Lie algebras of triangular polynomial derivations and an isomorphism criterion for their Lie factor algebras

Let $(S,\leqslant)$ be a partially ordered set, that is, the set $S$ admits a relation $\leqslant$ satisfying the following three conditions for all $a,b,c\in S$:

• (i)  
  $a\leqslant a$,
• (ii)  
  $a\leqslant b$ and $b\leqslant a$ imply that $a=b$,
• (iii)  
  $a\leqslant b$ and $b\leqslant c$ imply that $a\leqslant c$.

A partially ordered set $(S,\leqslant)$ is said to be Artinian if every non-empty subset $T$ of $S$ has a least element, call it $t$, that is, $t\leqslant t'$ for all $t'\in T$. A partially ordered set $(S,\leqslant)$ is totally ordered if for all elements $a,b\in S$, either $a\leqslant b$ or $b\leqslant a$. A bijection $f\colon S\to S'$ between partially ordered sets $(S,\leqslant)$ and $(S',\leqslant)$ is an isomorphism if $a\leqslant b$ in $S$ implies that $f(a)\leqslant f(b)$ in $S'$. We recall that ordinal numbers are the isomorphism classes of totally ordered Artinian sets (that is, well-ordered sets). The ordinal number (isomorphism class) of a totally ordered Artinian set $(S,\leqslant)$ is denoted by $\operatorname{ord}(S)$. The class of all ordinal numbers is denoted by $\text{W}$. The class $\text{W}$ is totally ordered by inclusion $\leqslant$ and is Artinian. It is endowed with an associative addition $+$ and an associative multiplication $\cdot$ that extend the addition and multiplication of the non-negative integers. Every positive integer $n$ is identified with $\operatorname{ord}(1 < 2 < \dotsb < n)$. We put $\omega:= \operatorname{ord}(\mathbb{N},\leqslant)$. (More details on ordinal numbers can be found in [5].)

Definition 3.1. Let $(S,\leqslant)$ be a partially ordered set. The uniserial dimension $\operatorname{u.dim}(S)$ of $S$ is the supremum of the ordinal numbers $\operatorname{ord}(\mathcal{I})$, where $\mathcal{I}$ runs through all Artinian totally ordered subsets of $S$.

For a Lie algebra $\mathcal{G}$, let $\mathcal{J}_0(\mathcal{G})$ (resp. $\mathcal{J}(\mathcal{G})$) be the set of all ideals (resp. all non-zero ideals) of $\mathcal{G}$. Thus $\mathcal{J}_0(\mathcal{G})= \mathcal{J}(\mathcal{G})\cup\{0\}$. The sets $\mathcal{J}_0(\mathcal{G})$ and $\mathcal{J}(\mathcal{G})$ are partially ordered by inclusion. A Lie algebra $\mathcal{G}$ is said to be Artinian (resp. Noetherian) if the partially ordered set $\mathcal{J}(\mathcal{G})$ is Artinian (resp. Noetherian). This means that every descending (resp. ascending) chain of ideals stabilizes. A Lie algebra $\mathcal{G}$ is said to be uniserial if the partially ordered set $\mathcal{J}(\mathcal{G})$ is totally ordered. This means that for any two ideals $\mathfrak{a}$, $\mathfrak{b}$ of the Lie algebra $\mathcal{G}$ we have either $\mathfrak{a}\subseteq\mathfrak{b}$, or $\mathfrak{b}\subseteq\mathfrak{a}$.

Definition 3.2. Let $\mathcal{G}$ be an Artinian uniserial Lie algebra. The ordinal number $\operatorname{u.dim}(\mathcal{G}):= \operatorname{ord}(\mathcal{J}(\mathcal{G}))$ of the Artinian totally ordered set $\mathcal{J}(\mathcal{G})$ of non-zero ideals of $\mathcal{G}$ is called the uniserial dimension of $\mathcal{G}$. For an arbitrary Lie algebra $\mathcal{G}$, the uniserial dimension $\operatorname{u.dim}(\mathcal{G})$ is the supremum of $\operatorname{ord}(\mathcal{I})$, where $\mathcal{I}$ runs through all Artinian totally ordered sets of ideals.

If $\mathcal{G}$ is a Noetherian Lie algebra, then $\operatorname{u.dim}(\mathcal{G})\leqslant\omega$. Thus the uniserial dimension is a measure of the deviation from the Noetherian condition. The concept of uniserial dimension makes sense for any class of algebras (associative, Jordan, etc.).

Let $A$ be an algebra, $M$ an $A$-module, and $\mathcal{J}_l(A)$ (resp. $\mathcal{M}(M)$) the set of all non-zero left ideals of $A$ (resp. all non-zero submodules of $M$). These sets are partially ordered with respect to $\subseteq$. The left uniserial dimension of $A$ is defined as $\operatorname{u.dim}(A):=\operatorname{u.dim}(\mathcal{J}_l(A))$, and the uniserial dimension of $M$ is defined as $\operatorname{u.dim}(M):=\operatorname{u.dim}(\mathcal{M}(M))$.

We define an Artinian total ordering $\leqslant$ on the canonical basis $\mathcal{B}_n$ of $\mathfrak{u}_n$ by putting $X_{\alpha,i}>X_{\beta,j}$ if and only if either $i<j$, or $i=j$ and $\alpha_{n-1}=\beta_{n-1},\dots,\alpha_{m+1}=\beta_{m+1}$, $\alpha_m > \beta_m$ for some $m$.

Example 3.3. For $n\,{=}\,2$ we have $\partial_2 < x_1\partial_2 < x_1^2\partial_2 < \dotsb < \partial_1$. For $n\,{=}\,3$ we have

$$\begin{equation*} \begin{aligned} &\partial_3<x_1\partial_3<x_1^2\partial_3<\dotsb<x_2\partial_3 <x_1x_2\partial_3<x_1^2x_2\partial_3<\dotsb \\ &\qquad\qquad\dotsb<x_2^i\partial_3<x_1x_2^i\partial_3<x_1^2x_2^i \partial_3<\dotsb< \partial_2<x_1\partial_2<x_1^2\partial_2 <\dotsb<\partial_1. \end{aligned} \end{equation*}$$

The following lemma is a straightforward consequence of the definition of the ordering $<$. We write $0<X_{\alpha,i}$ for all $X_{\alpha,i}\in\mathcal{B}_n$.

Lemma 3.4. If $X_{\alpha,i}>X_{\beta,j}$, then

• 1)  
  $X_{\alpha+\gamma,i}>X_{\beta+\gamma,j}$ for all $\gamma\in\mathbb{N}^{i-1}$,
• 2)  
  $X_{\alpha-\gamma,i}>X_{\beta-\gamma,j}$ for all $\gamma\in\mathbb{N}^{i-1}$ such that $\alpha-\gamma\in\mathbb{N}^{i-1}$ and $\beta-\gamma\in\mathbb{N}^{j-1}$,
• 3)  
  ${[}\partial_k,X_{\alpha,i}]>[\partial_k,X_{\beta,j}]$ for all $k=1,\dots,i-1$ such that $\alpha_k\neq 0$,
• 4)  
  ${[}X_{\gamma,k},X_{\alpha,i}]>[X_{\gamma,k},X_{\beta,j}]$ for all $X_{\gamma,k}>X_{\alpha,i}$ such that ${[}X_{\gamma, k},X_{\alpha,i}]\neq 0$, that is, $\alpha_k\neq 0$.

Let $\Omega_n$ be the set of indices $\{(\alpha,i)\}$ that parametrizes the canonical basis $\{X_{\alpha,i}\}$ of $\mathfrak{u}_n$. The set $(\Omega_n,\leqslant)$ is an Artinian totally ordered set, where $(\alpha,i)\geqslant (\beta,j)$ if and only if $X_{\alpha,i}\geqslant X_{\beta,j}$. It is isomorphic to the Artinian totally ordered set $(\mathcal{B}_n,\leqslant)$ via the map $(\alpha,i)\mapsto X_{\alpha,i}$. We identify the partially ordered sets $(\Omega_n,\leqslant)$ and $(\mathcal{B}_n,\leqslant)$ by means of this isomorphism. Clearly,

$$\begin{equation} \operatorname{ord}(\mathcal{B}_n)=\operatorname{ord}(\Omega_n)= \omega^{n-1}+\omega^{n-2}+\dotsb +\omega +1, \end{equation} \tag{ 3.1 }$$

$\Omega_2\subset\Omega_3\subset\dotsb$ and $\mathcal{B}_2\subset\mathcal{B}_3\subset\dotsb\,$. We put ${[}1,\operatorname{ord}(\Omega_n)]:=\{\lambda\in\text{W}\mid 1\leqslant\lambda \leqslant\operatorname{ord}(\Omega_n)\}$. By (2.2), if ${[}X_{\alpha,i},X_{\beta,j}]\neq 0$, then

$$\begin{equation} [X_{\alpha,i},X_{\beta,j}]<\min\{X_{\alpha,i},X_{\beta,j}\}. \end{equation} \tag{ 3.2 }$$

By (3.2), the map

$$\begin{equation} \rho_n\colon [1,\operatorname{ord}(\Omega_n)]\to\mathcal{J} (\mathfrak{u}_n), \qquad\lambda\mapsto I_\lambda :=I_{\lambda, n}:= \bigoplus_{(\alpha,i)\leqslant\lambda}KX_{\alpha,i}, \end{equation} \tag{ 3.3 }$$

is a monomorphism of partially ordered sets ($\rho_n$ is an order-preserving injection). We shall prove that $\rho_n$ is a bijection (Theorem 3.6(1)) and, as a result, we will have a classification of the ideals of $\mathfrak{u}_n$. Each non-zero element $u$ of $\mathfrak{u}_n$ is a finite linear combination

$$\begin{equation*} u=\lambda X_{\alpha,i}+\mu X_{\beta,j}+\dotsb+\nu X_{\sigma,k}= \lambda X_{\alpha,i}+\dotsb, \end{equation*}$$

where $\lambda,\mu,\dots,\nu\in K^*$ and $X_{\alpha,i}>X_{\beta,j}>\dotsb>X_{\sigma,k}$. The elements $\lambda X_{\alpha,i}$ and $\lambda\in K^*$ are called the leading term and leading coefficient of $u$ respectively, and the ordinal number denoted by $\operatorname{ord}(X_{\alpha,i})=\operatorname{ord}(\alpha,i) \in[1,\operatorname{ord}(\Omega_n)]$ and defined as the ordinal number that represents the Artinian totally ordered set $\{(\beta,j)\in\Omega_n\mid (\beta,j)\leqslant(\alpha,i)\}$, is called the ordinal degree of $u$ and is denoted by $\operatorname{ord}(u)$ (we hope that this notation will not lead to confusion). The following assertions hold for all non-zero elements $u,v\in\mathfrak{u}_n$ and all $\lambda\in K^*$:

• (i)  
  $\operatorname{ord}(u+v)\leqslant\max\{\operatorname{ord}(u),\operatorname{ord}(v)\}$ provided that $u+v\neq 0$,
• (ii)  
  $\operatorname{ord}(\lambda u)=\operatorname{ord}(u)$,
• (iii)  
  $\operatorname{ord}([u,v])<\min\{\operatorname{ord}(u),\operatorname{ord}(v)\}$ provided that ${[}u,v]\neq 0$,
• (iv)  
  $\operatorname{ord}(\sigma (u))=\operatorname{ord}(u)$ for all automorphisms $\sigma$ of $\mathfrak{u}_n$ (Corollary 3.12).

The following lemma plays a crucial role in the proof of Theorem 3.6.

Lemma 3.5. Let $I$ be a non-zero ideal of $\mathfrak{u}_n$. Then $I=\bigcup_{0\neq u\in I}I_{\operatorname{ord}(u)}$. In particular, $(v)=I_{\operatorname{ord}(v)}$ for all non-zero elements $v$ of $\mathfrak{u}_n$, where $(v)$ is the ideal of $\mathfrak{u}_n$ generated by the element $v$.

Proof. It suffices to prove the second assertion (that is, $(v)=I_{\operatorname{ord}(v)}$) because then

$$\begin{equation*} I=\sum_{0\neq u\in I}(u)=\sum_{0\neq u\in I}I_{\operatorname{ord}(u)}= \bigcup_{0\neq u\in I}I_{\operatorname{ord}(u)}. \end{equation*}$$

To prove that $(v)=I_{\operatorname{ord}(v)}$, we use a double induction: first on $n\geqslant 2$ and then, for a fixed $n$, on $\lambda=\operatorname{ord}(v)$. We may assume without loss of generality that the leading coefficient of $v$ is equal to 1. Notice that $\operatorname{ord}(v)$ is not a limit ordinal.

A. Let $n=2$. If $\lambda= 1$, then $v=\partial_2$ and, therefore, $(v)=K\partial_2=I_1$, as required. Suppose that $\lambda > 1$ and the desired assertion holds for all non-limit ordinals $\lambda'$ such that $\lambda' < \lambda$.

Case 1: $\lambda=\operatorname{ord}(x_1^i\partial_2)$ for some $i\geqslant 1$, that is,

$$\begin{equation*} v= x_1^i\partial_2+\mu_{i-1}x_1^{i-1}\partial_2+\dotsb+\mu_0\partial_2 \end{equation*}$$

for some scalars $\mu_j\in K$. Put $\delta=\operatorname{ad}(\partial_1)$. Then $I_\lambda\supseteq (v)\supseteq\sum_{j=0}^i K\delta^j(v)=I_\lambda$, that is, $(v)=I_\lambda$.

Case 2: $\lambda=\operatorname{ord}(\partial_1)=\omega+1$, that is, $v=\partial_1+p\partial_2$ for some element $p\in P_1$. For all $i\geqslant 1$ we have $(v)\ni[v,x_1^i\partial_2]=ix_1^{i-1}\partial_2$. By Case 1, $\mathfrak{u}_{2,1}\subseteq(v)$ and, therefore, $\partial_1\in(v)$. Hence $(v)=\mathfrak{u}_2=I_{\omega+1}=I_{\operatorname{ord}(v)}$.

B. Suppose that $n>2$ and the desired assertion holds for all $n'$ such that $n'< n$. If $\lambda= 1$, then $v\in K^*\partial_n$ and, therefore, $(v)=K\partial_n=I_1$, as required. Suppose that $\lambda > 1$ and the assertion holds for all non-limit ordinals $\lambda'$ such that $\lambda' < \lambda$.

Case 1: $\lambda=\operatorname{ord}(x^\alpha\partial_n)$ for some $\alpha$, $0\neq\alpha\in\mathbb{N}^{n-1}$, that is, $v= x^\alpha\partial_n+ \dotsb$, where the dots stand for smaller terms. The following claim is a corollary of (2.2) and the definition of the total ordering on the canonical basis $\mathcal{B}_n$ of $\mathfrak{u}_n$.

For every non-limit ordinal $\lambda'$ such that $\lambda' < \lambda$, there are elements of $\mathcal{B}_n$, say, $a_1,\dots,a_s$, of type $X_{\beta,i}$, $i\neq n$, such that

$$\begin{equation*} \operatorname{ord}\bigl(\operatorname{ad}(a_1)\dotsb \operatorname{ad}(a_s)(v)\bigr)=\lambda'. \end{equation*}$$

By induction on $\lambda$, it follows from this claim that $(v)=I_{\operatorname{ord}(v)}$.

Case 2: $v=\partial_i+\dotsb$ for some $i$ such that $i < n$. For all elements $x^\beta\partial_{i+1}$ with $\beta\in\mathbb{N}^i$ and $\beta_i\neq 0$ we have

$$\begin{equation*} (v)\ni [v, x^\beta\partial_{i+1}]=\beta_i x^{\beta - e_i}\partial_{i+1}+\dotsb, \end{equation*}$$

whence $\operatorname{ord}([v,x^\beta\partial_{i+1}])= \operatorname{ord}(x^{\beta -e_i}\partial_{i+1})$. By induction, $(v)\supseteq\bigcup_{\mu < \lambda}I_\mu=:J$. But $I_\lambda\supseteq(v)=Kv+J=K\partial_i+J=I_\lambda$ and, therefore, $(v)=I_\lambda$.

Case 3: $v=x^\alpha\partial_i+\dotsb$ for some $i$ such that $i<n$ and $\alpha\in\mathbb{N}^{i-1}\setminus\{0\}$. Notice that

$$\begin{equation*} (v)\ni\prod_{j=1}^{i-1}\frac{\operatorname{ad}(\partial_j)^{\alpha_j}}{\alpha_j!}(v) =\partial_i+\dotsb . \end{equation*}$$

By Case 2, $I_{\operatorname{ord}(\partial_i)}\subseteq (v)$. Since $\mathfrak{u}_{n,i+1}\subseteq I_{\operatorname{ord}(\partial_i)}$, we have $\mathfrak{u}_{n,i+1}\subseteq (v)$. By (2.6), $\mathfrak{u}_i\simeq\mathfrak{u}_n/\mathfrak{u}_{n,i+1}$. By assumption, $i < n$. Consider the element $v+\mathfrak{u}_{n,i+1}\in\mathfrak{u}_i$. Then the desired assertion follows by induction on $i$. □

Given any $u,v\in\mathfrak{u}_n\setminus\{0\}$, we have $(u)\subseteq(v)$ if and only if $\operatorname{ord}(u)\leqslant\operatorname{ord}(v)$.

Theorem 3.6. 

• 1)  
  The map (3.3) is a bijection.
• 2)  
  The Lie algebra $\mathfrak{u}_n$ is uniserial, Artinian and non-Noetherian. Its uniserial dimension is given by $\operatorname{u.dim}(\mathfrak{u}_n)=\operatorname{ord}(\Omega_n)= \omega^{n-1}+\omega^{n-2}+\dotsb+\omega+1$.

Proof. Part 1 follows at once from Lemma 3.5. Part 2 follows from part 1. □

An ideal $\mathfrak{a}$ of a Lie algebra $\mathcal{G}$ is said to be proper (resp. cofinite) if $\mathfrak{a}\neq 0,\mathcal{G}$ (resp. $\operatorname{dim}_K(\mathcal{G}/\mathfrak{a}) < \infty$).

Corollary 3.7. 

• 1)  
  The ideal $\mathfrak{u}_{n,2}$ is the largest proper ideal of $\mathfrak{u}_n$.
• 2)  
  The ideal $\mathfrak{u}_{n,2}$ is the only proper cofinite ideal of $\mathfrak{u}_n$, and $\operatorname{dim}_K(\mathfrak{u}_n/\mathfrak{u}_{n,2})\,{=}\,1$.
• 3)  
  The centre $Z(\mathfrak{u}_n)=K\partial_n$ of $\mathfrak{u}_n$ is the least non-zero ideal of $\mathfrak{u}_n$.
• 4)  
  The ideals $I_s:=\sum_{i=0}^{s-1} Kx_1^i\partial_n$, $s=1,2,\dots$, are the only finite-dimensional ideals of $\mathfrak{u}_n$, and $\operatorname{dim}_K(I_s)=s$.

Proof. All these statements are easy corollaries of Theorem 3.6. □

In combination with Theorem 3.6, the following proposition describes the centralizers of all ideals of the Lie algebra $\mathfrak{u}_n$. Notice that the centralizer of an ideal of a Lie algebra is also an ideal.

Proposition 3.8. 

• 1)  
  We have

  $$\begin{equation*} \operatorname{Cen}_{\mathfrak{u}_n}(I_{\lambda,n})\,{=}\,\!\begin{cases} K\partial_n &\text{if}\quad\lambda=\operatorname{ord}(\Omega_n) =\omega^{n-1}+\dotsb+\omega +1, \\ I_{\omega^i,n}=P_i\partial_n&\text{if}\quad\lambda\in(\omega^{n-1}\!+\dotsb +\omega^{i+1}, \omega^{n-1}+\dotsb+ \omega^i], \\ &\qquad\qquad\qquad\qquad\qquad\qquad\ \quad i=1,\dots,n-2, \\ \displaystyle \bigoplus_{i=m+1}^nP_{i-1}\partial_i&\text{if} \quad\lambda\in(\omega^{m-1},\omega^m], \ m=1,\dots,n-1, \\ \mathfrak{u}_n&\text{if}\quad\lambda=1. \end{cases} \end{equation*}$$
• 2)  
  The set $\mathcal{C}(\mathfrak{u}_n)$ of centralizers of all ideals of $\mathfrak{u}_n$ contains precisely $2n\,{-}\,1$ elements, and the map $\mathcal{C}(\mathfrak{u}_n)\,{\to}\,\mathcal{C}(\mathfrak{u}_n)$, $C\,{\mapsto}\,\operatorname{Cen}_{\mathfrak{u}_n}(C)$, is a bijection. Moreover, it is an inclusion-reversing involution, that is, $\operatorname{Cen}_{\mathfrak{u}_n}(\operatorname{Cen}_{\mathfrak{u}_n}(C))\,{=}\,C$ for all $C\,{\in}\,\mathcal{C}(\mathfrak{u}_n)$.
• 3)  
  We have $\operatorname{Cen}_{\mathfrak{u}_n}(\mathfrak{u}_n)=K\partial_n$, $\operatorname{Cen}_{\mathfrak{u}_n}(K\partial_n)=\mathfrak{u}_n$, $\operatorname{Cen}_{\mathfrak{u}_n}(P_{n-1}\partial_n)=P_{n-1}\partial_n$, $\operatorname{Cen}_{\mathfrak{u}_n}(P_i\partial_n)\,{=}\,\bigoplus_{j=i+1}^nP_{j-1}\partial_j$, $\operatorname{Cen}_{\mathfrak{u}_n}(\bigoplus_{j=i+1}^nP_{j-1}\partial_j)\,{=}\, P_i\partial_n$ for $i\,{=}\,1,\dots,n- 2$.

Proof. Put $C_\lambda:=\operatorname{Cen}_{\mathfrak{u}_n}(I_{\lambda,n})$. If $\lambda=\operatorname{ord}(\Omega_n)$, then $I_{\lambda,n}=\mathfrak{u}_n$ and $C_\lambda=Z(\mathfrak{u}_n)=K\partial_n$ by Proposition 2.1(6). If $\lambda= 1$, then $I_{1, n}=K\partial_n=Z(\mathfrak{u}_n)$ and, therefore, $C_1=\mathfrak{u}_n$. We can assume that $\lambda\neq 1, \operatorname{ord}(\Omega_n)$. To prove part 1, we use induction on $n\geqslant 2$. For $n=2$, the only case to consider is when $\lambda\in(1,\omega]$. In this case,

$$\begin{equation*} I_{\lambda,2}=\begin{cases}\displaystyle \bigoplus_{i=0}^{\lambda -1} x_1^i \partial_2&\text{if}\quad1<\lambda <\omega, \\ P_1\partial_2&\text{if}\quad\lambda=\omega\\ \end{cases} \end{equation*}$$

and, therefore, $C_\lambda=P_1\partial_2$.

Suppose that $n>2$ and part 1 holds for all $n'<n$.

We claim that $C_{\omega^m}= \bigoplus_{i=m+1}^nP_{i-1}\partial_i$ for $m=1,\dots,n-1$.

Indeed, for $m {=} n-1$ we have $I_{\omega^{n-1}\!,n} {=} P_{n{-}1}\partial_n$ and our claim is just Lemma 2.5(2). Suppose that $m<n-1$. Then $I_{\omega^m, n}=P_m\partial_n\subset I_{\omega^{n-1}, n}$, whence $C_{\omega^m}\supseteq C_{\omega^{n-1}}=P_{n-1}\partial_n$. This means that the $\mathfrak{u}_n$-module $I_{\omega^m, n}$ is also a $(\mathfrak{u}_n/P_{n-1}\partial_n)$-module and a $\mathfrak{u}_{n-1}$-module since $\mathfrak{u}_{n-1}\simeq \mathfrak{u}_n/P_{n-1}\partial_n$. The map

$$\begin{equation*} I_{\omega^m, n}=P_m\partial_n\to I_{\omega^m, n-1}= P_m\partial_{n-1},\qquad p\partial_n\mapsto p\partial_{n-1}, \end{equation*}$$

is an isomorphism of $\mathfrak{u}_{n-1}$-modules (since $\partial_n\in Z(\mathfrak{u}_n)$ and $\partial_{n-1}\in Z(\mathfrak{u}_{n-1})$), and our claim follows by induction on $n$.

Suppose that $1<\lambda\leqslant\omega^{n-1}$, that is, $\lambda\in (\omega^{m-1},\omega^m]$ for some $m\in\{ 1,\dots,n-1\}$. We must show that $C_\lambda=C_{\omega^m}$ (see the claim). By the claim, the inclusions $I_{\omega^{m-1}, n}\subset I_{\lambda,n} \subseteq I_{\omega^m,n}$ imply that $C_{\omega^{m-1}}\supseteq C_\lambda\supseteq C_{\omega^m}$. Notice that $C_{\omega^{m-1}}= C_{\omega^m}\oplus P_{m-1}\partial_m$, $x_m\partial_n\in I_{\lambda,m}$ (since $\lambda\in(\omega^{m-1},\omega^m]$) and ${[}p\partial_m,x_m\partial_n]= p\partial_n\neq 0$ for all non-zero elements $p\in P_{m-1}$. It follows that $C_\lambda=C_{\omega^m}$, as required.

Suppose that $\omega^{n-1} < \lambda < \operatorname{ord}(\Omega_n)= \omega^{n-1}+\dotsb+\omega+1$, that is, $\lambda\in(\omega^{n-1}+\dotsb+ \omega^{i+1}, \omega^{n-1}+\dotsb+\omega^i]$ for some $i\in\{ 1,\dots,n-2\}$. Put $\lambda_i:=\omega^{n-1}+ \dotsb+ \omega^i$. Then $I_{\lambda,n}\subseteq I_{\lambda_i,n}\,{=}\,\bigoplus_{j=i+1}^nP_{j-1}\partial_j\,{=}\, \operatorname{Cen}_{\mathfrak{u}_n}(I_{\omega^i,n})$ by the claim. Therefore, $C_\lambda\supseteq I_{\omega^i,n}=P_i\partial_n$. It follows from the inclusion $I_{\lambda,n}\supset P_{n-1}\partial_n=I_{\omega^{n-1},n}$ that $C_\lambda\subseteq C_{\omega^{n-1}}=P_{n-1}\partial_n$ by the claim. Then the inclusion $\{\partial_{i+1},\dots, \partial_n\}\subseteq I_{\lambda,n}$ implies that $C_\lambda\subseteq\operatorname{Cen}_{P_{n-1}}(\partial_{i+1},\dots, \partial_n)\partial_n=P_i\partial_n=I_{\omega^i, n}$. This means that $C_\lambda= I_{\omega^i,n}$.

Part 2 follows from part 3. Part 3 follows from part 1. □

For a Lie algebra $\mathcal{G}$ over a field $K$ we define its hypercentral series $\{ Z^{(\lambda)}(\mathcal{G})\}_{\lambda\in \text{W} }$ recursively. We put $Z^{(0)}(\mathcal{G})\,{:=}\,Z(\mathcal{G})$. If $\lambda$ is not a limit ordinal, that is, $\lambda\,{=}\,\mu+1$ for some $\mu\,{\in}\,\text{W}$, then

$$\begin{equation*} Z^{(\lambda)}(\mathcal{G}):=\pi_\mu^{-1}(Z(\mathcal{G}/Z^{(\mu)}(\mathcal{G}))), \quad \text{where}\quad\pi_\mu\colon\mathcal{G}\to\mathcal{G}/Z^{(\mu)}(\mathcal{G}), \quad a\mapsto a+Z^{(\mu)}(\mathcal{G}). \end{equation*}$$

If $\lambda$ is a limit ordinal, then $Z^{(\lambda)}(\mathcal{G}) {:=} \bigcup_{\mu < \lambda}\!Z^{(\mu)}(\mathcal{G})$. If $\lambda {\leqslant} \mu$, then $Z^{(\lambda)}(\mathcal{G}) {\subseteq} Z^{(\mu)}(\mathcal{G})$. Thus, $\{ Z^{(\lambda)}(\mathcal{G})\} _{\lambda\in\text{W}}$ is an ascending chain of ideals of $\mathcal{G}$. We put $Z^{(\text{W})}(\mathcal{G}):= \bigcup_{\lambda\in\text{W}}Z^{(\lambda)}(\mathcal{G})$.

Definition 3.9. The minimal ordinal number $\lambda$ (if it exists) such that $Z^{(\lambda)}(\mathcal{G})=Z^{(\text{W})}(\mathcal{G})$, is called the central dimension of the Lie algebra $\mathcal{G}$ and is denoted by $\operatorname{c.dim}(\mathcal{G})$. If there is no such $\lambda$, we write $\operatorname{c.dim}(\mathcal{G})=\text{W}$. The concept of central dimension makes sense for any class of algebras (associative, Jordan etc.).

A Lie algebra $\mathcal{G}$ is central (that is, $Z(\mathcal{G})\,{=}\,0$) if and only if $\ \operatorname{c.dim}(\mathcal{G})\,{=}\,0$. Thus the central dimension measures the deviation from 'being central'. The following theorem describes the hypercentral series for the Lie algebra $\mathfrak{u}_n$ and gives $\operatorname{c.dim}(\mathfrak{u}_n)$.

Theorem 3.10. The hypercentral series $\{ Z^{(\lambda)}(\mathfrak{u}_n)\}_{\lambda\in \text{W}}$ stabilizes precisely at step $\operatorname{ord}(\Omega_n)= \omega^{n-1}+\omega^{n-2}+\dotsb+\omega+1$, that is, $\operatorname{c.dim}(\mathfrak{u}_n)=\operatorname{ord}(\Omega_n)$. Moreover, $Z^{(\lambda)}(\mathfrak{u}_n)=I_\lambda$ for every $\lambda\in[1,\operatorname{ord}(\Omega_n)]$. In particular, $Z^{(\operatorname{c.dim}(\mathfrak{u}_n))}(\mathfrak{u}_n)=\mathfrak{u}_n$.

Proof. It suffices to show that $Z^{(\lambda)}=I_\lambda$ for all $\lambda\in[1,\operatorname{ord}(\Omega_n)]$, where $Z^{(\lambda)}:=Z^{(\lambda)}(\mathfrak{u}_n)$. We use induction on $\lambda$. The case $\lambda= 1$ follows from Proposition 2.1(6): $Z^{(1)}=Z(\mathcal{G})=K\partial_n=I_1$. Suppose that $\lambda > 1$ and the desired equality holds for all ordinals $\lambda'<\lambda$.

If $\lambda$ is a limit ordinal, then

$$\begin{equation*} Z^{(\lambda)}=\bigcup_{\mu<\lambda}Z^{(\mu)}=\bigcup_{\mu<\lambda}I_\mu=I_\lambda. \end{equation*}$$

Suppose that $\lambda$ is not a limit ordinal, that is, $\lambda=\mu+1$ for some ordinal number $\mu$.

Case 1: $\lambda\leqslant\omega^{n-1}$. Then we necessarily have $\lambda<\omega^{n-1}$ since $\lambda$ is not a limit ordinal but $\omega^{n-1}$ is. Clearly, $I_\lambda\subseteq Z^{(\lambda)}$ by (2.2) (since $Z^{(\mu)}=I_\mu$ by induction). Suppose that $I_\lambda\neq Z^{(\lambda)}$. Then necessarily $I_{\lambda +1}\subseteq Z^{(\lambda)}$ (by Theorem 3.6(1)). If $\lambda=X_{\alpha,i}$ for some $\alpha\in\mathbb{N}^{n-1}$ (recall that we identify $(\Omega_n,\leqslant)$ and $(\mathcal{B}_n,\leqslant)$), then $\lambda +1= X_{\alpha +e_1, n}$ and $X_{\alpha +e_1, n}\in I_{\lambda+1}$, but ${[}\partial_1,X_{\alpha +e_1,n}]=(\alpha_1+1)X_{\alpha, n}\not\in Z^{(\mu)}$, a contradiction. Therefore, $Z^{(\lambda)}=I_\lambda$.

Case 2: $\lambda>\omega^{n-1}$. Notice that $Z^{(\lambda)}\supseteq I_\lambda\varsupsetneq I_{\omega^{n-1}}= \mathfrak{u}_{n,n}$ and $\mathfrak{u}_n/\mathfrak{u}_{n,n}\simeq \mathfrak{u}_{n-1}$ by (2.6). We now complete the argument by induction on $n$. The base of the induction $(n=2)$ is covered by Cases 1, 2 for $n=2$ above. □

An ideal $I$ of a Lie algebra $\mathcal{G}$ is said to be characteristic if it is invariant under all automorphisms of $\mathcal{G}$, that is, $\sigma (I)=I$ for all $\sigma\in\operatorname{Aut}_K(\mathcal{G})$. Clearly, an ideal $I$ is characteristic if and only if $\sigma (I)\subseteq I$ for all $\sigma\in\operatorname{Aut}_K(\mathcal{G})$.

Corollary 3.11. All ideals of the Lie algebra $\mathfrak{u}_n$ are characteristic.

Proof. By definition, the members of the hypercentral series $\{ Z^{(\lambda)}(\mathfrak{u}_n)\}$ are characteristic ideals of $\mathfrak{u}_n$. But Theorem 3.10 shows that these are all the ideals of $\mathfrak{u}_n$. We can also deduce this corollary from Theorem 3.6(2). □

Corollary 3.12. For all non-zero elements $u\in\mathfrak{u}_n$ and all automorphisms $\sigma$ of $\mathfrak{u}_n$ we have $\operatorname{ord}(\sigma(u))=\operatorname{ord}(u)$.

Proof. Notice that

$$\begin{equation} \operatorname{ord}(u)=\min\bigl\{\lambda\in[1,\operatorname{ord}(\Omega_n)] \mid u\in I_\lambda\bigr\}. \end{equation} \tag{ 3.4 }$$

The statement now follows from the fact that all the ideals $I_\lambda$ are characteristic (Corollary 3.11).

Let $\mathfrak{a}$ be an ideal of a Lie algebra $\mathcal{G}$. Then the ideal $U(\mathcal{G})\mathfrak{a} U(\mathcal{G})$ of the universal enveloping algebra $U(\mathcal{G})$ generated by $\mathfrak{a}$ is equal to $U(\mathcal{G})\mathfrak{a}= \mathfrak{a} U(\mathcal{G})$. The chain

$$\begin{equation*} I_1\subset\dotsb\subset I_\lambda\subset\dotsb\subset\mathfrak{u}_n= I_{\operatorname{ord}(\Omega_n)},\qquad\lambda\in [1,\operatorname{ord}(\Omega_n)], \end{equation*}$$

of ideals of $\mathfrak{u}_n$ yields a chain of subalgebras and a chain of ideals of $U_n$:

$$\begin{equation*} U(I_1)\subset\dotsb\subset U(I_\lambda)\subset\dotsb\subset U_n, \qquad U_nI_1\subset\dotsb\subset U_nI_\lambda\subset\dotsb\subset U_n. \end{equation*}$$

We put $U_{n,\lambda }'{:=}\chi_n(U(I_\lambda))$ and $I_\lambda'{:=} \chi_n(U_nI_\lambda)$, where $\chi_n$ is the algebra homomorphism (2.11). Then

$$\begin{equation} {U_{n,1 }'\subseteq\dotsb\subseteq U_{n,\lambda }'\subseteq\dotsb\subseteq U_n'=\chi_n (U_n)} \end{equation} \tag{ 3.5 }$$

is a chain of subalgebras of $U_n'$, and

$$\begin{equation} I_{n,1 }'\subseteq\dotsb\subseteq I_{n,\lambda }'\subseteq\dotsb\subseteq U_n' \end{equation} \tag{ 3.6 }$$

is a chain of ideals of $U_n'$.

Proposition 3.13. 

• 1)  
  All the inclusions in (3.5) are strict. The algebra $U_n'$ is not finitely generated.
• 2)  
  All the inclusions in (3.6) are strict. In particular, $U_n'$ is neither left Noetherian nor right Noetherian and does not satisfy the ascending chain condition for ideals.

Proof. 1. We use induction on $n$. Suppose that $n=2$. For every $i\in [1,\omega)$, the algebra $U_{2,i}'$ is generated by the commuting elements $x_1^j\partial_2$ $(0\leqslant j\leqslant i-1)$ since

$$\begin{equation*} x_1^j\partial_2\cdot x_1^k\partial_2=x_1^{j+k}\partial_2^2\qquad\forall\,j,k. \end{equation*}$$

These equalities mean that $U_{2,i}'$ is isomorphic to the monoid algebra $K\mathcal{M}_i$ (via $x_1^j\partial_2\,{\mapsto}\,(j,1)$), where $\mathcal{M}_i$ is the submonoid of $(\mathbb{N}^2, +)$ generated by the elements $(j,1)$, $0\leqslant j\leqslant i-1$. It follows that

$$\begin{equation*} x_1^{i-1}\partial_2\in U_{2,i}'\setminus U_{2,i-1}',\qquad i=2,3,\dots\,. \end{equation*}$$

This means that the inclusions $U_{2,1}'\subset U_{2,2}'\subset \dotsb\subset U_{2,i}'\subset\dotsb\subset U_{2,\omega}'=\bigcup_{i\geqslant 1}U_{2,i}'$ are strict and, therefore, the algebra $U_{2,\omega}'$ is not finitely generated. We have $\partial_1\in U_2'\setminus U_{2,\omega }'$, $U_2'=U_{2,\omega}'[\partial_1;\operatorname{ad}\ (\partial_1)]$ is the skew polynomial algebra and ${[}\partial_1, U_{2,i}']\subseteq U_{2,i}'$ for all $i\geqslant 1$. Therefore part 1 holds for $n=2$.

Suppose that $n > 2$ and part 1 holds for all $n'<n$. Let $\lambda\in [1,\omega^{n-1})$. Then $\lambda+ 1$ is not a limit ordinal. Notice that ${[}1,\omega^{n-1})\subseteq[1,\operatorname{ord}(\Omega_n))$. Hence $\lambda+1=(\alpha,n)$. The elements $\{ x^{\beta}\partial_n\mid\beta\in\mathbb{N}^{n-1}\}$ commute. Moreover,

$$\begin{equation*} x^{\beta}\partial_n\cdot x^\gamma\partial_n=x^{\beta +\gamma} \partial_n^2,\qquad \beta,\gamma\in\mathbb{N}^{n-1}. \end{equation*}$$

Therefore the algebra $U_{n,\lambda}'$ is commutative and is isomorphic to the monoid algebra $K\mathcal{M}_{n,\lambda}$ (via $x^{\beta}\partial_n\mapsto(\beta,1)$), where $\mathcal{M}_{n,\lambda}$ is the submonoid of $(\mathbb{N}^n,+)$ generated by the elements $\{(\beta, 1)\mid(\beta,n)\leqslant\lambda\}$. It follows that

$$\begin{equation*} x^\alpha\partial_n\in U_{n,\lambda +1}'\setminus U_{n,\lambda }' \qquad \forall\,\lambda\in [1,\omega^{n-1}). \end{equation*}$$

This means that

$$\begin{equation*} U_{n,1}'\subset U_{n,2}'\subset\dotsb\subset U_{n,\lambda }'\subset \dotsb \subset U_{n,\omega^{n-1} }'=\bigcup_{\lambda\in [1,\omega^{n-1})}U_{n,\lambda}' \end{equation*}$$

(the inclusions are strict). Let $\lambda\geqslant\omega^{n-1}$. By Theorem 2.6(1) and (3.3),

$$\begin{equation} U_n'/I_{n,\omega^{n-1}}'\simeq U_{n-1}',\quad (U_{n,\omega^{n-1}+\mu}'+I_{n,\omega^{n-1}}')/I_{n,\omega^{n-1}}' \simeq U_{n-1,\mu}', \ \ \mu\in[1,\operatorname{ord}(\Omega_{n-1})]. \end{equation} \tag{ 3.7 }$$

By induction on $n$, part 1 holds.

2. We make free use of the facts established in the proof of part 1. Arguing by induction on $n$, we first take $n=2$. For every $i\in [1,\omega)$ the ideal $I_{2,i}'$ of $U_2'$ is generated by the commuting elements $x_1^j\partial_2$, where $0\leqslant j\leqslant i-1$. By Corollary 2.7, the set $W_2'=\{\partial_1^i\partial_2^j,\partial_1^ix_1^k\partial_2\cdot\partial_2^j \mid i,j\in\mathbb{N},k\geqslant 1\}$ is a $K$-basis of $U_2'$. For each $i\in [1,\omega)$,

$$\begin{equation*} I_{2,i}'=\sum_{j=0}^{i-1}U_2'x_1^j\partial_2=\bigoplus_{i\geqslant 0} \bigoplus_{k\geqslant 0,\,j\geqslant 2}K\partial_1^ix_1^k\partial_2^j\oplus \bigoplus_{i\geqslant 0}\bigoplus_{k=0}^{i-1}K\partial_1^ix_1^k\partial_2. \end{equation*}$$

It follows that

$$\begin{equation*} x_1^{i-1}\partial_2\in I_{2,i}'\setminus I_{2,i-1}',\qquad i=2,3,\dots\,. \end{equation*}$$

This means that

$$\begin{equation*} I_{2,1}'\subset I_{2,2}'\subset\dotsb\subset I_{2,i}'\subset \dotsb \subset I_{2,\omega}'=\bigcup_{i\geqslant 1}I_{2,i}'. \end{equation*}$$

Since $1\in U_2'\setminus I_{2,\omega +1}'$ and $\partial_1\in I_{2,\omega+1}'\setminus I_{2,\omega}'$, part 2 holds for $n=2$.

Suppose that $n>2$ and part 2 holds for all $n'<n$. Let $\lambda\in [1,\omega^{n-1})$. Then $\lambda+ 1$ is not a limit ordinal. Notice that ${[}1,\omega^{n-1})\subseteq[1,\operatorname{ord}(\Omega_n))$. Hence $\lambda+1=(\alpha,n)$. It follows from the equality $I_{n,\lambda +1}'= U_n'\chi_n(I_{\lambda +1})=\sum_{(\beta,n)\leqslant \lambda +1}U_n'x^\beta\partial_n$ and (2.2) that

$$\begin{equation} I_{n,\lambda +1}'\cap\biggl(\bigoplus_{\beta\in\mathbb{N}^{n-1}}K x^{\beta}\partial_n\biggr) =\bigoplus_{(\beta,n)\leqslant\lambda +1}Kx^{\beta}\partial_n. \end{equation} \tag{ 3.8 }$$

Therefore,

$$\begin{equation*} x^\alpha\partial_n\in I_{n,\lambda +1}'\setminus I_{n,\lambda }' \qquad \forall\,\lambda\in [1,\omega^{n-1}). \end{equation*}$$

This means that

$$\begin{equation*} I_{n,1}'\subset I_{n,2}'\subset\dotsb\subset I_{n,\lambda }' \subset\dotsb \subset I_{n,\omega^{n-1} }'=\bigcup_{\lambda\in[1,\omega^{n-1})}I_{n,\lambda}'. \end{equation*}$$

Let $\lambda >\omega^{n-1}$. In view of (3.7) we also have

$$\begin{equation} I_{n,\omega^{n-1}+\mu}'/I_{n,\omega^{n-1}}'\simeq I_{n-1,\mu}' \qquad \forall\,\mu\in[1,\operatorname{ord}(\Omega_{n-1})]. \end{equation} \tag{ 3.9 }$$

By induction on $n$, part 2 holds. □

Let

$$\begin{equation*} \operatorname{gl}_n(K)=\bigoplus_{i,j=1}^nKE_{ij} \end{equation*}$$

be the Lie algebra of $n\times n$ matrices, where $E_{ij}$ are the elementary matrices. For every $n\geqslant 2$, let $\operatorname{UT}_n(K)=\bigoplus_{i\leqslant j}KE_{ij}$ be the Lie algebra of upper triangular $n\times n$ matrices, $\operatorname{UT}_n(K)\subseteq\operatorname{gl}_n(K)$. The $K$-linear map

$$\begin{equation*} \operatorname{UT}_n(K)\to\mathfrak{u}_n,\qquad E_{ij}\mapsto x_i\partial_j, \end{equation*}$$

is a Lie algebra monomorphism. We identify the Lie algebra $\operatorname{UT}_n(K)$ with its image in $\mathfrak{u}_n$.

The Heisenberg Lie algebra $\mathcal{H}_n$ is the $(2n+1)$-dimensional Lie algebra with $K$-basis $X_1,\dots,X_n$, $Y_1,\dots,Y_n, Z$ such that $Z$ is a central element of $\mathcal{ H}_n$ and

$$\begin{equation*} {[}Y_i, X_j]=\delta_{ij}Z,\quad [X_i,X_j]=0,\quad [Y_i,Y_j]=0,\qquad i,j=1,\dots,n, \end{equation*}$$

where $\delta_{ij}$ is the Kronecker delta. The $K$-linear map

$$\begin{equation*} \mathcal{H}_{n-1}\to\mathfrak{u}_n,\quad X_i\mapsto x_i\partial_n,\quad Y_i\mapsto \partial_i, \quad Z\mapsto\partial_n,\qquad i=1,\dots,n-1, \end{equation*}$$

is a Lie algebra monomorphism. We identify $\mathcal{H}_{n-1}$ with its image in $\mathfrak{u}_n$. Since $\mathcal{H}_{i-1}\subseteq\mathfrak{u}_i\subseteq\mathfrak{u}_n$ for $i=2,\dots,n-1$, $\mathfrak{u}_n$ contains the $\mathcal{H}_i$, $i=1,\dots,n-1$. For all positive integers $i$ and $j$ with $i\neq j$ we have $\mathcal{H}_i\cap\mathcal{H}_j= \sum_{k=1}^{\min\{ i,j\}}K\partial_k$.

Let $A$ be an algebra or a Lie algebra and $M$ an $A$-module. Then $\operatorname{ann}_A(M):=\{a\in A\mid aM=0\}$ is called the annihilator of $M$. This is an ideal of $A$. A module is said to be faithful if its annihilator is 0.

To say that $P_n$ is a $\mathfrak{u}_n$-module is the same as to say that $P_n$ is a $U_n'$-module. This obvious observation is important because it enables us to use the relations of the Weyl algebra $A_n$ in various computations with $U_n'$ (since $U_n'\subseteq A_n$). For every $n\geqslant 2$, the Lie algebra $\mathfrak{u}_n$ is a Lie subalgebra of $\mathfrak{u}_{n+1}=\mathfrak{u}_n\oplus P_n\partial_{n+1}$, where $P_n\partial_{n+1}$ is an ideal of $\mathfrak{u}_{n+1}$ and ${[}P_n\partial_{n+1},P_n\partial_{n+1}]= 0$. In particular, $P_n\partial_{n+1}$ is a left $\mathfrak{u}_n$-module, where the action of $\mathfrak{u}_n$ on $P_n\partial_{n+1}$ is given by the rule $uv:=[u,v]$ for all $u\in\mathfrak{u}_n$ and $v\in P_n\partial_{n+1}$. We recall that the polynomial algebra $P_n$ is a left $\mathfrak{u}_n$-module.

Lemma 3.14. 

• 1)  
  The $K$-linear map $P_n\to P_n\partial_{n+1}$, $p\mapsto p\partial_{n+1}$, is an isomorphism of $\mathfrak{u}_n$-modules.
• 2)  
  The $\mathfrak{u}_n$-module $P_n$ is indecomposable and uniserial. Moreover, $\operatorname{u.dim}(P_n) {=}\omega^n$ and $\operatorname{ann}_{\mathfrak{u}_n}(P_n)= 0$.
• 3)  
  The set $\{ P_{\lambda,n}:=\bigoplus_{\alpha\in\mathbb{N}^n, (\alpha,n+1)\leqslant\lambda}Kx^\alpha\mid\lambda\in[1,\omega^n]\}$ is the set of all non-zero $\mathfrak{u}_n$-submodules of $P_n$. We have $P_{\lambda,n}\subset P_{\mu,n}$ if and only if $\lambda<\mu$. Moreover, $\operatorname{u.dim}(P_{\lambda,n})=\lambda$ for all $\lambda\in [1,\omega^n]$.
• 4)  
  The $\mathfrak{u}_n$-submodules of $P_n$ are pairwise non-isomorphic indecomposable uniserial $\mathfrak{u}_n$-modules.

Proof. To prove part 1, we notice that the map indicated is a bijection and a $\mathfrak{u}_n$-homomorphism.

Let us prove part 3. By part 1, the $\mathfrak{u}_n$-module $P_n$ can be identified with the ideal $P_n\partial_{n+1}$ of $\mathfrak{u}_{n+1}$. Under this identification, every $\mathfrak{u}_n$-submodule of $P_n$ becomes an ideal of $\mathfrak{u}_{n+1}$ in $P_n\partial_{n+1}$, and vice versa. Part 3 now follows from part 1 and the classification of ideals of $\mathfrak{u}_{n+1}$ (Theorem 3.6(1)).

We now prove part 2. By part 3, the $\mathfrak{u}_n$-module $P_n$ is uniserial and, therefore, indecomposable and $\operatorname{u.dim}(P_n)= \omega^n$. The Weyl algebra $A_n$ is a simple algebra, whence $\operatorname{ann}_{A_n}(P_n)= 0$. Then we have $\operatorname{ann}_{\mathfrak{u}_n}(P_n)= \mathfrak{u}_n\cap\operatorname{ann}_{A_n}(P_n)= 0$ because $\mathfrak{u}_n\subseteq A_n$.

Let us prove part 4. By part 3 we have $\operatorname{u.dim}(P_{\lambda,n})=\lambda$. Hence $P_{\lambda, n}\not\simeq P_{\mu,n}$ for all $\lambda\neq\mu$. The rest is obvious (see part 2). □

The following corollary describes the annihilators of all the $\mathfrak{u}_n$-submodules of $P_n$. In particular, it classifies the faithful ones.

Corollary 3.15. 

• 1)  
  The $\mathfrak{u}_n$-submodule $P_{\lambda,n}$ of $P_n$ is a faithful submodule if and only if $\lambda\in [\omega^{n-1}+1,\omega^n]$.
• 2)  
  $\operatorname{ann}_{\mathfrak{u}_n}(P_{\omega^{n-1}, n})= P_{n-1}\partial_n$.
• 3)  
  $\operatorname{ann}_{\mathfrak{u}_n}(P_{\lambda,n})= \begin{cases} 0&\textit{if}\quad\lambda\in(\omega^{n-1},\omega^n], \\ \displaystyle \bigoplus_{i=m+1}^nP_{i-1}\partial_i&\textit{if}\quad \lambda\in(\omega^{m-1},\omega^m], \\ &\qquad\qquad\qquad m=1,\dots,n-1,\\ \mathfrak{u}_n &\textit{if}\quad\lambda=1.\end{cases}$

Proof. We first prove part 2. The inclusion $\supseteq$ is obvious. The opposite inclusion $\subseteq$ follows since $P_{n-1}$ is a faithful $\mathfrak{u}_{n-1}$-module (Lemma 3.14(2)), $\mathfrak{u}_{n-1}\simeq\mathfrak{u}_n/ P_{n-1}\partial_n$ and $P_{n-1}=P_{\omega^{n-1}, n}$ is a $(\mathfrak{u}_n/P_{n-1}\partial_n)$-module.

To prove part 1, we notice that if $N$ is a submodule of $M$, then $\operatorname{ann}(N)\supseteq\operatorname{ann}(M)$. In view of this fact and part 2, to finish the proof of part 1, it suffices to show that the $\mathfrak{u}_n$-module $P_{\omega^{n-1}+1, n}$ is faithful. Since $P_{\omega^{n-1}, n}\subseteq P_{\omega^{n-1}+1, n}$, we have $\mathfrak{a}:=\operatorname{ann}_{\mathfrak{u}_n}(P_{\omega^{n-1}+1, n})\subseteq \operatorname{ann}_{\mathfrak{u}_n}(P_{\omega^{n-1}, n})=P_{n-1}\partial_n$. Since $x_n\in P_{\omega^{n-1}+1, n}$ and $p\partial_n *(x_n)=p\neq 0$ for all non-zero elements $p\in P_{n-1}$, we have $\mathfrak{a}= 0$.

Let us prove part 3. We use induction on $n\geqslant 2$. The base of the induction is $n=2$, and there are three cases to consider: $\lambda\in (\omega,\omega^2]$, $\lambda=(1,\omega]$ and $\lambda= 1$. The first case is part 1, the last case is obvious, and in the second,

$$\begin{equation*} \operatorname{ann}_{\mathfrak{u}_2}(P_{m,2})=\operatorname{ann}_{\mathfrak{u}_2} \biggl(\bigoplus_{i=0}^{m-1}Kx_1^i\biggr)= K[x_1]\partial_2. \end{equation*}$$

Suppose that $n>2$ and part 3 holds for all $n'<n$. If $\lambda\in(\omega^{n-1},\omega^n]$, then the $\mathfrak{u}_n$-module $P_{\lambda,n}$ is faithful by part 1. If $\lambda= 1$, then $\operatorname{ann}_{\mathfrak{u}_n}(P_{1, n})=\operatorname{ann}_{\mathfrak{u}_n}(K) = \mathfrak{u}_n$. If $1<\lambda\leqslant\omega^{n-1}$, that is, $\lambda\in(\omega^{m-1},\omega^m]$ for some $m\in\{1,\dots,n-1\}$, then $P_{\lambda,n}\subseteq P_{n-1}$. By part 2 we have $\mathfrak{a}:=\operatorname{ann}_{\mathfrak{u}_n}(P_{\lambda,n})\supseteq \operatorname{ann}_{\mathfrak{u}_n}(P_{\omega^{n-1}, n})= P_{n-1}\partial_n$. Since $\mathfrak{u}_{n-1}\simeq\mathfrak{u}_n/P_{n-1}\partial_n$, the inclusion $P_{\lambda,n}\subseteq P_{n-1}$ is an inclusion of $\mathfrak{u}_{n-1}$-modules. Moreover, the $\mathfrak{u}_{n-1}$-submodule $P_{\lambda,n}$ of $P_{n-1}$ can be identified with the $\mathfrak{u}_{n-1}$-submodule $P_{\lambda,n-1}$ of $P_{n-1}$. The result now follows by induction on $n$. □

Corollary 3.16. If $n\geqslant 3$, then the ideal $I_{\omega^{n-2}+1}=Kx_{n-1}\partial_n+ \sum_{\alpha\in\mathbb{N}^{n-2}}Kx^\alpha\partial_n$ is the least ideal $I$ of the Lie algebra $\mathfrak{u}_n$ which is a faithful $\mathfrak{u}_{n-1}$-module (that is, $\operatorname{ann}_{\mathfrak{u}_{n-1}}(I)= 0$).

Proof. By Lemma 3.14(1) and Corollary 3.15(1), the $\mathfrak{u}_{n-1}$-module $I_{\omega^{n-2}+1}$ is faithful, but its predecessor $I_{\omega^{n-2}}$ is not since

$$\begin{equation*} \operatorname{ann}_{\mathfrak{u}_{n-1}}(I_{\omega^{n-2}})= \operatorname{ann}_{\mathfrak{u}_{n-1}} \biggl(\sum_{\alpha\in\mathbb{N}^{n-2}}Kx^\alpha\partial_n\biggr)\ni\partial_{n-1}. \end{equation*}$$

□

The inclusion $\mathfrak{u}_n\subseteq\mathfrak{u}_{n+1}= \mathfrak{u}_{n}\oplus P_{n-1}\partial_n$ of Lie algebras respects the total orderings on the bases $\mathcal{B}_{n}$ and $\mathcal{B}_{n+1}$. The isomorphism $P_n\to P_n\partial_{n+1}$, $p\mapsto p\partial_{n+1}$ of $\mathfrak{u}_n$-modules (Lemma 3.14(1)) induces a total ordering on the monomials $\{x^\alpha\}_{\alpha\in\mathbb{N}^n}$ of the polynomial algebra $P_n$ by the following rule: $x^\alpha >x^\beta$ if and only if $X_{\alpha,n+1}>X_{\beta,n+1}$, that is, if and only if $\alpha_n=\beta_n$, $\alpha_{n-1}=\beta_{n-1}$, $\dots$, $\alpha_{m+1}=\beta_{m+1}$ and $\alpha_m > \beta_m$ for some $m\in\{1,\dots,n\}$. This is the so-called reverse lexicographic ordering on $\{ x^\alpha\}_{\alpha\in\mathbb{N}^n}$ or on $\mathbb{N}^n$ ($\alpha>\beta$ if and only if $x^\alpha>x^\beta$). By Lemma 3.4(3),(4) and Lemma 3.14(1), if $x^\alpha>x^\beta$ (where $\alpha,\beta\in\mathbb{N}^n$), then

• (i)  
  $\partial_k*x^\alpha >\partial_k*x^\beta$ for all $k=1,\dots,n$ such that $\alpha_k\neq 0$,
• (ii)  
  $X_{\gamma,k} *x^\alpha >X_{\gamma,k} *x^\beta$ for all $k=1,\dots,n-1$ and $\gamma\in\mathbb{N}^{k-1}$ such that $X_{\gamma,k} *x^\alpha\neq 0$, that is, $\alpha_k\neq 0$.

Let $\mathcal{M}(P_n)$ be the set of all non-zero submodules of the $\mathfrak{u}_n$-module $P_n$. This set is totally ordered with respect to $\subseteq$. By Lemma 3.14(3), the map

$$\begin{equation} \kappa_n\colon[1,\omega^n]\to\mathcal{M}(P_n),\qquad\lambda\mapsto P_{\lambda,n}, \end{equation} \tag{ 3.10 }$$

is an isomorphism of totally ordered sets. Every non-zero polynomial $p\in P_n$ can be written uniquely as a sum

$$\begin{equation*} \lambda_\alpha x^\alpha+\sum_{\alpha > \beta}\lambda_\beta x^\beta, \qquad\lambda_\alpha\in K^*,\quad\lambda_\beta\in K. \end{equation*}$$

The elements $\lambda_\alpha x^\alpha$ and $\lambda_\alpha$ are called the leading term and leading coefficient of $p$ respectively (with respect to the total ordering $>$). The ordinal number

$$\begin{equation*} \operatorname{ord}(p):=\operatorname{ord}([1,\alpha]) \end{equation*}$$

(where ${[}1,\alpha]\subseteq[1,\omega^n]$) is called the ordinal degree of $p$. The following assertions hold for all non-zero polynomials $p,q\in P_n$ and all $\lambda\in K^*$:

• (i)  
  $\operatorname{ord}(p+q)\leqslant\max\{\operatorname{ord}(p), \operatorname{ord}(q)\}$ provided that $p+q\neq 0$,
• (ii)  
  $\operatorname{ord}(\lambda p)=\operatorname{ord}(p)$,
• (iii)  
  $\operatorname{ord}(u*p)<\operatorname{ord}(p)$ for all $u\in\mathfrak{u}_n$ such that $u*p\neq 0$.

For every ordinal $\lambda\in[1,\omega^n)$ there is a unique representation

$$\begin{equation*} \lambda=\alpha_n\omega^{n-1}+\alpha_{n-1}\omega^{n-2}+\dotsb+\alpha_2\omega+\alpha_1, \end{equation*}$$

where $\alpha_i\in\mathbb{N}$ and not all $\alpha_j$ are equal to zero (notice the shift by 1 in the subscripts of the coefficients $\alpha_i$). Then

$$\begin{equation} P_{\lambda,n}=\begin{cases} \oplus\bigl\{Kx^\beta\mid\beta\in\mathbb{N}^n,\,x^\beta\leqslant x_1^{-1}x^\alpha\bigr\} &\text{if}\quad\alpha_1\neq 0, \\ \oplus\bigl\{Kx^\beta\mid\beta\in\mathbb{N}^n, \,x^\beta<x^\alpha\bigr\} &\text{if}\quad\alpha_1= 0, \end{cases} \end{equation} \tag{ 3.11 }$$

where $x^\alpha=\prod_{i=1}^n x_i^{\alpha_i}$. Notice that $\alpha_1\neq 0$ if and only if $\lambda$ is not a limit ordinal. The vector space $P_{\lambda,n}$ possesses a largest monomial if and only if $\lambda$ is not a limit ordinal, and in this case $x_1^{-1}x^\alpha$ is the largest monomial in $P_{\lambda,n}$. The ordinal number $\lambda\in [1,\omega^n)$ can be written uniquely as a sum

$$\begin{equation*} \lambda=\alpha_m\omega^{m-1}+\alpha_{m-1}\omega^{m-2}+\dotsb+ \alpha_j\omega^{j-1},\qquad\alpha_m\neq 0,\quad\alpha_j\neq 0, \end{equation*}$$

where $\alpha_i\in\mathbb{N}$, $1\leqslant m\leqslant n$ and $1\leqslant j\leqslant m$. The positive integers $\alpha_m$ and $\alpha_j$ are called the multiplicity and the comultiplicity of the ordinal number $\lambda$. The non-negative integers $m-1$ and $j-1$ are called the degree and the codegree of $\lambda$.

The following lemma is obvious.

Lemma 3.17. Suppose that $\lambda\in[1,\omega^n)$, that is, $\lambda=\alpha_m\omega^{m-1}+ \alpha_{m-1}\omega^{m-2}+\dotsb\dotsb+ \alpha_j\omega^{j-1}$ with $\alpha_i\in\mathbb{N}$, $\alpha_m\neq 0$, $\alpha_j\neq 0$, $m\leqslant n$ and $j\geqslant 1$. Then

$$\begin{equation*} \begin{aligned} P_{\lambda,n}&=\sum_{i=0}^{\alpha_m-1}x_m^iP_{m-1} + x_m^{\alpha_m}\sum_{0\leqslant i\leqslant\alpha_{m-1}-1}x_{m-1}^iP_{m-2}+\dotsb \\ &\qquad\qquad\qquad\qquad\qquad\dots+x_m^{\alpha_m} x_{m-1}^{\alpha_{m-1}} \dotsb x_{k+1}^{\alpha_{k+1}}\sum_{0\leqslant i\leqslant\alpha_k-1}x_k^iP_{k-1}+\dotsb \\ &\qquad\qquad\qquad\qquad\qquad\dots+x_m^{\alpha_m} x_{m-1}^{\alpha_{m-1}} \dotsb x_{j+1}^{\alpha_{j+1}} \sum_{0\leqslant i\leqslant\alpha_j-1}x_j^iP_{j-1}. \end{aligned} \end{equation*}$$

In particular, ${P_{\alpha_m\omega^{m-1},n}=\sum_{i=0}^{\alpha_m-1}x_m^iP_{m-1}}$.

Remark 3.18. If $\alpha_k= 0$, then the corresponding summand is absent.

Let $V$ be a vector space over $K$. A linear map $\varphi\colon V\to V$ is called a Fredholm map/operator if it has finite-dimensional kernel and cokernel. Then the number

$$\begin{equation*} \operatorname{ind}(\varphi):=\operatorname{dim}_K(\operatorname{ker}(\varphi)) -\operatorname{dim}_K(\operatorname{coker}(\varphi)) \end{equation*}$$

is called the index of $\varphi$. Let $\mathcal{F}(V)$ be the set of all Fredholm linear maps on $V$. It is actually a monoid since

$$\begin{equation} \operatorname{ind}(\varphi\psi)=\operatorname{ind}(\varphi)+\operatorname{ind}(\psi) \qquad\forall\,\varphi,\psi\in\mathcal{F}(V). \end{equation} \tag{ 3.12 }$$

Let $V$ be a vector space with a countable $K$-basis $\{e_i\}_{i\in\mathbb{N}}$ and let $\partial$ be a $K$-linear map on $V$ given by the rule $\partial e_i=e_{i-1}$ for all $i\in\mathbb{N}$, where $e_{-1}:= 0$. For example, $V=K[x]$, $e_i:=\frac{x^i}{i!}$ and $\partial=\frac{d}{dx}$. The subalgebra of $\operatorname{End}_K(V)$ generated by the map $\partial$ is the polynomial algebra $K[\partial]$. Since $\partial$ is locally nilpotent, the algebra $\operatorname{End}_K(V)$ contains the algebra $K[[\partial]]=\{\sum_{i=0}^\infty\lambda_i\partial^i\mid\lambda_i\in K\}$ of formal power series in $\partial$. The set $K[[\partial]]^*= \{\sum_{i=0}^\infty\lambda_i\partial^i\in K[[\partial]]\mid\lambda_0\in K^*\}$ is the group of units of the algebra $K[[\partial]]$. The vector space $V$ is a $K[\partial]$-module.

The following lemma is trivial.

Lemma 3.19. 

• 1)  
  We have

  $$\begin{equation*} \begin{aligned} \operatorname{End}_{K[\partial]}(V)&=\biggl\{\varphi\in\operatorname{End}_K(V), \, \varphi (e_i)=\sum_{j=0}^i\lambda_je_{i-j}= \biggl(\sum_{j\geqslant 0}\lambda_j\partial^j\biggr)(e_i),\,i\in\mathbb{N}\Bigm|\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\lambda_i\in K,\,i\in\mathbb{N}\biggr\} =K[[\partial]], \\ \operatorname{Aut}_{K[\partial]}(V)&=\biggl\{\varphi\in\operatorname{End}_K(V), \, \varphi(e_i)=\sum_{j=0}^i\lambda_je_{i-j},\,i\in\mathbb{N}\Bigm|\\ &\qquad\qquad\qquad\qquad\qquad\qquad \lambda_0\neq 0,\, \lambda_i\in K, \,i\in\mathbb{N}\biggr\}=K[[\partial]]^*. \end{aligned} \end{equation*}$$
• 2)  
  In particular, the non-zero elements of $\operatorname{End}_{K[\partial]}(V)$ are surjective Fredholm maps.
• 3)  
  If $a=\sum_{i\geqslant d}\lambda_i\partial^i\in K[[\partial]]$ and $\lambda_d\neq 0$, then $d=\operatorname{ind}(a)= \operatorname{dim}_K(\operatorname{ker}(a))$.

Let $V$ be a vector space with $K$-basis $\{e_i\}_{i\in\mathbb{N}}$. We define a $K$-linear map $x$ on $V$ by putting $xe_i:=e_{i+1}$ for all $i\in\mathbb{N}$. For example, $V=K[x]$, $e_i:= x^i$ and $x\colon K[x]\to K[x]$, $p\mapsto xp$. The polynomial algebra $K[x]$ is a subalgebra of the algebra $\operatorname{End}_K(V)$.

The following lemma is obvious.

Lemma 3.20. 

• 1)  
  We have $\operatorname{End}_{K[x]}(V)=\{p\colon V\to V, \,a\mapsto pa\mid p\in K[x]\}\simeq K[x]$ (via $p\mapsto p$). In particular, the non-zero elements of $\operatorname{End}_{K[x]}(V)$ are injective maps.
• 2)  
  We have $\operatorname{Aut}_{K[x]}(V)=K^*$.
• 3)  
  For all $p\in\operatorname{End}_{K[x]}(V)=K[x]$ we have $\deg (p)=-\operatorname{ind}(p)= \operatorname{dim}_K(\operatorname{coker}(p))$.

The following proposition describes the algebra of all $\mathfrak{u}_n$-homomorphisms (and its group of units) of the $\mathfrak{u}_n$-module $P_n$. Clearly, $K[x_n]\subseteq P_n$ is an embedding of $K[\partial_n]$-modules and $K[[\frac{d}{dx_n}]]=\operatorname{End}_{K[\partial_n]}(K[x_n])$ (Lemma 3.19(1)). The $K$-derivation $\frac{d}{dx_n}$ of the polynomial algebra $K[x_n]$ is also denoted by $\partial_n$.

Proposition 3.21. 

• 1)  
  The map $\operatorname{End}_{\mathfrak{u}_n}(P_n)\to \operatorname{End}_{K[\partial_n]}(K[x_n])=K[[\frac{d}{d x_n}]]$, $\varphi\mapsto\varphi|_{K[x_n]}$, is an isomorphism of $K$-algebras with inverse map $\varphi'\mapsto\varphi$, where $\varphi(x^\beta x_n^i):=X_{\beta,n}\varphi'(\frac{x_n^{i+1}}{i+1})$ for all $\beta\in\mathbb{N}^{n-1}$ and $i\in\mathbb{N}$.
• 2)  
  The map $\operatorname{Aut}_{\mathfrak{u}_n}(P_n)\to \operatorname{Aut}_{K[\partial_n]}(K[x_n])=K[[\frac{d}{dx_n}]]^*$, $\varphi\mapsto\varphi|_{K[x_n]}$, is an isomorphism of groups with the same inverse map as in part 1.
• 3)  
  Every non-zero element $\varphi\in \operatorname{End}_{\mathfrak{u}_n}(P_n)$ is a surjective map with kernel $\operatorname{ker}(\varphi)= \bigoplus_{i=0}^{d-1}P_{n-1}x_n^i$, where $d=\operatorname{ind}(\varphi|_{K[x_n]})= \operatorname{dim}_K(\operatorname{ker}(\varphi|_{K[x_n]}))$.
• 4)  
  For all integers $d\geqslant 1$ we have $(\frac{\partial} {\partial x_n})^d\in\operatorname{End}_{\mathfrak{u}_n}(P_n)$ and $\operatorname{ker}_{P_n}(\frac{\partial}{\partial x_n})^d= \bigoplus_{i=0}^{d-1}P_{n-1}x_n^i$. In particular, $(\frac{\partial}{\partial x_n})^d\colon P_n/\bigoplus_{i=0}^{d-1} P_{n-1}x_n^i\to P_n$ is an isomorphism of $\mathfrak{u}_n$-modules.

Proof. We first prove part 1. By Lemma 3.19, $\operatorname{End}_{K[\partial_n]}(K[x_n])=K[[\frac{d}{dx_n}]]$. Since $K[x_n]=\bigcap_{i=1}^{n-1}\operatorname{ker}_{P_n}(\partial_i)$ and the maps $\partial_1,\dots,\partial_n$ commute, the restriction map $\varphi\mapsto \varphi|_{K[x_n]}$ is a well-defined homomorphism of $K$-algebras. For all $\beta\in\mathbb{N}^{n-1}$ and $i\in\mathbb{N}$,

$$\begin{equation*} \varphi(x^\beta x_n^i)=\varphi\biggl(X_{\beta,n}*\frac{x_n^{i+1}}{i+1}\biggr)= X_{\beta,n}\varphi\biggl(\frac{x_n^{i+1}}{i+1}\biggr). \end{equation*}$$

Therefore the restriction map is a monomorphism. To prove that it is surjective, we must show that for any map $\varphi'\in\operatorname{End}_{K[\partial_n]}(K[x_n])$, its extension $\varphi$ defined as in the statement of part 1 is a $\mathfrak{u}_n$-homomorphism. The map $\varphi$ is $K$-linear. Hence we must only check that $\varphi X_{\alpha,i }=X_{\alpha,i}\varphi$ for all $i=1,\dots,n$ and $\alpha\in \mathbb{N}^{i-1}$.

Case 1: $i<n$. For all $\beta\in\mathbb{N}^{n-1}$ and $j\in\mathbb{N}$ we have

$$\begin{equation*} \begin{aligned} \varphi X_{\alpha,i}(x^\beta\cdot x^j_n)&=\varphi (x^\alpha\partial_i*x^\beta x_n^j)=\varphi(\beta_ix^{\alpha+\beta-e_i}x_n^j) =\beta_i X_{\alpha+\beta-e_i,n}\varphi'\biggl(\frac{x_n^{j+1}}{j+1}\biggr), \\ X_{\alpha,i}\varphi(x^\beta\cdot x_n^j)&=x^\alpha\partial_i*\biggr(x^\beta\partial_n *\varphi' \biggl(\frac{x_n^{j+1}}{j+1}\biggr)\biggr)= \beta_i x^{\alpha+\beta-e_i}\partial_n*\varphi' \biggl(\frac{x_n^{j+1}}{j+1}\biggr)\\ &\qquad+x^\beta\partial_n*x^\alpha\partial_i*\varphi'\biggl(\frac{x_n^{j+1}}{j+1}\biggr) =\beta_i x^{\alpha+\beta-e_i}\partial_n*\varphi'\biggl(\frac{x_n^{j+1}}{j+1}\biggr)+0\\ &=\beta_i X_{\alpha+\beta-e_i,n}\varphi'\biggl(\frac{x_n^{j+1}}{j+1}\biggr). \end{aligned} \end{equation*}$$

Case 2: $i=n$. Let $\beta\in\mathbb{N}^{n-1}$ and $j\in\mathbb{N}$. Suppose that $j\geqslant 1$. Then

$$\begin{equation*} \begin{gathered} \varphi X_{\alpha,n}(x^\beta x^j_n)=\varphi(x^{\alpha +\beta}j x_n^{j-1})= X_{\alpha+\beta,n}\varphi'(x_n^j), \\ \begin{aligned} X_{\alpha,n}\varphi(x^\beta x_n^j)&=x^\alpha\partial_n* \biggl(x^\beta\partial_n*\varphi'\biggl(\frac{x_n^{j+1}}{j+1}\biggr)\biggr) =x^{\alpha+\beta}\partial_n*\partial_n*\varphi' \biggl(\frac{x_n^{j+1}}{j+1}\biggr)\\ &=X_{\alpha+\beta,n}\varphi'\biggl(\partial_n*\frac{x_n^{j+1}}{j+1}\biggr) =X_{\alpha +\beta,n}\varphi'(x_n^j). \end{aligned} \end{gathered} \end{equation*}$$

Suppose that $j= 0$. Then

$$\begin{equation*} \begin{gathered} \varphi X_{\alpha,n}(x^\beta)=\varphi(0)=0, \\ \begin{aligned} X_{\alpha,n}\varphi (x^\beta)&=x^\alpha\partial_n*x^\beta\partial_n*\varphi'(x_n) =x^{\alpha+\beta}\partial_n*\partial_n*\varphi'(x_n)\\ &={X_{\alpha+\beta,n}\varphi'(\partial_n*x_n)=X_{\alpha+\beta,n}\varphi'(1)=0} \end{aligned} \end{gathered} \end{equation*}$$

(since $\varphi'(1)\in K$).

Part 2 follows from part 1. Part 3 follows from part 1 and Lemma 3.19. Part 4 follows from part 1. □

Let $S$ be a subset of $\mathbb{N}^n$. The vector space $P_n(S):=\bigoplus_{\alpha\in S}Kx^\alpha$ is called a monomial subspace of the polynomial algebra $P_n$ with support $S$. We put $P_n(\varnothing):= 0$ by definition. Clearly, $\bigcap_{i\in I}P_n(S_i)=P_n(\bigcap_{i\in I}S_i)$.

Example 3.22. For every integer $i=1,\dots,n$ and any ordinal number $\lambda\in[1,\omega^n)$, the vector space $\{p\in P_n\mid\frac{\partial p}{\partial x_i}\in P_{\lambda,n}\}$ is a monomial subspace of $P_n$. Hence so is their intersection

$$\begin{eqnarray} &&P_{\lambda,n}':=\biggl\{p\in P_n\biggm|\frac{\partial p}{\partial x_i} \in P_{\lambda,n},\,i=1,\dots,n\biggr\}\\ &&\qquad\qquad=\bigoplus\biggl\{Kx^\alpha\biggm| \alpha\in\mathbb{N}^n,\frac{\partial x^\alpha}{\partial x_i}\in P_{\lambda, n}, \,i=1,\dots,n\biggr\}. \end{eqnarray}$$

Clearly, $P_{\lambda,n}\subseteq P_{\lambda,n}'$. The following theorem describes the vector space $P_{\lambda,n}'$ and shows that the inclusion is always strict and $\operatorname{dim}_K(P_{\lambda,n}'/P_{\lambda,n})<\infty$.

Theorem 3.23. Suppose that $\lambda\in[1,\omega^n)$, that is, $\lambda=\alpha_n\omega^{n-1}+\alpha_{n-1}\omega^{n-2} +\dotsb\dots+ \alpha_i\omega^{i-1}+\dotsb+\alpha_j\omega^{j-1}$, where $\alpha_i\in\mathbb{N}$, $\alpha_j\neq 0$ and $j\geqslant 1$. Then the following assertions hold.

• 1)  
  $P_{\lambda,n}'=P_{\lambda,n}\oplus\bigoplus_{i=j}^nK\theta_i$, where $\theta_i=\begin{cases}x^\alpha=\prod_{k=1}^nx_k^{\alpha_k}&\textit{if}\quad i=j,\\ x_i\prod_{k=i}^nx_k^{\alpha_k}&\textit{if}\quad j<i\leqslant n.\end{cases}$
• 2)  
  $1\leqslant\operatorname{dim}_K(P_{\lambda,n}'/P_{\lambda,n})=n-j+1\leqslant n$.
• 3)  
  $\lambda$ is a non-limit ordinal (that is, $\alpha_1\,{\neq}\,0$) if and only if $\operatorname{dim}_K(P_{\lambda,n}'/P_{\lambda,n})\,{=}\,n$.
• 4)  
  We have $1\leqslant\operatorname{dim}_K(P_{\lambda +1,n}'/P_{\lambda,n}')= j\leqslant n-1$, and the set $\{x_ix^\alpha+P_{\lambda,n}'\mid i=1,\dots,j\}$ is a basis of the vector space $P_{\lambda +1,n}'/P_{\lambda,n}'$.
• 5)  
  The vector spaces $\{P_{\lambda,n}'\mid\lambda\in[1,\omega^n)\}$ are distinct. In particular, if $\lambda<\mu$, then $P_{\lambda,n}'\varsubsetneq P_{\mu,n}'$.

Proof. We first prove part 1. Let $R$ be the right-hand side of the equality stated in part 1. Then $P_{\lambda,n}'\supseteq R$ (by Lemma 3.17). In particular, the set $P_{\lambda,n}'\setminus P_{\lambda,n}$ is non-empty.

Case 1: $\lambda$ is a non-limit ordinal, that is, $j= 1$ $(\alpha_1\neq 0)$. In this case,

$$\begin{equation*} P_{\lambda,n}=\oplus\{ Kx^\alpha\mid x^\beta\leqslant x^{\alpha'}\} \end{equation*}$$

(see (3.11)), where $x^{\alpha'}:= x_1^{-1}x^\alpha$, that is, $\alpha_1'=\alpha_1-1$ and $\alpha_i'=\alpha_i$ for $i=2,\dots,n$. Suppose that $x^\beta\in P_{\lambda,n}'\setminus P_{\lambda,n}$. Then $x^\beta > x^{\alpha'}$ and there is an integer $i$, $1\leqslant i\leqslant n$, such that $\beta_j=\alpha_j$ for all $j>i$ and $\beta_i=\alpha_i'+1$. Then necessarily $\beta_k= 0$ for all $k< i$ (since $\frac{\partial x^\beta}{\partial x_k}\leqslant x^{\alpha'}$ for all $k<i$) and, therefore, $x^\beta=\theta_i$.

Case 2: $\lambda$ is a limit ordinal, that is, $j> 1$ $(\alpha_1= 0)$. In this case,

$$\begin{equation*} P_{\lambda,n}=\oplus\{Kx^\gamma\mid x^\gamma <x^\alpha\} \end{equation*}$$

(see (3.11)). Suppose that $x^\beta\not\in P_{\lambda,n}$. Then $x^\beta\in P_{\lambda,n}'$ if and only if $\frac{\partial x^\beta}{\partial x_i} < x^\alpha$ for all $i=1,\dots,n$ (we recall that $P_{\lambda,n}=\oplus\{Kx^\gamma\mid x^\gamma<x^\alpha\}$) or, in other words, if and only if the following conditions hold:

• (i)  
  $\frac{\partial x^\beta}{\partial x_i}\leqslant x^\alpha$ for all $i=1,\dots,n$,
• (ii)  
  $\frac{\partial x^\beta}{\partial x_i}\not\in K^* x^\alpha$ for all $i=1,\dots,n$.

Condition (i) is Case 1 for $\lambda +1$. Therefore,

$$\begin{equation*} \begin{aligned} P_{\lambda,n}'&=P_{\lambda,n}'\cap P_{\lambda +1,n}'= P_{\lambda,n}'\cap\biggl(P_{\lambda +1,n}\oplus \bigoplus_{i=1}^nK\theta_i'\biggr)\\ &=P_{\lambda,n}'\cap \biggl(P_{\lambda,n}\oplus Kx^\alpha\oplus\bigoplus_{i=1}^nK\theta_i'\biggr) =P_{\lambda,n}\oplus Kx^\alpha\oplus P_{\lambda,n}'\cap \bigoplus_{i=1}^nK\theta_i'\\ &=P_{\lambda,n}\oplus Kx^\alpha \oplus\bigoplus_{i=1}^n P_{\lambda,n}'\cap K\theta_i', \end{aligned} \end{equation*}$$

where

$$\begin{equation*} \theta_i':=\begin{cases} \displaystyle x_i\prod_{k=i}^nx_k^{\alpha_k} &\text{if}\quad i>j,\\ x_ix^\alpha &\text{if}\quad i\leqslant j.\\ \end{cases} \end{equation*}$$

Notice that $\theta_j=x^\alpha$ and $\theta_i'=\theta_i\in P_{\lambda,n}'$ for $j < i\leqslant n$. Condition (ii) excludes precisely the elements $\{\theta_i'\mid i\leqslant j\}$, that is, $\bigoplus_{i=1}^n(P_{\lambda,n}'\cap K\theta_i')= \bigoplus_{j<i\leqslant n}K\theta_i'=\bigoplus_{j<i\leqslant n}K\theta_i$. The proof of part 1 is complete.

Part 2 follows from part 1. Part 3 follows from part 2.

We now prove part 4. The ordinal number $\lambda +1$ is not a limit ordinal. By part 3 we have $\operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda +1,n})=n$. Notice that $P_{\lambda,n}\subset P_{\lambda+1,n}\subset P_{\lambda+1,n}'$ and $\operatorname{dim}_K(P_{\lambda+1,n}/P_{\lambda,n})= 1$. By part 1,

$$\begin{equation*} \operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda,n})= \operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda+1,n})+ \operatorname{dim}_K(P_{\lambda+1,n}/P_{\lambda,n})=n+1. \end{equation*}$$

Finally, by part 2,

$$\begin{equation*} \operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda,n}')\,{=} \operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda,n})- \operatorname{dim}_K(P_{\lambda,n}'/P_{\lambda,n})\,{=}\,n+1-(n-j+1)\,{=}\,j \end{equation*}$$

and $1\leqslant j=\operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda,n}')\leqslant n-1$. The elements $\theta_i':=x_ix^\alpha$, $i=1,\dots,j$, are the elements $\theta_i$, $i=1,\dots,j$, in part 1, but for the ordinal $\lambda+1$ instead of $\lambda$. Clearly, $\theta_i'\in P_{\lambda+1, n}' \setminus P_{\lambda,n}'$ for $i=1,\dots,j$. Therefore the elements $\{\theta_i'+P_{\lambda,n}\mid i=1,\dots,j\}$ are $K$-linearly independent in the vector space $P_{\lambda+1,n}'/P_{\lambda,n}'$ since the vector spaces $P_{\lambda+1,n}'$ and $P_{\lambda,n}'$ are monomial. These elements form a basis of the vector space $P_{\lambda+1,n}'/P_{\lambda,n}'$ since $j=\operatorname{dim}_K(P_{\lambda+1,n}'/P_{\lambda,n}')$.

Part 5 follows from part 4. □

For example, for all positive integers $i$ and $j$ we have

$$\begin{equation*} P_{i\omega^{n-1},n}'=P_{i\omega^{n-1},n}\oplus Kx_n^i,\qquad P_{i\omega^{n-1}+j,n}' =P_{i\omega^{n-1}+j,n}\oplus Kx_1^jx_n^i\oplus\bigoplus_{2\leqslant k\leqslant n}K x_k x_n^i. \end{equation*}$$

Example 3.24. Consider the following conditions:

$$\begin{equation} P_{i-1}\frac{\partial p }{\partial x_i}\subseteq P_{\lambda,n},\qquad i=1,\dots,n. \end{equation} \tag{ 3.13 }$$

For every integer $i=1,\dots,n$, any element $\alpha\in\mathbb{N}^{i-1}$ and any ordinal number $\lambda\in [1,\omega^n)$, the vector space $\{p\in P_n\mid x^\alpha\frac{\partial p}{\partial x_i}\in P_{\lambda,n}\}$ is a monomial subspace of $P_n$. Hence so is their intersection

$$\begin{equation*} P_{\lambda, n}'':=\{ p\in P_n\mid p\text{ satisfies }(3.13)\} =\oplus\{Kx^\beta\mid\beta\in\mathbb{N}^n,\ x^\beta\text{ satisfies }(3.13)\}. \end{equation*}$$

For all ordinal numbers $\lambda\in [1,\omega^n)$,

$$\begin{equation*} P_{\lambda,n}\subseteq P_{\lambda,n}''\subseteq P_{\lambda,n}'. \end{equation*}$$

The first inclusion follows since $P_{\lambda,n}$ is a $\mathfrak{u}_n$-module. If $\lambda\,{\leqslant}\,\mu$, then $P_{\lambda,n}'' {\subseteq} P_{\mu,n}''$. The following corollary shows that these inclusions are strict and gives a $K$-basis for every vector space $P_{\lambda,n}''$.

Corollary 3.25. Suppose that $\lambda\in [1,\omega^n)$, that is, $\lambda=\alpha_m\omega^{m-1} +\alpha_{m-1}\omega^{m-2}+\dotsb\dotsb+ \alpha_j\omega^{j-1}$, where $\alpha_i\in\mathbb{N}$, $\alpha_m\neq 0$, $\alpha_j\neq 0$, $m\leqslant n$ and $j\geqslant 1$. Then the following assertions hold.

• 1)  
  $P_{\lambda,n}''=P_{\lambda,n}\oplus Kx^\alpha=P_{\lambda+1,n}$, where $x^\alpha=\prod_{k=j}^mx_k^{\alpha_k}$.
• 2)  
  The vector spaces $\{ P_{\lambda,n}''\mid\lambda\in[1,\omega^n)\}$ are distinct. In particular, if $\lambda<\mu$, then $P_{\lambda,n}''\varsubsetneq P_{\mu,n}''$.
• 3)  
  $\operatorname{dim}_K(P_{\lambda,n}''/P_{\lambda,n})= 1$.
• 4)  
  $\operatorname{dim}_K(P_{\lambda+1,n}''/P_{\lambda,n}'')= 1$.

Proof. Part 1 follows at once from Lemma 3.17, the inclusions $P_{\lambda,n}\subseteq P_{\lambda,n}''\subseteq P_{\lambda,n}'$ and Theorem 3.23.

Parts 2–4 follow from part 1. □