{n},\dots,G_{1}](x)=0$ by (10). This shows that $x\in\operatorname{UNil}^{\text{\rm L}}(\Delta)$, as desired. ∎

Lemma 3.3**.**

Let $\Delta$ be a finite subset of $\mathcal{L}_{\mathbf{k}}(V)$. Then $\operatorname{Nil}^{\text{\rm L}}(\Delta)=\operatorname{UNil}^{\text{\rm L}}(\Delta)$. In particular, $\Delta$ is Lie-locally nilpotent if and only if it is uniformly Lie-locally nilpotent.

Proof 3.4**.**

It suffices to show that $\operatorname{Nil}^{\text{\rm L}}(\Delta)\subseteq\operatorname{UNil}^{\text{\rm L}}(\Delta)$. Let $x\in\operatorname{Nil}^{\text{\rm L}}(\Delta)\setminus\{0\}$. Consider the set $S$ of finite sequences $(F_{0},F_{1}\dots,F_{N})$ of elements of $\Delta$ for which there exist $m\geq 0$ and $(G_{1},\dots,G_{m})\in\Delta^{m}$ satisfying $[G_{m},\dots,G_{1},F_{N},\dots,F_{0}](x)\neq 0$. Let $E\subseteq S\times S$ be the set of ordered pairs of the form

[TABLE]

Then $(S,E)$ is a directed tree and the empty sequence $()\in S$ is the root of this tree (see 1). For each vertex $v\in S$, the number of $v^{\prime}\in S$ satisfying $(v,v^{\prime})\in E$ is at most $|\Delta|$, which is finite. The fact that $x\in\operatorname{Nil}^{\text{\rm L}}(\Delta)$ implies that there is no infinite path \big{(}(),(F_{0}),(F_{0},F_{1}),\dots\big{)} in this tree. It follows from Thm 1.1 that $S$ is a finite set, and this implies that $x\in\operatorname{UNil}^{\text{\rm L}}(\Delta)$. ∎

Proposition 3.5**.**

Let $\Delta$ be a subset of $\mathcal{L}_{\mathbf{k}}(V)$ and let $\tilde{\Delta}$ be the Lie subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$ generated by $\Delta$. Then the following hold.

1. (a)

Let $x\in V$. If $N$ is a positive integer satisfying

[TABLE]

then the same $N$ satisfies

[TABLE] 2. (b)

$\operatorname{UNil}^{\text{\rm L}}(\tilde{\Delta})=\operatorname{UNil}^{\text{\rm L}}(\Delta)$** 3. (c)

If $\Delta$ is uniformly Lie-locally nilpotent then so is $\tilde{\Delta}$.

Proof 3.6**.**

(a) Let $x\in V$ and let $N$ be a positive integer satisfying (11). Let $n\geq N$. To prove (a), it suffices to show that

[TABLE]

Consider the set \Delta_{\square}=\big{\{}\,[F_{k},\dots,F_{1}]\,\mid\,k\geq 1\text{ and }(F_{1},\dots,F_{k})\in\Delta^{k}\,\big{\}} and note that $\Delta\subseteq\Delta_{\square}\subseteq\tilde{\Delta}$ and (by Lemma 1.2(b)) $\tilde{\Delta}=\operatorname{Span}_{\mathbf{k}}(\Delta_{\square})$. The first step in the proof of (12) consists in proving

[TABLE]

Let $(G_{1},\dots,G_{n})\in(\Delta_{\square})^{n}$. For each $i\in\{1,\dots,n\}$, write

[TABLE]

where $k_{i}\geq 1$ and $F_{i,j}\in\Delta$. Then Lemma 1.2(a) implies that $[G_{n},\dots,G_{1}]=\sum_{i}\lambda_{i}[W_{i,k},\dots,W_{i,1}]$ where $k=k_{1}+\cdots+k_{n}\geq n$ and (for each $i$) $\lambda_{i}\in\mathbf{k}$ and $(W_{i,1},\dots,W_{i,k})$ is a permutation of $(F_{1,1},\dots,F_{1,k_{1}},F_{2,1},\dots,F_{n,k_{n}})\in\Delta^{k}$. Since $k\geq n\geq N$, (11) implies that $[W_{i,k},\dots,W_{i,1}](x)=0$ for all $i$, so $[G_{n},\dots,G_{1}](x)=0$, proving (13).

Now let us prove (12). Let $(H_{1},\dots,H_{n})\in\tilde{\Delta}^{n}$. For each $i$ we have $H_{i}=\sum_{j}\lambda_{i,j}G_{i,j}$ with $\lambda_{i,j}\in\mathbf{k}$ and $G_{i,j}\in\Delta_{\square}$. Thus $[H_{n},\dots,H_{1}]$ is a linear combination of terms of the form $[G_{n,j_{n}},\dots,G_{1,j_{1}}]$. By (13), $[G_{n,j_{n}},\dots,G_{1,j_{1}}](x)=0$ for every choice of $(j_{1},\dots,j_{n})$, so $[H_{n},\dots,H_{1}](x)=0$. This proves (12), so (a) is proved. Assertions (b) and (c) follow from (a). ∎

Theorem 3.7**.**

Let $(A,\cdot)$ be an algebra over a field $\mathbf{k}$. If $\Delta$ is a subset of $\operatorname{Der}_{\mathbf{k}}(A)$ then the sets $\operatorname{Nil}^{\text{\rm L}}(\Delta)$ and $\operatorname{UNil}^{\text{\rm L}}(\Delta)$ are subalgebras of $(A,\cdot)$.

Proof 3.8**.**

Consider a sequence $\mathbf{D}=(D_{0},D_{1},\dots)\in\Delta^{\mathbb{N}}$. For each $N\in\mathbb{N}$, let $K_{N}(\mathbf{D})$ be the intersection of the kernels of the derivations $[G_{m},\dots,G_{1},D_{N},\dots,D_{0}]$, for all $m\geq 0$ and $(G_{1},\dots,G_{m})\in\Delta^{m}$. Since $K_{0}(\mathbf{D})\subseteq K_{1}(\mathbf{D})\subseteq K_{2}(\mathbf{D})\subseteq\cdots$ are subalgebras of $A$, so is $K(\mathbf{D})=\bigcup_{N=0}^{\infty}K_{N}(\mathbf{D})$. Then $\operatorname{Nil}^{\text{\rm L}}(\Delta)=\bigcap_{\mathbf{D}\in\Delta^{\mathbb{N}}}K(\mathbf{D})$ is a subalgebra of $A$.

For each $N\in\mathbb{N}$, let $K_{N}$ be the intersection of the $\ker[D_{n},\dots,D_{1}]$ for all $n\geq N$ and $(D_{1},\dots,D_{n})\in\Delta^{n}$. Then $K_{0}\subseteq K_{1}\subseteq K_{2}\subseteq\cdots$ is a chain of subalgebras of $A$, so $\operatorname{UNil}^{\text{\rm L}}(\Delta)=\bigcup_{N=0}^{\infty}K_{N}$ is a subalgebra of $A$. ∎

4 Nilpotency conditions for algebras

Throughout paragraphs 4–4.3, $(A,\cdot)$ is an algebra over a field $\mathbf{k}$ in the sense of 1(a). Starting in 4, we restrict our attention to the special case where $A$ is either an associative or a Lie algebra.

{Definition}

Given a subset $H$ of $A$, let us define the subsets $\operatorname{Nil}^{\prime}(H)$ and $\operatorname{UNil}^{\prime}(H)$ of $A$ as follows. If $H=\varnothing$, we set $\operatorname{Nil}^{\prime}(H)=A=\operatorname{UNil}^{\prime}(H)$. If $H\neq\varnothing$,

• $\bullet$

$\operatorname{Nil}^{\prime}(H)$ is the set of all $x\in A$ satisfying:

for every sequence $(a_{0},a_{1},\dots)\in H^{\mathbb{N}}$ there exists $n\in\mathbb{N}$ such that $a_{n}\cdots a_{0}\cdot x=0$;222Recall that $a_{n}\cdots a_{0}\cdot x=0$ means $a_{n}\cdot(a_{n-1}\cdots(a_{1}\cdot(a_{0}\cdot x))\dots)=0$, by the right-associativity convention (1).

• $\bullet$

$\operatorname{UNil}^{\prime}(H)$ is the set of all $x\in A$ satisfying:

there exists $n\in\mathbb{N}$ such that for every $(a_{1},\dots,a_{n})\in H^{n}$ we have $a_{n}\cdots a_{1}\cdot x=0$.

Define $\deg^{\prime}_{H}:A\to\mathbb{N}\cup\{\pm\infty\}$ by declaring that $\deg^{\prime}_{H}(0)=-\infty$, $\deg^{\prime}_{H}(x)=\infty$ for all $x\in A\setminus\operatorname{UNil}^{\prime}(H)$, and if $x\in\operatorname{UNil}^{\prime}(H)\setminus\{0\}$ then $\deg^{\prime}_{H}(x)$ is

[TABLE]

If $H=\varnothing$ then we set $\deg^{\prime}_{H}(0)=-\infty$ and $\deg^{\prime}_{H}(x)=0$ for all $x\in A\setminus\{0\}$.

Let us now connect these notions to the formalism of Section 2.

{nothing*}

Define the map $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)$ by stipulating that, given $a\in A$, $\varphi(a):A\to A$ is the $\mathbf{k}$-linear map $x\mapsto a\cdot x$. Note that $\varphi$ is a $\mathbf{k}$-linear map, but not necessarily a homomorphism of algebras. If $H$ is a subset of $A$ then $\varphi(H)$ is a subset of $\mathcal{L}_{\mathbf{k}}(A)$, so we may consider (as in Sec. 2) the subsets $\operatorname{Nil}(\varphi(H))$ and $\operatorname{UNil}(\varphi(H))$ of $A$ and the map $\deg_{\varphi(H)}:A\to\mathbb{N}\cup\{\pm\infty\}$. Then we have

[TABLE]

Indeed, the right-associativity convention implies that a_{n}\cdots a_{0}\cdot x=\big{(}\varphi(a_{n})\circ\cdots\circ\varphi(a_{0})\big{)}(x) for all $x,a_{0},\dots,a_{n}\in A$, and (14) immediately follows.

Corollary 4.1**.**

If $H$ is a finite subset of $A$ then $\operatorname{Nil}^{\prime}(H)=\operatorname{UNil}^{\prime}(H)$.

Proof 4.2**.**

Apply Lemma 2.1 to the finite set $\varphi(H)$ and use (14). ∎

Lemma 4.3**.**

The sets $\operatorname{Nil}^{\prime}(A)$ and $\operatorname{UNil}^{\prime}(A)$ are left ideals of $A$. Moreover, for each $n\in\mathbb{N}$ the set Z_{n}=\big{\{}\,x\in A\,\mid\,\deg^{\prime}_{A}(x)<n\,\big{\}} is a left ideal of $A$.

Proof 4.4**.**

We have $\{0\}=Z_{0}\subseteq Z_{1}\subseteq Z_{2}\subseteq\cdots$ and $\operatorname{UNil}^{\prime}(A)=\bigcup_{n=0}^{\infty}Z_{n}$; so, to prove the Lemma, it suffices to show that $\operatorname{Nil}^{\prime}(A)$ and all $Z_{n}$ are left ideals of $A$. It is clear that $\operatorname{Nil}^{\prime}(A)$ and $Z_{n}$ are $\mathbf{k}$-subspaces of $A$.

Let $x\in A$ and $y\in\operatorname{Nil}^{\prime}(A)$. To prove that $x\cdot y\in\operatorname{Nil}^{\prime}(A)$, we have to show that for every $(x_{0},x_{1},\dots)\in A^{\mathbb{N}}$ there exist an $n$ such that $x_{n}\cdots x_{0}\cdot(x\cdot y)=0$. Note that $x_{n}\cdots x_{0}\cdot(x\cdot y)=x_{n}\cdots x_{0}\cdot x\cdot y$ by the right-associativity convention. So the claim is true, because $(x,x_{0},x_{1},\dots)\in A^{\mathbb{N}}$ and $y\in\operatorname{Nil}^{\prime}(A)$. So $x\cdot y\in\operatorname{Nil}^{\prime}(A)$, showing that $\operatorname{Nil}^{\prime}(A)$ is a left ideal of $A$.

Note that $Z_{0}=\{0\}$ is indeed a left ideal of $A$. Let $n>0$, $x\in A$ and $y\in Z_{n}$. Then $x_{n}\cdots x_{1}\cdot y=0$ for all $(x_{1},\dots,x_{n})\in A^{n}$. In particular, $x_{n}\cdots x_{2}\cdot x\cdot y=0$ for all $(x_{2},\dots,x_{n})\in A^{n-1}$. Since $x_{n}\cdots x_{2}\cdot(x\cdot y)=x_{n}\cdots x_{2}\cdot x\cdot y=0$, this shows that $x\cdot y\in Z_{n-1}\subseteq Z_{n}$. So $Z_{n}$ is a left ideal of $A$. ∎

Cor. 4.1 and Lemma 4.3 are valid for all algebras $A$. We can say more if we assume that $A$ is either associative or Lie.

{observation}

Let $A$ be a $\mathbf{k}$-algebra and consider the map $\varphi$ of 4.

1. (a)

If $A$ is an associative algebra then $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)$ is a homomorphism of associative algebras. 2. (b)

If $A$ is a Lie algebra then $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}$ is a homomorphism of Lie algebras and $\varphi(A)\subseteq\operatorname{Der}_{\mathbf{k}}(A)$.

Proof 4.5**.**

Part (a) is clear and (b) is the same as 1(i,ii) (we have $\varphi=\operatorname{ad}$). ∎

Corollary 4.6**.**

Let $(A,\cdot)$ be either an associative algebra or a Lie algebra, let $H$ be a subset of $A$ and let $\hat{H}$ be the subalgebra of $(A,\cdot)$ generated by $H$.

1. (a)

$\operatorname{Nil}^{\prime}(H)=\operatorname{Nil}^{\prime}(\hat{H})$, $\operatorname{UNil}^{\prime}(H)=\operatorname{UNil}^{\prime}(\hat{H})$ and $\deg^{\prime}_{H}=\deg^{\prime}_{\hat{H}}$. 2. (b)

If $(A,\cdot)$ is a Lie algebra then $\operatorname{Nil}^{\prime}(H)$ and $\operatorname{UNil}^{\prime}(H)$ are Lie subalgebras of $A$ and $\deg^{\prime}_{H}(x\cdot y)\leq\deg^{\prime}_{H}(x)+\deg^{\prime}_{H}(y)$ for all $x,y\in\operatorname{UNil}^{\prime}(H)$.

Proof 4.7**.**

(a) Let $\mathbf{k}$ be the field over which $A$ is an algebra. Consider the algebra $(\mathcal{L}_{\mathbf{k}}(A),\star)$, where for $F,G\in\mathcal{L}_{\mathbf{k}}(A)$ we define $F\star G=F\circ G$ (resp. $F\star G=F\circ G-G\circ F$) if we are proving the case where $A$ is associative (resp. $A$ is Lie). Let $\widehat{\varphi(H)}$ be the subalgebra of $(\mathcal{L}_{\mathbf{k}}(A),\star)$ generated by $\varphi(H)$. Since $\varphi:(A,\cdot)\to(\mathcal{L}_{\mathbf{k}}(A),\star)$ is a homomorphism of algebras by Obs. 4, we have $\widehat{\varphi(H)}=\varphi(\hat{H})$, so $\operatorname{Nil}^{\prime}(H)=\operatorname{Nil}(\varphi(H))=\operatorname{Nil}(\widehat{\varphi(H)})=\operatorname{Nil}(\varphi(\hat{H}))=\operatorname{Nil}^{\prime}(\hat{H})$, where the second equality is Prop. 2.4 and the first and last equalities are (14). It is clear that $\operatorname{UNil}^{\prime}(H)=\operatorname{UNil}^{\prime}(\hat{H})$ and $\deg^{\prime}_{H}=\deg^{\prime}_{\hat{H}}$ follow by the same argument, so (a) is proved.

(b) If $(A,\cdot)$ is a Lie algebra then $\varphi(H)\subseteq\operatorname{Der}_{\mathbf{k}}(A,\cdot)$ by Obs. 4, so Thm 2.8 implies that $\operatorname{Nil}(\varphi(H))$ and $\operatorname{UNil}(\varphi(H))$ are Lie subalgebras of $(A,\cdot)$ and that $\deg_{\varphi(H)}(x\cdot y)\leq\deg_{\varphi(H)}(x)+\deg_{\varphi(H)}(y)$ for all $x,y\in\operatorname{UNil}(\varphi(H))$. In view of (14), this is the desired conclusion. ∎

{notation}

Let $A$ be either an associative algebra or a Lie algebra. Given a subset $H$ of $A$, define s(H)=\sup\big{\{}\,\deg^{\prime}_{H}(x)\,\mid\,x\in H\,\big{\}}\in\mathbb{N}\cup\{\pm\infty\}, where we agree that $s(\varnothing)=-\infty$. Note that if $H\subseteq\{0\}$ then $s(H)=-\infty$, and if $H\nsubseteq\{0\}$ then s(H)=\sup\big{\{}\,n\in\mathbb{N}\,\mid\,x_{n}\cdots x_{0}\neq 0\text{ for some (x_{0},\dots,x_{n})\in H^{n+1}}\,\big{\}}.

Lemma 4.8**.**

Let $A$ be either an associative algebra or a Lie algebra, let $H$ be a subset of $A$ and let $\hat{H}$ be the subalgebra of $A$ generated by $H$.

1. (a)

If $H\subseteq\operatorname{Nil}^{\prime}(H)$ then $\hat{H}\subseteq\operatorname{Nil}^{\prime}(\hat{H})$. 2. (b)

If $H\subseteq\operatorname{UNil}^{\prime}(H)$ then $\hat{H}\subseteq\operatorname{UNil}^{\prime}(\hat{H})$. 3. (c)

$s(H)=s(\hat{H})$**

Proof 4.9**.**

(a) Assume that $H\subseteq\operatorname{Nil}^{\prime}(H)$. Cor. 4.6 gives $\operatorname{Nil}^{\prime}(H)=\operatorname{Nil}^{\prime}(\hat{H})$, so $H\subseteq\operatorname{Nil}^{\prime}(\hat{H})$ and hence $H\subseteq\hat{H}\cap\operatorname{Nil}^{\prime}(\hat{H})$. Applying Lemma 4.3 to the algebra $\hat{H}$ shows that $\hat{H}\cap\operatorname{Nil}^{\prime}(\hat{H})$ is a left ideal of $\hat{H}$ and hence a subalgebra of $\hat{H}$; since that subalgebra contains $H$, it must then contain $\hat{H}$, so $\hat{H}\subseteq\operatorname{Nil}^{\prime}(\hat{H})$. This proves (a). We obtain a proof of (b) by replacing $\operatorname{Nil}^{\prime}$ by $\operatorname{UNil}^{\prime}$ everywhere.

(c) It is clear that $s(H)\leq s(\hat{H})$ and that equality holds whenever $s(H)\notin\mathbb{N}$. Assume that $s(H)=n\in\mathbb{N}$. Since $\deg^{\prime}_{H}=\deg^{\prime}_{\hat{H}}$ by Cor. 4.6, we have $\deg^{\prime}_{\hat{H}}(x)\leq n$ for all $x\in H$. Thus $H\subseteq Z_{n+1}$, where we define Z_{j}=\big{\{}\,x\in\hat{H}\,\mid\,\deg_{\hat{H}}^{\prime}(x)<j\,\big{\}}. Lemma 4.3 implies that $Z_{n+1}$ is a left ideal (and hence a subalgebra) of $\hat{H}$, so $\hat{H}\subseteq Z_{n+1}$. It follows that $s(\hat{H})\leq n$, so $s(\hat{H})=s(H)$. ∎

Corollary 4.10**.**

Let $A$ be either an associative algebra or a Lie algebra and let $H$ be a generating set for $A$. Then the following hold.

1. (a)

If for every sequence $(a_{0},a_{1},\dots)\in H^{\mathbb{N}}$ there exists an $n$ such that $a_{n}\cdots a_{0}=0$, then for every sequence $(a_{0},a_{1},\dots)\in A^{\mathbb{N}}$ there exists an $n$ such that $a_{n}\cdots a_{0}=0$. 2. (b)

Suppose that $n$ is a positive integer such that for every $(a_{1},\dots,a_{n})\in H^{n}$ we have $a_{n}\cdots a_{1}=0$. Then we have $a_{n}\cdots a_{1}=0$ for all $(a_{1},\dots,a_{n})\in A^{n}$.

Proof 4.11**.**

We have $\hat{H}=A$ by assumption, where $\hat{H}$ denotes the subalgebra of $A$ generated by $H$. Part (a) asserts that $H\subseteq\operatorname{Nil}^{\prime}(H)\Rightarrow\operatorname{Nil}^{\prime}(A)=A$; since $\hat{H}=A$, the claim follows from Lemma 4.8(a). Part (b) asserts that if $n$ satisfies $s(H)<n-1$, then $s(A)<n-1$; Lemma 4.8(c) implies that $s(H)=s(\hat{H})=s(A)$, so the claim is true. ∎

{Definition}

Let $(A,\cdot)$ be either an associative algebra or a Lie algebra. Consider the map $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)$ of 4. We say that

• $\bullet$

$A$ is nilpotent (N) if there exists an $n\geq 1$ such that for all $(x_{1},\dots,x_{n})\in A^{n}$ we have $x_{n}\cdots x_{1}=0$;

• $\bullet$

$A$ is sequentially nilpotent (SN) if for every infinite sequence $(x_{0},x_{1},\dots)$ of elements of $A$, there exists an $n\geq 0$ such that $x_{n}\cdots x_{0}=0$;

• $\bullet$

$A$ is locally nilpotent (LN) if every finitely generated subalgebra of $A$ is nilpotent;

• $\bullet$

$A$ is nil (nil) if for each $x\in A$ the map $\varphi(x):A\to A$ is nilpotent;

• $\bullet$

$A$ is locally nil (Lnil) if for each $x\in A$ the map $\varphi(x):A\to A$ is locally nilpotent.

{Remark}

We found nothing in the literature about (SN) or (Lnil); as far as we know, these two conditions have not been considered previously. The definitions of (N), (LN) and (nil) given in Def. 4 are compatible with standard usage of terminology. Regarding (nil), note the following.

1. (1)

Prop. 4.12(d) shows that an associative algebra $A$ satisfies (nil) if and only if every element of $A$ is nilpotent (which is the standard meaning of “nil” for associative algebras). 2. (2)

Our definition of “nil” for Lie algebras is the standard one. Note that a Lie algebra that is nil is also said to be Engel. However, an associative algebra $A$ is said to be Engel if the Lie algebra $A_{\text{\rm L}}$ is Engel. So “Engel” and “nil” have the same meaning for Lie algebras but not for associative algebras.

{Remark}

It follows from Lemma 1.2 that if an associative or Lie algebra is both finitely generated and nilpotent, then it is finite dimensional (as a vector space over $\mathbf{k}$). Consequently, if $A$ is (LN) then every finitely generated subalgebra of $A$ is nilpotent and finite dimensional.

Proposition 4.12**.**

Let $A$ be either an associative algebra or a Lie algebra.

1. (a)

*We have the following implications for $A$: ** [TABLE] (SN)(LN)(N)(nil)(Lnil) .

---

2. (b)

If $A$ is finitely generated then (N) $\Leftrightarrow$ (SN) $\Leftrightarrow$ (LN). 3. (c)

If $A$ is finite dimensional then (LN) $\Leftrightarrow$ (SN) $\Leftrightarrow$ (N) $\Leftrightarrow$ (nil) $\Leftrightarrow$ (Lnil). 4. (d)

If $A$ is associative then (SN) $\Rightarrow$ (nil) and moreover

[TABLE]

Proof 4.13**.**

(a), (SN) $\Rightarrow$ (LN). Assume that $A$ satisfies (SN) and let $H$ be a finite subset of $A$; we have to show that $\hat{H}$ is nilpotent, where $\hat{H}$ denotes the subalgebra of $A$ generated by $H$. The assumption that $A$ satisfies (SN) implies that $H\subseteq\operatorname{Nil}^{\prime}(H)$. Since $H$ is finite we have $\operatorname{UNil}^{\prime}(H)=\operatorname{Nil}^{\prime}(H)$ by Cor. 4.1, so $H\subseteq\operatorname{UNil}^{\prime}(H)$ and consequently $\deg_{H}^{\prime}(x)<\infty$ for each $x\in H$. Since $H$ is finite, it follows that $s(H)<\infty$. Then Lemma 4.8 implies that $s(H)=s(\hat{H})$, so $s(\hat{H})<\infty$, so $\hat{H}$ is nilpotent. So $A$ satisfies (LN).

(LN) $\Rightarrow$ (Lnil). Suppose that $A$ is (LN). To show that $A$ is (Lnil), it suffices to show that for all $x,y\in A$ there exists an $n>0$ such that $\varphi(x)^{n}(y)=0$. Consider $x,y\in A$. As $A$ is (LN), the subalgebra $A_{0}$ of $A$ generated by $\{x,y\}$ is nilpotent, so there exists $n>0$ such that $x_{n}\cdots x_{0}=0$ for all $(x_{0},\dots,x_{n})\in A_{0}^{n+1}$, i.e., \big{(}\varphi(x_{n})\circ\cdots\circ\varphi(x_{1})\big{)}(x_{0})=0 for all $(x_{0},\dots,x_{n})\in A_{0}^{n+1}$. As $x,y\in A_{0}$ it follows that $\varphi(x)^{n}(y)=0$. So $A$ is (Lnil).

The implications (Lnil) $\Leftarrow$ (nil) $\Leftarrow$ (N) $\Rightarrow$ (SN) are clear, so (a) is proved.

If $A$ is finitely generated then (LN) $\Rightarrow$ (N) is clear, by definition of (LN); so (b) is clear.

Let us prove (d) before (c). Suppose that $A$ is associative.

If $A$ satisfies (Lnil) then, for every choice of $x,y\in A$, there exists $n$ such that $\varphi(x)^{n}(y)=0$; so for every $x\in A$ there exists $n$ such that $x^{n+1}=\varphi(x)^{n}(x)=0$. This shows that (Lnil) implies that every element of $A$ is nilpotent.

Since $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)$ is a homomorphism of associative algebras, it is clear that if every element of $A$ is nilpotent then every element of $\varphi(A)$ is nilpotent, i.e., $A$ satisfies (nil).

This proves the equivalences in (d). Now suppose that $A$ satisfies (SN). Given any $x\in A$, applying the property (SN) to the sequence $(x,x,x,\dots)\in A^{\mathbb{N}}$ implies that there exists $n$ such that $x^{n}=0$. So (SN) implies that every element of $A$ is nilpotent, and the proof of (d) is complete.

(c) By (a), it suffices to show that (Lnil) $\Rightarrow$ (nil) $\Rightarrow$ (N) when $A$ is finite dimensional.

If $A$ is finite dimensional then every locally nilpotent linear map $A\to A$ is in fact nilpotent, so (Lnil) implies (nil).

For finite dimensional Lie algebras, the fact that (nil) implies (N) is Engel’s theorem on abstract Lie algebras, see page 36 of [Jac79].

Any finite dimensional associative $\mathbf{k}$-algebra $A$ is (isomorphic to) a subalgebra of a matrix algebra $M_{n}(\mathbf{k})$, for some $n$. If $A$ satisfies (nil), then part (d) implies that every element of $A$ is a nilpotent matrix. It is then well known that $A$ is a nilpotent algebra (see for instance Thm 1, page 33 of [Jac79]). So (nil) implies (N) in the associative case as well, and this proves (c). ∎

{nothing*}

By Ex. 6, (SN) does not imply (nil) for Lie algebras.

{nothing}

It is known that (nil) does not imply (LN). Indeed, the following statements are valid over an arbitrary field and are consequences of more precise results due to Golod ([Gol64], [Gol68], cited in [Kos90, p. 6] and [Row88, Thm 6.2.9]):

1. (a)

For each integer $m\geq 2$, there exists an infinite dimensional associative algebra $A$ generated by $m$ elements such that every $(m-1)$-subset of $A$ is nilpotent.333A subset $H$ of an associative algebra $A$ is nilpotent if there exists $n$ such that $h_{n}\cdots h_{1}=0$ for all $(h_{1},\dots,h_{n})\in H^{n}$. By Cor. 4.10, $H$ is nilpotent if and only if $\bar{H}$ is nilpotent, where $\bar{H}$ is the subalgebra of $A$ generated by $H$. Moreover, $\bigcap_{n=1}^{\infty}A^{n}=\{0\}$. 2. (b)

There exists a $3$-generated Lie algebra that satisfies (nil) but not (N).

In (a), the fact that every $(m-1)$-subset of $A$ is nilpotent implies that $A$ is nil and the fact that $A$ is finitely generated and infinite dimensional implies that $A$ is not nilpotent (Rem. 4).

{nothing*}

If (P) $\in\{\text{(N)},\text{(SN)},\text{(LN)},\text{(nil)},\text{(Lnil)}\}$ and $A$ is an associative algebra satisfying (P), does it follow that $A_{\text{\rm L}}$ satisfies (P)? Example 4 shows that there exists an associative algebra $A$ such that $A$ satisfies (SN) but $A_{L}$ doesn’t. The following Lemma gives positive results.

Lemma 4.14**.**

Let $A$ be an associative algebra.

1. (a)

Let (P)$\in\{\text{\rm(N)},\text{\rm(nil)},\text{\rm(LN)},\text{\rm(Lnil)}\}$. If $A$ satisfies (P), then $A_{L}$ satisfies (P). 2. (b)

If $A$ satisfies (SN), then $A_{L}$ satisfies (LN).

Proof 4.15**.**

(a) (i) Define A_{n}=\operatorname{Span}_{\mathbf{k}}\big{\{}\,a_{n}\cdots a_{1}\,\mid\,(a_{1},\dots,a_{n})\in A^{n}\,\big{\}}. If $A$ satisfies (N) then $A_{n}=0$ for $n$ large enough; since $[a_{n},\dots,a_{1}]\in A_{n}$ for all $(a_{1},\dots,a_{n})\in A^{n}$, it follows that $A_{L}$ satisfies (N).

(ii) Suppose that $A$ satisfies (Lnil). Then Prop. 4.12(d) implies that each element of $A$ is nilpotent. We claim that $A_{L}$ satisfies (nil). Indeed, let $x\in A_{L}$; to prove the assertion, we have to show that $\operatorname{ad}(x):A_{L}\to A_{L}$ is nilpotent. There exists $m$ such that $x^{m}=0$ in $A$. For any $y\in A_{L}$,

[TABLE]

As $x^{i}yx^{j}=0$ for all $i,j$ satisfying $i+j=2m-1$, we have \big{(}\operatorname{ad}(x)^{2m-1}\big{)}(y)=0 for all $y\in A_{L}$. So $\operatorname{ad}(x)^{2m-1}=0$ and consequently $A_{L}$ satisfies (nil).

(iii) Suppose that $A$ satisfies (LN). Consider a finite subset $H$ of $A_{L}$; to show that $A_{L}$ is (LN), it suffices to show that $\tilde{H}$ is nilpotent, where $\tilde{H}$ denotes the subalgebra of $A_{L}$ generated by $H$. Note that $H\subseteq A$ (since $A=A_{L}$ as sets) and consider the subalgebra $\bar{H}$ of $A$ generated by $H$. Then $\bar{H}$ satisfies (N), so part (i) of the proof shows that $(\bar{H})_{L}$ satisfies (N). Since $\tilde{H}\subseteq\bar{H}$, $\tilde{H}$ is a subalgebra of the nilpotent Lie algebra $(\bar{H})_{L}$; so $\tilde{H}$ is nilpotent, which shows that $A_{L}$ satisfies (LN). This proves (a).

(b) Since $A$ satisfies (SN), it satisfies (LN) by Prop. 4.12, so $A_{L}$ satisfies (LN) by part (a). ∎

{Example}

Let $V$ be a vector space of dimension $|\mathbb{N}|$ over a field $\mathbf{k}$. Then there exists an associative subalgebra $A$ of $\mathcal{L}_{\mathbf{k}}(V)$ satisfying:

1. (a)

$A$ is (SN) but the Lie algebra $A_{L}$ is not (SN). 2. (b)

$A$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$.

Proof 4.16**.**

Let $(e_{i})_{i\in\mathbb{N}}$ be a basis of $V$, let I=\big{\{}\,(i,j)\in\mathbb{N}^{2}\,\mid\,i\geq j\,\big{\}} and, for each $(i,j)\in I$, define the linear map $T_{i,j}:V\to V$ by

[TABLE]

Let A=\operatorname{Span}_{\mathbf{k}}\big{\{}\,T_{i,j}\,\mid\,(i,j)\in I\,\big{\}}\subseteq\mathcal{L}_{\mathbf{k}}(V). Note that if $(i_{1},j_{1}),(i_{2},j_{2})\in I$ then

1. $(\alpha)$

if $j_{1}\leq i_{2}$ then $T_{i_{2},j_{2}}\circ T_{i_{1},j_{1}}=0$; 2. $(\beta)$

if $j_{1}>i_{2}$ then $(i_{1},j_{2})\in I$ and $T_{i_{2},j_{2}}\circ T_{i_{1},j_{1}}=T_{i_{1},j_{2}}\in A$.

So $A$ is a subalgebra of $\mathcal{L}_{\mathbf{k}}(V)$.

Proof that $A$ is (SN). Consider the subset \Delta=\big{\{}\,T_{i,j}\,\mid\,(i,j)\in I\,\big{\}} of $A$. By Cor. 4.10(a), it suffices to show that for every sequence $(S_{0},S_{1},\dots)\in\Delta^{\mathbb{N}}$ there exists $n$ such that $S_{n}S_{n-1}\cdots S_{0}=0$. In other words, it suffices to show that for every sequence $(T_{u_{0},v_{0}},T_{u_{1},v_{1}},T_{u_{2},v_{2}},\dots)\in\Delta^{\mathbb{N}}$ (where $(u_{\nu},v_{\nu})\in I$ for all $\nu\in\mathbb{N}$) there exists $n$ such that $T_{u_{n},v_{n}}T_{u_{n-1},v_{n-1}}\cdots T_{u_{0},v_{0}}=0$. Note that the above statement $(\alpha)$ implies that if $(i_{1},j_{1}),(i_{2},j_{2})\in I$ are such that $T_{i_{2},j_{2}}\circ T_{i_{1},j_{1}}\neq 0$, then $i_{2}<i_{1}$ and $j_{2}<j_{1}$. Consequently, if $T_{u_{n},v_{n}}T_{u_{n-1},v_{n-1}}\cdots T_{u_{0},v_{0}}\neq 0$ then in particular $u_{n}<u_{n-1}<\cdots<u_{0}$. So there must exist an $n$ such that $T_{u_{n},v_{n}}T_{u_{n-1},v_{n-1}}\cdots T_{u_{0},v_{0}}=0$.

Proof that $A_{L}$ is not (SN). Consider the sequence $(S_{0},S_{1},S_{2},\dots)\in A^{\mathbb{N}}$, where we define $S_{i}=T_{i,i}$ for all $i$. Note that $S_{i}S_{j}=0$ whenever $i\geq j$. This implies that $[S_{1},S_{0}]=-S_{0}S_{1}$, $[S_{2},S_{1},S_{0}]=[S_{2},-S_{0}S_{1}]=S_{0}S_{1}S_{2}$, and by induction we find $[S_{n},\dots,S_{0}]=(-1)^{n}S_{0}\cdots S_{n}$ for all $n\geq 0$. As $(S_{0}\cdots S_{n})(e_{n+1})=e_{0}$, we have $S_{0}S_{1}\cdots S_{n}\neq 0$ and hence $[S_{n},\dots,S_{0}]\neq 0$ (for all $n\geq 0$). So $A_{L}$ is not (SN) and (a) is proved.

(b) Let $k\in\mathbb{N}$; let us show that $e_{k}\in\operatorname{UNil}(\Delta)$. Indeed, if $T_{i,j}$ is any element of $\Delta$ such that $T_{i,j}(e_{k})\neq 0$, then $i<k$. It follows that if $(T_{u_{1},v_{1}},\dots,T_{u_{n},v_{n}})\in\Delta^{n}$ satisfies $(T_{u_{n},v_{n}}\circ\cdots\circ T_{u_{1},v_{1}})(e_{k})\neq 0$ then $k>u_{1}>\cdots>u_{n}\geq 0$ ($T_{u_{n},v_{n}}\circ\cdots\circ T_{u_{1},v_{1}}\neq 0$ implies $u_{1}>\cdots>u_{n}$ by the preceding paragraph), so $n\leq k$, proving that $e_{k}\in\operatorname{UNil}(\Delta)$. It follows that $\operatorname{UNil}(\Delta)=V$. As $\bar{\Delta}=A$, Prop. 2.4 implies that $\operatorname{UNil}(\Delta)=\operatorname{UNil}(A)$, so $\operatorname{UNil}(A)=V$, so (b) is proved. ∎

5 Locally nilpotent sets of derivations

In this section we assume that $B$ is an algebra over a field $\mathbf{k}$ (in the sense of 1(a)) and we consider the Lie subalgebra $\operatorname{Der}_{\mathbf{k}}(B)$ of $\mathcal{L}_{\mathbf{k}}(B)_{\text{\rm L}}$ (see 1).

A derivation $D\in\operatorname{Der}_{\mathbf{k}}(B)$ is said to be locally nilpotent if it is locally nilpotent as a linear map $B\to B$. We write $\operatorname{LND}(B)$ for the set of locally nilpotent derivations of $B$, i.e., $\operatorname{LND}(B)=\operatorname{LN}(\mathcal{L}_{\mathbf{k}}B)\cap\operatorname{Der}_{\mathbf{k}}(B)$. For any subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$, the sets $\operatorname{UNil}(\Delta)\subseteq\operatorname{Nil}(\Delta)$ are subalgebras of $B$ (by Thm 2.8). If $\operatorname{Nil}(\Delta)=B$, we say that $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. If $\operatorname{UNil}(\Delta)=B$, we say that $\Delta$ is a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

This section explores Questions 1 and 2, which are stated in the Introduction. In particular, we give a series of examples showing that both questions have negative answers when $B$ is the commutative polynomial ring in $|\mathbb{N}|$ variables over an arbitrary field $\mathbf{k}$. The next section studies the case where $B$ is assumed to satisfy some finiteness condition.

Lemma 5.1**.**

Let $V$ be a vector space over a field $\mathbf{k}$ and $B$ a commutative polynomial ring over $\mathbf{k}$ satisfying $\operatorname{trdeg}_{\mathbf{k}}(B)=\dim_{\mathbf{k}}(V)$. Then there exist an injective $\mathbf{k}$-linear map $\nu:V\to B$ and an injective homomorphism of Lie algebras $\psi:\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}\to\operatorname{Der}_{\mathbf{k}}(B)$ such that $\mathbf{k}[\operatorname{im}\nu]=B$ and

[TABLE]

Moreover, the following statements are true for every subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$:

1. (a)

$\Delta\subseteq\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$* if and only if $\psi(\Delta)\subseteq\operatorname{LND}(B)$.* 2. (b)

$\Delta$* is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ if and only if $\psi(\Delta)$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.* 3. (c)

$\Delta$* is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ if and only if $\psi(\Delta)$ is a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.*

Proof 5.2**.**

Let $(x_{i})_{i\in I}$ be a family of indeterminates over $\mathbf{k}$ such that $B=\mathbf{k}[(x_{i})_{i\in I}]$. Then there exists a basis $(e_{i})_{i\in I}$ of $V$ indexed by the same set $I$. Consider the $\mathbf{k}$-linear map $\nu:V\to B$ given by $\nu(e_{i})=x_{i}$ for all $i\in I$. Then $\nu$ is injective and $\nu(V)=B_{1}$, where we define B_{1}=\operatorname{Span}_{\mathbf{k}}\big{\{}\,x_{i}\,\mid\,i\in I\,\big{\}}. We might as well assume that $V=B_{1}$ and that $\nu$ is the inclusion map $B_{1}\hookrightarrow B$. For each $F\in\mathcal{L}_{\mathbf{k}}(B_{1})$, there exists a unique $D_{F}\in\operatorname{Der}_{\mathbf{k}}(B)$ satisfying $D_{F}(v)=F(v)$ for all $v\in B_{1}$. Consider the map $\psi:\mathcal{L}_{\mathbf{k}}(B_{1})\to\operatorname{Der}_{\mathbf{k}}(B)$, $F\mapsto D_{F}$. If $a,b\in\mathbf{k}$ and $F,G\in\mathcal{L}_{\mathbf{k}}(B_{1})$ then the derivations $D_{aF+bG}$ and $aD_{F}+bD_{G}$ have the same restriction to $B_{1}$, and hence must be equal; so $\psi$ is a $\mathbf{k}$-linear map (and is clearly injective). Similarly, the derivations $D_{F\circ G-G\circ F}$ and $D_{F}\circ D_{G}-D_{G}\circ D_{F}$ have the same restriction to $B_{1}$, and hence must be equal. Thus $\psi$ is a homomorphism of Lie algebras $\mathcal{L}_{\mathbf{k}}(B_{1})_{\text{\rm L}}\to\operatorname{Der}_{\mathbf{k}}(B)$ and (15) is true. To prove (b) and (c), consider a subset $\Delta$ of $\mathcal{L}_{\mathbf{k}}(V)$. Since $\operatorname{Nil}(\Delta)$ and $\operatorname{UNil}(\Delta)$ are linear subspaces of $V$ and $(x_{i})_{i\in I}$ spans $V$, we have:

[TABLE]

Since $\operatorname{Nil}(\psi(\Delta))$ and $\operatorname{UNil}(\psi(\Delta))$ are subalgebras of $B=\mathbf{k}[(x_{i})_{i\in I}]$ by Thm 2.8, and contain $\mathbf{k}$ by Rem. A special case: sets of derivations, we have:

[TABLE]

It is easily verified that for each $i\in I$ we have

[TABLE]

Consequently, we have $\operatorname{Nil}(\Delta)=V\Leftrightarrow\operatorname{Nil}(\psi(\Delta))=B$ and $\operatorname{UNil}(\Delta)=V\Leftrightarrow\operatorname{UNil}(\psi(\Delta))=B$, i.e., assertions (b) and (c) are true. Assertion (a) follows from (b). Indeed, the condition $\Delta\subseteq\operatorname{LN}(\mathcal{L}_{\mathbf{k}}V)$ is equivalent to “for each $F\in\Delta$, $\{F\}$ is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$;” by (b), this is equivalent to “for each $F\in\Delta$, $\{\psi(F)\}$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$,” which is itself equivalent to $\psi(\Delta)\subseteq\operatorname{LND}(B)$. ∎

In view of Question 2, it is natural to ask whether an arbitrary Lie algebra $L$ can be embedded as a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ (for some $B$) in such a way that $L\subseteq\operatorname{LND}(B)$. A preliminary question is whether $L$ can be embedded in $\operatorname{Der}_{\mathbf{k}}(B)$ at all (for some $B$). The answer is affirmative:

Proposition 5.3**.**

Let $L$ be a Lie algebra over a field $\mathbf{k}$. Then there exists a commutative polynomial ring $B$ over $\mathbf{k}$ such that $L$ is isomorphic to a Lie-subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. Moreover, if $L$ is finite dimensional then we can choose $B$ to be of finite transcendence degree over $\mathbf{k}$.

Proof 5.4**.**

It is known444See for instance p. 6 of [Jac79]. that there exists a vector space $V$ over $\mathbf{k}$ such that $L$ is isomorphic to a subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$. It is also known that if $\dim L<\infty$ then $V$ can be chosen to be finite dimensional (if $\operatorname{char}\mathbf{k}=0$ this is called Ado’s Theorem [Ado49]; the general case is due to Iwasawa [Iwa48]). Consider a polynomial ring $B$ over $\mathbf{k}$ such that $\operatorname{trdeg}_{\mathbf{k}}(B)=\dim_{\mathbf{k}}(V)$. By Lemma 5.1, $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$ is isomorphic to a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. As $L$ is isomorphic to a Lie subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$, we are done. ∎

The following is a preliminary step to Ex. 5.

{Example}

If $\mathbf{k}$ is a field and $V$ is a $\mathbf{k}$-vector space of dimension $|\mathbb{N}|$, then there exists an associative subalgebra $A$ of $\mathcal{L}_{\mathbf{k}}(V)$ satisfying:

1. (a)

$A$ is the free associative algebra on a countably infinite set; 2. (b)

$A$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$.

Proof 5.5**.**

Let $\mathbf{k}$ be a field, $\mathcal{V}=\{x_{1},x_{2},x_{3},\dots\}$ a countably infinite set of indeterminates and $\mathbf{k}\langle\mathcal{V}\rangle$ the polynomial ring over $\mathbf{k}$ in the noncommutative variables $x_{1},x_{2},x_{3},\dots$. By a nonempty monomial, we mean a nonempty finite product of elements of $\mathcal{V}$, i.e., a product $x_{i_{1}}\cdots x_{i_{n}}$ with $n\geq 1$. Let $W\subset\mathbf{k}\langle\mathcal{V}\rangle$ be the set of nonempty monomials and consider the associative algebra $\mathscr{A}=\operatorname{Span}_{\mathbf{k}}W$. Note that $\mathscr{A}$ is the associative subalgebra $\bar{\mathcal{V}}$ of $\mathbf{k}\langle\mathcal{V}\rangle$ generated by $\mathcal{V}$; so $\mathscr{A}$ is the free associative algebra on $\mathcal{V}$.

We say that a nonempty monomial $x_{i_{1}}\cdots x_{i_{n}}\in W$ is admissible if it satisfies $n>i_{n}$. Let $W_{0}$ be the set of all admissible monomials and consider the $\mathbf{k}$-subspace $\mathscr{A}_{0}=\operatorname{Span}_{\mathbf{k}}(W_{0})$ of $\mathscr{A}$. Then $V=\mathscr{A}/\mathscr{A}_{0}$ is a vector space over $\mathbf{k}$ of dimension $|\mathbb{N}|$. In fact $\mathscr{A}_{0}$ is a left ideal of $\mathscr{A}$, so $V$ is also a left $\mathscr{A}$-module; so if $a\in\mathscr{A}$ and $x\in V$ then $ax\in V$ is defined, and if also $b\in\mathscr{A}$ then $(ab)x=a(bx)$. If $a\in\mathscr{A}$, let $\mu(a):V\to V$ be the $\mathbf{k}$-linear map $x\mapsto ax$ (for $x\in V$). Then $\mu:\mathscr{A}\to\mathcal{L}_{\mathbf{k}}(V)$ is a homomorphism of associative $\mathbf{k}$-algebras.

We claim that $\mu$ is injective. To see this, consider $a\in\mathscr{A}\setminus\{0\}$. Write $a=\lambda_{1}w_{1}+\cdots+\lambda_{n}w_{n}$ where $w_{1},\dots,w_{n}$ are distinct elements of $W$, $n\geq 1$ and $\lambda_{1},\dots,\lambda_{n}\in\mathbf{k}^{*}$. We have $w_{1}=x_{i_{1}}\cdots x_{i_{m}}$; then the monomial $w_{1}x_{m+1}\in W$ is not admissible. Since $ax_{m+1}=\lambda_{1}w_{1}x_{m+1}+\cdots+\lambda_{n}w_{n}x_{m+1}$ where $w_{1}x_{m+1}$, …, $w_{n}x_{m+1}$ are distinct elements of $W$ and $w_{1}x_{m+1}\notin W_{0}$, $ax_{m+1}\notin\mathscr{A}_{0}$, so \big{(}\mu(a)\big{)}(x)=ax\neq 0 where $x=x_{m+1}+\mathscr{A}_{0}\in\mathscr{A}/\mathscr{A}_{0}$, showing that $\mu(a)\neq 0$. So $\mu$ is injective. Define $A=\mu(\mathscr{A})$, then $A$ is an associative subalgebra of $\mathcal{L}_{\mathbf{k}}(V)$ and satisfies (a).

We claim that $A$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$ (meaning $\operatorname{UNil}(A)=V$). Since \big{\{}\,w+\mathscr{A}_{0}\,\mid\,w\in W\,\big{\}} is a spanning set for the vector space $V$, in order to prove the claim it suffices to show that $w+\mathscr{A}_{0}\in\operatorname{UNil}(A)$ for each $w\in W$. So consider $w\in W$ and write $w=x_{i_{1}}\cdots x_{i_{m}}$. Then for every $(w_{1},w_{2},\dots,w_{i_{m}})\in W^{i_{m}}$ we have $w_{i_{m}}\cdots w_{2}\cdot w_{1}\cdot w\in W_{0}$ (because $w_{1},\dots,w_{i_{m}}$ are nonempty monomials); it follows that $a_{i_{m}}\cdots a_{1}\cdot w\in\mathscr{A}_{0}$ for all $(a_{1},a_{2},\dots,a_{i_{m}})\in\mathscr{A}^{i_{m}}$, which implies that $(F_{i_{m}}\circ\cdots\circ F_{1})(w+\mathscr{A}_{0})=0$ for all $(F_{1},\dots,F_{i_{m}})\in A^{i_{m}}$. Thus $w+\mathscr{A}_{0}\in\operatorname{UNil}(A)$ and consequently $\operatorname{UNil}(A)=V$. ∎

The next example is interesting because (a) says that $L$ is as non-nilpotent as a Lie algebra can be, while (b) says that it is as locally nilpotent as a subset of $\operatorname{Der}_{\mathbf{k}}(B)$ can be. So this answers Question 2 in the negative.

{Example}

If $\mathbf{k}$ is a field and $B$ is the commutative polynomial algebra in $|\mathbb{N}|$ variables over $\mathbf{k}$, then there exists a Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

$L$ is the free Lie algebra on a countably infinite set; 2. (b)

$L$ is a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

Proof 5.6**.**

*Write $B=\mathbf{k}[x_{1},x_{2},x_{3},\dots]$ and let $V=\operatorname{Span}_{\mathbf{k}}\{x_{1},x_{2},x_{3},\dots\}\subseteq B$. By Ex. 5, there exists an associative subalgebra $A$ of $\mathcal{L}_{\mathbf{k}}(V)$ satisfying:

$\bullet$   $A$ is the free associative algebra on a countably infinite set;

$\bullet$   $A$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$.

Consider a countably infinite subset $S$ of $A$ such that $A$ is the free associative algebra on $S$. Let $\bar{S}$ (resp.  $\tilde{S}$) be the associative subalgebra of $\mathcal{L}_{\mathbf{k}}(V)$ (resp. the Lie subalgebra of $\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}$) generated by $S$, then $\tilde{S}\subseteq\bar{S}=A$ and $\tilde{S}$ is the free Lie algebra on $S$. Consider the injective homomorphism of Lie algebras $\psi:\mathcal{L}_{\mathbf{k}}(V)_{\text{\rm L}}\to\operatorname{Der}_{\mathbf{k}}(B)$ of 5.1 and define $L=\psi(\tilde{S})$. Then $L$ is the free Lie algebra on a countably infinite set, and is a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. We have $\operatorname{UNil}(\tilde{S})=\operatorname{UNil}(\bar{S})=\operatorname{UNil}(A)=V$, so $\tilde{S}$ is a uniformly locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(V)$; then part (c) of 5.1 implies that $L$ is a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. ∎*

The following gives a negative answer to Question 1.

{Example}

If $B$ is the commutative polynomial algebra in $|\mathbb{N}|$ variables over a field $\mathbf{k}$ then there exists a Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

$\operatorname{Nil}(L)\neq B$ and if $\operatorname{char}\mathbf{k}=0$ then $\operatorname{Nil}(L)=\mathbf{k}$; 2. (b)

every finitely generated subalgebra $L_{0}$ of $L$ satisfies $\operatorname{UNil}(L_{0})=B$; 3. (c)

the Lie algebra $L$ satisfies (LN).

Note that (b) implies that $L\subseteq\operatorname{LND}(B)$.

Proof 5.7**.**

We use the notation $B=\mathbf{k}[x_{0},x_{1},x_{2},\dots]$. For each $n\in\mathbb{N}$, define $D_{n}\in\operatorname{Der}_{\mathbf{k}}(B)$ by

[TABLE]

Let $L$ be the Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ generated by $\{D_{0},D_{1},D_{2},\dots\}$. Let $n\in\mathbb{N}$; then $x_{n}\notin\operatorname{Nil}(L)$ because the sequence $(D_{n},D_{n+1},\dots)\in L^{\mathbb{N}}$ is such that $(D_{N}\circ\cdots\circ D_{n+1}\circ D_{n})(x_{n})=x_{N+1}\neq 0$ for all $N\geq n$. In particular, $\operatorname{Nil}(L)\neq B$.

Assume that $\operatorname{char}\mathbf{k}=0$. To prove $\operatorname{Nil}(L)=\mathbf{k}$, it suffices to show that

[TABLE]

Let $f\in B\setminus\mathbf{k}$. There is a unique $n\in\mathbb{N}$ such that $f\in\mathbf{k}[x_{n},x_{n+1},\dots]\setminus\mathbf{k}[x_{n+1},x_{n+2},\dots]$. Let $R=\mathbf{k}[x_{n+1},x_{n+2},\dots]$; then $f=P(x_{n})$ for some $P(T)\in R[T]\setminus R$. Since $\operatorname{char}R=0$ and $R\subseteq\ker(D_{n})$, we have $D_{n}(f)=P^{\prime}(x_{n})D_{n}(x_{n})=x_{n+1}P^{\prime}(x_{n})\in B\setminus\mathbf{k}$. This proves (16), so $\operatorname{Nil}(L)=\mathbf{k}$ and (a) is proved.

We now drop the assumption on $\operatorname{char}\mathbf{k}$ (so $\mathbf{k}$ is an arbitrary field until the end of the proof). Let $L_{0}$ be a finitely generated Lie subalgebra of $L$. If we choose $n$ large enough and define $\Delta=\{D_{0},D_{1},\dots,D_{n}\}$ then $L_{0}\subseteq\tilde{\Delta}$, where $\tilde{\Delta}$ is the Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ generated by $\Delta$. Let us fix $n$ with this property. Note that $\operatorname{UNil}(L_{0})\supseteq\operatorname{UNil}(\tilde{\Delta})$. So, to prove (b) and (c), it suffices to show that

1. (b*′*)

$\operatorname{UNil}(\tilde{\Delta})=B$*   (c**′**) $\tilde{\Delta}$ is a nilpotent Lie algebra.*

Given $\delta\in\Delta=\{D_{0},D_{1},\dots,D_{n}\}$, we have $\delta(x_{i})\in\{0,x_{i+1}\}$ for all $i\in\mathbb{N}$ and $\delta(x_{i})=0$ for all $i>n$. It follows that $(\delta_{n+2}\circ\cdots\circ\delta_{1})(x_{j})=0$ for all $(\delta_{1},\delta_{2},\dots,\delta_{n+2})\in\Delta^{n+2}$ and all $j\in\mathbb{N}$, i.e., $\delta_{n+2}\circ\cdots\circ\delta_{1}=0$ for all $(\delta_{1},\delta_{2},\dots,\delta_{n+2})\in\Delta^{n+2}$. This certainly implies that $\operatorname{UNil}(\Delta)=B$ (hence $\operatorname{UNil}(\tilde{\Delta})=B$ by Prop. 2.4), and it also implies (by Cor. 4.10) that $\bar{\Delta}$ satisfies (N), where $\bar{\Delta}$ is the associative subalgebra of $\mathcal{L}_{\mathbf{k}}(B)$ generated by $\Delta$. By Lemma 4.14, we obtain that $(\bar{\Delta})_{\text{\rm L}}$ satisfies (N), and since $\tilde{\Delta}$ is a subalgebra of $(\bar{\Delta})_{\text{\rm L}}$ it follows that $\tilde{\Delta}$ satisfies (N). So (b′) and (c′) are proved and we are done. ∎

In view of the above Example, one may ask whether Question 1 always has an affirmative answer when $L$ is finitely generated. The answer is negative:

{Example}

Let $B$ be the commutative polynomial algebra in $|\mathbb{N}|$ variables over a field $\mathbf{k}$. Then for each integer $m\geq 2$ there exists an $m$-generated Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

1. (a)

$L$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. 2. (b)

Every $(m-1)$-generated Lie subalgebra of $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ and a nilpotent Lie algebra. 3. (c)

$L\subseteq\operatorname{LND}(B)$ and $L$ is a nil Lie algebra.

Proof 5.8**.**

Let $m\geq 2$. By the result of Golod stated in 4(a), there exists an $m$-generated associative $\mathbf{k}$-algebra $A$ such that $A$ is not nilpotent but every $(m-1)$-subset of $A$ is nilpotent (so $A$ is nil). Consider the map $\varphi:A\to\mathcal{L}_{\mathbf{k}}(A)$ of 4. We noted in 4 that $\varphi$ is a homomorphism of associative algebras. Since the $\mathbf{k}$-vector space $A$ has dimension $|\mathbb{N}|$, we have $\dim_{\mathbf{k}}(A)=\operatorname{trdeg}_{\mathbf{k}}(B)$; so we may consider an injective $\mathbf{k}$-linear map $\nu:A\to B$ and an injective homomorphism of Lie algebras $\psi:\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}\to\operatorname{Der}_{\mathbf{k}}(B)$ satisfying the conditions of Lemma 5.1 (with $V=A$); in particular, $B=\mathbf{k}[\operatorname{im}\nu]$ and the diagram in part (i) of (17) commutes for every $F\in\mathcal{L}_{\mathbf{k}}(A)$. Let $H$ be a generating set of $A$ with $|H|=m$, let $\Delta=\varphi(H)\subset\mathcal{L}_{\mathbf{k}}(A)$, let $\Lambda=\psi(\Delta)\subset\operatorname{Der}_{\mathbf{k}}(B)$ and let $L$ be the Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ generated by $\Lambda$. Let us prove that $L$ satisfies the desired conditions.

[TABLE]

Let $\bar{\Delta}$ (resp. $\tilde{\Delta}$) be the subalgebra of $\mathcal{L}_{\mathbf{k}}(A)$ (resp. of $\mathcal{L}_{\mathbf{k}}(A)_{\text{\rm L}}$) generated by $\Delta$. Since $A$ is generated by $H$ and $\varphi$ is a homomorphism of associative algebras, $\varphi(A)=\bar{\Delta}$. Since $\psi$ is an injective homomorphism of Lie algebras, it restricts to an isomorphism $\tilde{\Delta}\to L$. See part (ii) of (17).

Since $A$ is finitely generated but not nilpotent, Prop. 4.12(b) implies that $A$ is not (SN). So there exists an infinite sequence $(a,a_{0},a_{1},a_{2},\dots)\in A^{\mathbb{N}}$ such that $a_{n}\cdots a_{0}\cdot a\neq 0$ for all $n\in\mathbb{N}$. If we define $F_{i}=\varphi(a_{i})$ for all $i\in\mathbb{N}$ then $(F_{0},F_{1},\dots)\in(\bar{\Delta})^{\mathbb{N}}$ satisfies $(F_{n}\circ\cdots\circ F_{0})(a)\neq 0$ for all $n$, so $a\notin\operatorname{Nil}(\bar{\Delta})=\operatorname{Nil}(\Delta)$ (cf. Prop. 2.4) and hence $\Delta$ is not a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(A)$. By Lemma 5.1(a), $\Lambda=\psi(\Delta)$ is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. In particular, (a) is true.

To prove (b), consider an $(m-1)$-subset $\Lambda^{\prime}$ of $L$ and let $L^{\prime}$ be the subalgebra of $L$ generated by $\Lambda^{\prime}$. There exists an $(m-1)$-subset $H^{\prime}$ of $A$ such that $\psi(\varphi(H^{\prime}))=\Lambda^{\prime}$. By our choice of $A$, $H^{\prime}$ is a nilpotent subset of $A$; it follows that $\Delta^{\prime}=\varphi(H^{\prime})$ is a nilpotent subset of $\bar{\Delta}$, i.e., there exists $n>0$ such that

[TABLE]

In particular, $\Delta^{\prime}$ is a locally nilpotent subset of $\mathcal{L}_{\mathbf{k}}(A)$; by Lemma 5.1(b), $\Lambda^{\prime}=\psi(\Delta^{\prime})$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$, so Prop. 2.4 implies that $L^{\prime}$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. On the other hand, (18) and Cor. 4.10 imply that the associative algebra $\overline{\Delta^{\prime}}$ is nilpotent. By Lemma 4.14, it follows that the Lie algebra $(\overline{\Delta^{\prime}})_{\text{\rm L}}$ is nilpotent. Since $\widetilde{\Delta^{\prime}}$ is a subalgebra of $(\overline{\Delta^{\prime}})_{\text{\rm L}}$, $\widetilde{\Delta^{\prime}}$ is nilpotent. Since $\psi$ maps $\widetilde{\Delta^{\prime}}$ onto $L^{\prime}$, we obtain that $L^{\prime}$ is nilpotent. This proves (b).

We have $L\subseteq\operatorname{LND}(B)$ by (b). Since $\bar{\Delta}$ is a homomorphic image of the nil algebra $A$, $\bar{\Delta}$ is nil (as an associative algebra). So the Lie algebra $(\bar{\Delta})_{\text{\rm L}}$ is nil, by Lemma 4.14. As $\tilde{\Delta}$ is a subalgebra of $(\bar{\Delta})_{\text{\rm L}}$, it is a nil Lie algebra. Since $L\cong\tilde{\Delta}$, $L$ is nil. So (c) is proved. ∎

Corollary 5.9**.**

Let $B$ be the commutative polynomial algebra in $|\mathbb{N}|$ variables over a field $\mathbf{k}$.

1. (a)

There exists an infinite subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

• $\bullet$

$\Delta$* is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$;*

• $\bullet$

every finite subset of $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. 2. (b)

For each integer $m\geq 2$, there exists an $m$-subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$ satisfying:

• $\bullet$

$\Delta$* is not a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$;*

• $\bullet$

every proper subset of $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

Proof 5.10**.**

The following is an obvious consequence of Prop. 2.4:

1. $(*)$

For any subset $\Delta$ of $\operatorname{Der}_{\mathbf{k}}(B)$, $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ if and only if the Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ generated by $\Delta$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.

Proof of (a). Consider the Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ given in Ex. 5 and let $\Delta$ be any generating set of $L$. Then, by $(*)$, $\Delta$ satisfies (a).

Proof of (b). Let $m\geq 2$. Consider the Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$ given by Ex. 5. Let $\Delta$ be a generating set of $L$ with $|\Delta|=m$. Then, by $(*)$, $\Delta$ satisfies (b). ∎

6 The case of derivation-finite algebras

{Definition}

Let $B$ be an algebra over a field $\mathbf{k}$. We say that $B$ is derivation-finite if there exists a finite subset $X$ of $B$ satisfying:

[TABLE]

For instance, if $B$ is a finitely generated $\mathbf{k}$-algebra then it is derivation-finite. If $\operatorname{char}\mathbf{k}=0$ and $B$ is a commutative integral domain of finite transcendence degree over $\mathbf{k}$ then $B$ is derivation-finite.

This section re-examines Question 2 under the assumption that $B$ is a derivation-finite algebra over a field $\mathbf{k}$. We begin with an example.

{Example}

Consider the polynomial ring $B=\mathbf{k}[X_{1},\dots,X_{n}]$ where $n\geq 2$ and $\mathbf{k}$ is a field of characteristic zero. Let $L$ be the set of all $D\in\operatorname{Der}_{\mathbf{k}}(B)$ satisfying $D(X_{1})\in\mathbf{k}$ and $D(X_{i})\in\mathbf{k}[X_{1},\dots,X_{i-1}]$ for all $i=2,\dots,n$. Note that $L$ is a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$. We claim:

1. (a)

$L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. 2. (b)

$L$ is not a uniformly locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. 3. (c)

The Lie algebra $L$ satisfies (SN) but not (nil).

Proof 6.1**.**

The proof of (a) uses the following trivial observation: if $b\in B$ is such that $D(b)\in\operatorname{Nil}(L)$ for all $D\in L$, then $b\in\operatorname{Nil}(L)$. We have $D(X_{1})\in\mathbf{k}\subseteq\operatorname{Nil}(L)$ for all $D\in L$, so $X_{1}\in\operatorname{Nil}(L)$, so $\mathbf{k}[X_{1}]\subseteq\operatorname{Nil}(L)$ by Thm 2.8. Let $i\in\{2,\dots,n\}$ be such that $\mathbf{k}[X_{1},\dots,X_{i-1}]\subseteq\operatorname{Nil}(L)$. Since $D(X_{i})\in\mathbf{k}[X_{1},\dots,X_{i-1}]\subseteq\operatorname{Nil}(L)$ for all $D\in L$, $X_{i}\in\operatorname{Nil}(L)$ and hence $\mathbf{k}[X_{1},\dots,X_{i}]\subseteq\operatorname{Nil}(L)$ by Thm 2.8. It follows by induction that $\operatorname{Nil}(L)=B$, so (a) is true.

(b) Given any $m>0$ let $D_{m}=X_{1}^{m}\,\frac{\partial}{\partial X_{2}}$, and let $E=\frac{\partial}{\partial X_{1}}$; then $\{D_{m},E\}\subset L$ and (since $\operatorname{char}\mathbf{k}=0$) $(E^{m}\circ D_{m})(X_{2})\neq 0$, so $X_{2}\notin\operatorname{UNil}(L)$.

(c) Let us check that $L$ is not a nil Lie algebra. Let $E=\frac{\partial}{\partial X_{1}}\in L$. We claim that $\operatorname{ad}(E):L\to L$ is not nilpotent. Indeed, let $m>0$ and let us prove that $\operatorname{ad}(E)^{m}\neq 0$. It suffices to show that $\operatorname{ad}(E)^{m}(D_{m})=[E,\dots,E,D_{m}]$ is not zero (where $D_{m}=X_{1}^{m}\,\frac{\partial}{\partial X_{2}}$ and where there are $m$ “$E$” in the bracket); so it suffices to show that $[E,\dots,E,D_{m}](X_{2})\neq 0$. Since $E(X_{2})=0$, we have $[E,\dots,E,D_{m}](X_{2})=(E^{m}\circ D_{m})(X_{2})\neq 0$. So $\operatorname{ad}(E)$ is not nilpotent and consequently the Lie algebra $L$ is not nil.

To show that the Lie algebra $L$ is (SN), we use the following notation. Let $B_{0}=\mathbf{k}$ and $B_{j}=\mathbf{k}[X_{1},\dots,X_{j}]$ for all $j=1,\dots,n$. If $j\in\{1,\dots,n\}$ and $f\in B_{j-1}$, define $D_{f}^{j}=f\frac{\partial}{\partial X_{j}}\in L$. Let $H$ be the set of $D_{f}^{j}$, for all pairs $(j,f)$ such that $j\in\{1,\dots,n\}$ and $f\in B_{j-1}$ (thus $H\subseteq L$). Consider $D_{f}^{j},D_{g}^{k}\in H$ (where $f\in B_{j-1}$ and $g\in B_{k-1}$). A straightforward calculation gives $[D_{f}^{j},D_{g}^{k}]=0$ when $j=k$, and $[D_{f}^{j},D_{g}^{k}]=D^{k}_{D_{f}^{j}(g)}$ if $j<k$. Since $D_{f}^{j}(g)=0$ when $j=k$, we get

[TABLE]

We have $L=\operatorname{Span}_{\mathbf{k}}(H)$, so in particular $H$ is a generating set for the Lie algebra $L$. By Cor. 4.10(a), to show that $L$ is (SN) it suffices to show that for every sequence $(d_{0},d_{1},d_{2},\dots)\in H^{\mathbb{N}}$ there exists an $m$ such that $[d_{m},\dots,d_{0}]=0$. So consider $(d_{0},d_{1},d_{2},\dots)\in H^{\mathbb{N}}$. Then for each $i$ we have $d_{i}=D_{f_{i}}^{j_{i}}$, where $j_{i}\in\{1,\dots,n\}$ and $f_{i}\in\mathbf{k}[X_{1},\dots,X_{j_{i}-1}]$. Choose $\nu\in\mathbb{N}$ such that j_{\nu}=\max\big{\{}\,j_{i}\,\mid\,i\in\mathbb{N}\,\big{\}}. Then (19) implies that

[TABLE]

We have $[d_{m+1},d_{m},\dots,d_{0}]=[d_{m+1},[d_{m},\dots,d_{0}]]=[d_{m+1},D^{j_{\nu}}_{g_{m}}]=[D^{j_{m+1}}_{f_{m+1}},D^{j_{\nu}}_{g_{m}}]\\ =D^{j_{\nu}}_{D^{j_{m+1}}_{f_{m+1}}(g_{m})}$, so $g_{m+1}=D^{j_{m+1}}_{f_{m+1}}(g_{m})=d_{m+1}(g_{m})$ (for all $m\geq\nu$). Thus $g_{m}=(d_{m}\circ\cdots\circ d_{\nu+1})(g_{\nu})$ for all $m>\nu$. Since $(d_{i})_{i=\nu+1}^{\infty}$ is an infinite sequence in $H$, and $H$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$, it follows that for $m$ large enough we have $g_{m}=0$ and hence $[d_{m},\dots,d_{0}]=0$. This shows that the Lie algebra $L$ is (SN). ∎

Until the end of the section, we assume:

[TABLE]

For each element $D$ of the Lie algebra $\operatorname{Der}_{\mathbf{k}}(B)$, we may consider the map $\operatorname{ad}(D):\operatorname{Der}_{\mathbf{k}}(B)\to\operatorname{Der}_{\mathbf{k}}(B)$, $E\mapsto[D,E]$ (this is the map defined in 1). Then we have the following.

Lemma 6.2**.**

If $D\in\operatorname{LND}(B)$ then the map $\operatorname{ad}(D):\operatorname{Der}_{\mathbf{k}}(B)\to\operatorname{Der}_{\mathbf{k}}(B)$ is locally nilpotent.

Proof 6.3**.**

We have to show that for each $E\in\operatorname{Der}_{\mathbf{k}}(B)$ there exists $N>0$ such that $\operatorname{ad}(D)^{N}(E)=0$. Let $E\in\operatorname{Der}_{\mathbf{k}}(B)$. Choose a nonempty finite subset $X$ of $B$ such that

[TABLE]

Choose $n>0$ such that $D^{n}(x)=0$ for all $x\in X$. Since

[TABLE]

is a finite subset of $B$, there exists $m>0$ such that $D^{m}(y)=0$ for all $y\in Y$; then $(D^{m}\circ E\circ D^{i})(x)=0$ for all $(i,x)\in\{0,\dots,n-1\}\times X$. Let $F=\operatorname{ad}(D)^{m+n-1}(E)=[D,D,\dots,D,E]$. Since $F\in\operatorname{Der}_{\mathbf{k}}(B)$, in order to show that $F=0$ it suffices to show that $F(x)=0$ for all $x\in X$. Let $x\in X$; then $F(x)$ is a linear combination (over $\mathbf{k}$) of terms of the form $(D^{j}\circ E\circ D^{i})(x)$ where $i,j\geq 0$ and $i+j=m+n-1$. In each one of these terms we have either $i\geq n$ or ($i<n$ and $j\geq m$), so $(D^{j}\circ E\circ D^{i})(x)=0$. It follows that $F(x)=0$ and hence that $\operatorname{ad}(D)^{m+n-1}(E)=0$. ∎

The next facts give some positive answers to Question 2. Observe that the conclusion of Cor. 6.4(a) cannot be strenghtened from (Lnil) to (nil), by Ex. 6.

Corollary 6.4**.**

Let $L$ be a Lie subalgebra of $\operatorname{Der}_{\mathbf{k}}(B)$ such that $L\subseteq\operatorname{LND}(B)$.

1. (a)

$L$* is locally nil (Lnil).* 2. (b)

If $L$ is finite dimensional then it is a nilpotent Lie algebra.

Proof 6.5**.**

Assertion (a) follows immediately from Lemma 6.2. Part (b) follows from the fact (Prop. 4.12) that (Lnil) $\Leftrightarrow$ (N) for finite dimensional Lie algebras. ∎

Proposition 6.6**.**

For a finitely generated Lie subalgebra $L$ of $\operatorname{Der}_{\mathbf{k}}(B)$, the following are equivalent:

1. (a)

$L$* is a nilpotent Lie algebra;* 2. (b)

$L$* is a Lie-locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$;* 3. (c)

$L$* is a uniformly Lie-locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$.*

Moreover, if $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$ then (a–c) are satisfied.

Proof 6.7**.**

Choose a finite subset $\Delta$ of $L$ such that $\tilde{\Delta}=L$. Choose a nonempty finite subset $X$ of $B$ such that

[TABLE]

It is obvious that (a) implies (b). If (b) is true then $\Delta$ is a finite Lie-locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$; by Lemma 3.3, $\Delta$ is uniformly Lie-locally nilpotent; so $L=\tilde{\Delta}$ is uniformly Lie-locally nilpotent by Prop. 3.5, and this shows that (b) implies (c).

Assume that (c) is true. Since $X$ is a finite set and $L$ is uniformly Lie-locally nilpotent, there exists $N\in\mathbb{N}$ satisfying

[TABLE]

In (20), the fact that $[D_{n},\dots,D_{1}]$ is a derivation that annihilates each element of $X$ implies that $[D_{n},\dots,D_{1}]=0$. So we have $[D_{N},\dots,D_{1}]=0$ for all $(D_{1},\dots,D_{N})\in L^{N}$, i.e., $L$ is nilpotent. So (c) implies (a) and consequently the three conditions are equivalent.

Assume that $L$ is a locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$. Then $\Delta$ is a finite locally nilpotent subset of $\operatorname{Der}_{\mathbf{k}}(B)$, so $\Delta$ is uniformly locally nilpotent by Lemma 2.1, so $L=\tilde{\Delta}$ is uniformly locally nilpotent by Prop. 2.4, so $L$ is uniformly Lie-locally nilpotent by Lemma 3.1, so (a–c) are satisfied. ∎